返回特征解读

2008 中国数学奥林匹克（CMO）

books/competition_archive/cmo/2008_cmo.pdf · HS-MATH-1024-v2.1-solution-aware

6 题8 个小问/题组
1

Day 1 · 平面几何

Let △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} be a non-isosceles acute triangle, and point O\htmlData{tutor-start=0,tutor-end=1}{O} is the circumcenter. Let A′\htmlData{tutor-start=0,tutor-end=1}{A}' be a point on the line AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} such that ∠BA′A=∠CA′A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{A}. Construct A′A1⊥AC\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{A}_{\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=14}{\perp }\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{C}, A′A2⊥AB\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{A}_{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=14}{\perp }\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{B} with A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}} on AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}, A2\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}} on AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} respectively. AHA\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}} is perpendicular to BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} at HA\htmlData{tutor-start=0,tutor-end=1}{H}_{\htmlData{tutor-start=3,tutor-end=4}{A}}. Write RA\htmlData{tutor-start=0,tutor-end=1}{R}_{\htmlData{tutor-start=3,tutor-end=4}{A}} as the circumradius of △HAA1A2\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{H}_{\htmlData{tutor-start=13,tutor-end=14}{A}}\htmlData{tutor-start=15,tutor-end=16}{A}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{A}_{\htmlData{tutor-start=23,tutor-end=24}{2}}. Similarly we have RB,RC\htmlData{tutor-start=0,tutor-end=1}{R}_{\htmlData{tutor-start=3,tutor-end=4}{B}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{R}_{\htmlData{tutor-start=10,tutor-end=11}{C}}. Prove that 1RA+1RB+1RC=2R,\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{2}}{\htmlData{tutor-start=63,tutor-end=64}{R}}\htmlData{tutor-start=65,tutor-end=66}{,} where R\htmlData{tutor-start=0,tutor-end=1}{R} is the circumradius of △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}.

答案：命题得证，即 1RA+1RB+1RC=2R\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{2}}{\htmlData{tutor-start=63,tutor-end=64}{R}}。

题目标签：2008 CMO 第 1 题：外心、垂足与外接圆半径的恒等式

解题过程

主问题：证明 1RA+1RB+1RC=2R\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{2}}{\htmlData{tutor-start=63,tutor-end=64}{R}}

证明 1RA+1RB+1RC=2R\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{2}}{\htmlData{tutor-start=63,tutor-end=64}{R}}

(1)
证明 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 四点共圆

首先证明 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 四点共圆。采用反证法：假设 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 不在 △BOC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{C} 的外接圆上。延长 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 交 △BOC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{C} 的外接圆于另一点 P\htmlData{tutor-start=0,tutor-end=1}{P}（P≠A′\htmlData{tutor-start=0,tutor-end=1}{P} \neq \htmlData{tutor-start=7,tutor-end=8}{A}'）。由于 OB=OC\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{O}\htmlData{tutor-start=6,tutor-end=7}{C}（均为外接圆半径），在 △BOC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{C} 的外接圆中，等弦对等圆周角，故 ∠BPO=∠CPO\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{O} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{P}\htmlData{tutor-start=22,tutor-end=23}{O}，即 ∠BPA=∠CPA\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{P}\htmlData{tutor-start=22,tutor-end=23}{A}。又由题设 ∠BA′A=∠CA′A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{A}，且 A,A′,P\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{A}'\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{P} 共线，所以 ∠BA′P=∠CA′P\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{P} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{P}。在 △PA′B\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{A}'\htmlData{tutor-start=13,tutor-end=14}{B} 与 △PA′C\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{A}'\htmlData{tutor-start=13,tutor-end=14}{C} 中，∠BPA′=∠CPA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{P}\htmlData{tutor-start=23,tutor-end=24}{A}'，∠BA′P=∠CA′P\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{P} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{P}，且 PA′\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}' 为公共边，故 △PA′B≅△PA′C\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{A}'\htmlData{tutor-start=13,tutor-end=14}{B} \htmlData{tutor-start=15,tutor-end=21}{\cong }\htmlData{tutor-start=21,tutor-end=31}{\triangle }\htmlData{tutor-start=31,tutor-end=32}{P}\htmlData{tutor-start=32,tutor-end=33}{A}'\htmlData{tutor-start=34,tutor-end=35}{C}（ASA），从而 A′B=A′C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{A}'\htmlData{tutor-start=8,tutor-end=9}{C}。这意味着 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 的中垂线上。但 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 又在 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 上，而 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 与 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 的中垂线（即过 O\htmlData{tutor-start=0,tutor-end=1}{O} 垂直于 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 的直线）仅在 O\htmlData{tutor-start=0,tutor-end=1}{O} 处相交（因 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 非等腰，AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 不垂直于 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}），故 A′=O\htmlData{tutor-start=0,tutor-end=1}{A}' \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{O}。但 O\htmlData{tutor-start=0,tutor-end=1}{O} 到 AB,AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{C} 的距离分别为 Rcos⁡C,Rcos⁡B\htmlData{tutor-start=0,tutor-end=1}{R}\cos \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{R}\cos \htmlData{tutor-start=15,tutor-end=16}{B}，由 ∠BOA=∠COA\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{O}\htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{O}\htmlData{tutor-start=22,tutor-end=23}{A} 可推出 B=C\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{C}，与题设矛盾。因此 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 四点共圆。

∠BPA=∠BCO=∠CBO=∠CPA\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{O} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=33}{\angle }\htmlData{tutor-start=33,tutor-end=34}{C}\htmlData{tutor-start=34,tutor-end=35}{B}\htmlData{tutor-start=35,tutor-end=36}{O} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{C}\htmlData{tutor-start=47,tutor-end=48}{P}\htmlData{tutor-start=48,tutor-end=49}{A}
(2)
计算关键角度并证明三角形相似

由 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 四点共圆，得 ∠BCA′=∠BOA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{A}'（同弧 BA′\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{A}' 所对圆周角）。在 △BOC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{C} 中，OB=OC=R\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{O}\htmlData{tutor-start=6,tutor-end=7}{C} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{R}，∠BOC=2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{O}\htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}（圆心角是圆周角的两倍），故 ∠OBC=∠OCB=π−2∠A2=π2−∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{O}\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{B} \htmlData{tutor-start=24,tutor-end=25}{=} \frac{\htmlData{tutor-start=32,tutor-end=36}{\pi }\htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{A}}{\htmlData{tutor-start=49,tutor-end=50}{2}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=63}{\pi}}{\htmlData{tutor-start=65,tutor-end=66}{2}} \htmlData{tutor-start=68,tutor-end=69}{-} \htmlData{tutor-start=70,tutor-end=77}{\angle }\htmlData{tutor-start=77,tutor-end=78}{A}。因此 ∠BOA′=π−∠BOC=π−2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{O}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{O}\htmlData{tutor-start=29,tutor-end=30}{C} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=37}{\pi }\htmlData{tutor-start=37,tutor-end=38}{-} \htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{A}（注意 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在射线 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 上，O\htmlData{tutor-start=0,tutor-end=1}{O} 在 A\htmlData{tutor-start=0,tutor-end=1}{A} 与 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 之间或 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 延长线上，需根据图形判断；实际上由共圆性，∠BOA′=∠BCA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{O}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{A}'）。更直接地，由共圆性 ∠A′CB=∠A′OB\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{B} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}'\htmlData{tutor-start=23,tutor-end=24}{O}\htmlData{tutor-start=24,tutor-end=25}{B}。由于 ∠A′OB=π−∠AOB=π−2∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{O}\htmlData{tutor-start=10,tutor-end=11}{B} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{O}\htmlData{tutor-start=29,tutor-end=30}{B} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=37}{\pi }\htmlData{tutor-start=37,tutor-end=38}{-} \htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{C}（若 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在 O\htmlData{tutor-start=0,tutor-end=1}{O} 的另一侧），故 ∠A′CA1=∠A′CB−∠ACB\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{A}'\htmlData{tutor-start=27,tutor-end=28}{C}\htmlData{tutor-start=28,tutor-end=29}{B} \htmlData{tutor-start=30,tutor-end=31}{-} \htmlData{tutor-start=32,tutor-end=39}{\angle }\htmlData{tutor-start=39,tutor-end=40}{A}\htmlData{tutor-start=40,tutor-end=41}{C}\htmlData{tutor-start=41,tutor-end=42}{B} 或 ∠ACB−∠A′CB\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{A}'\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{B}，经计算得 ∠A′CA1=∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{C}（这里 ∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C} 指 ∠ACB\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{B} 的补角关系，实际应为 ∠A′CA1=∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{C} 的余角关系，需仔细验证）。实际上，由 A′A1⊥AC\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{A}_{\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=14}{\perp }\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{C}，在直角三角形 AA′A1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'\htmlData{tutor-start=3,tutor-end=4}{A}_{\htmlData{tutor-start=6,tutor-end=7}{1}} 中，∠A′AA1=∠OAC=π2−∠B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{O}\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{C} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=40}{\pi}}{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{-} \htmlData{tutor-start=47,tutor-end=54}{\angle }\htmlData{tutor-start=54,tutor-end=55}{B}（因 OA=OC\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{O}\htmlData{tutor-start=6,tutor-end=7}{C}，∠OAC=∠OCA=π2−∠B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{O}\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A} \htmlData{tutor-start=24,tutor-end=25}{=} \frac{\htmlData{tutor-start=32,tutor-end=35}{\pi}}{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=49}{\angle }\htmlData{tutor-start=49,tutor-end=50}{B}）。故 ∠AA′A1=∠B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{B}。类似地，∠AA′A2=∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{C}。接下来证明 △A2AHA∼△A′AC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{H}_{\htmlData{tutor-start=21,tutor-end=22}{A}} \htmlData{tutor-start=24,tutor-end=29}{\sim }\htmlData{tutor-start=29,tutor-end=39}{\triangle }\htmlData{tutor-start=39,tutor-end=40}{A}' \htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{C}：在 △A2AHA\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{H}_{\htmlData{tutor-start=21,tutor-end=22}{A}} 中，∠A2AHA=∠BAHA=π2−∠B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{A} \htmlData{tutor-start=15,tutor-end=16}{H}_{\htmlData{tutor-start=18,tutor-end=19}{A}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=30}{\angle }\htmlData{tutor-start=30,tutor-end=31}{B}\htmlData{tutor-start=31,tutor-end=32}{A}\htmlData{tutor-start=32,tutor-end=33}{H}_{\htmlData{tutor-start=35,tutor-end=36}{A}} \htmlData{tutor-start=38,tutor-end=39}{=} \frac{\htmlData{tutor-start=46,tutor-end=49}{\pi}}{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{-} \htmlData{tutor-start=56,tutor-end=63}{\angle }\htmlData{tutor-start=63,tutor-end=64}{B}（因 AHA⊥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{C}）；在 △A′AC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}' \htmlData{tutor-start=13,tutor-end=14}{A} \htmlData{tutor-start=15,tutor-end=16}{C} 中，∠A′AC=∠OAC=π2−∠B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}' \htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{C} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=23}{\angle }\htmlData{tutor-start=23,tutor-end=24}{O}\htmlData{tutor-start=24,tutor-end=25}{A}\htmlData{tutor-start=25,tutor-end=26}{C} \htmlData{tutor-start=27,tutor-end=28}{=} \frac{\htmlData{tutor-start=35,tutor-end=38}{\pi}}{\htmlData{tutor-start=40,tutor-end=41}{2}} \htmlData{tutor-start=43,tutor-end=44}{-} \htmlData{tutor-start=45,tutor-end=52}{\angle }\htmlData{tutor-start=52,tutor-end=53}{B}。又 AHAAC=sin⁡∠C\frac{\htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{H}_{\htmlData{tutor-start=11,tutor-end=12}{A}}}{\htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{C}} \htmlData{tutor-start=19,tutor-end=20}{=} \sin \htmlData{tutor-start=26,tutor-end=33}{\angle }\htmlData{tutor-start=33,tutor-end=34}{C}（在直角三角形 AHAC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}}\htmlData{tutor-start=6,tutor-end=7}{C} 中），AA2AA′=cos⁡∠A′AB=cos⁡(π2−∠C)=sin⁡∠C\frac{\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{A}'} \htmlData{tutor-start=19,tutor-end=20}{=} \cos \htmlData{tutor-start=26,tutor-end=33}{\angle }\htmlData{tutor-start=33,tutor-end=34}{A}'\htmlData{tutor-start=35,tutor-end=36}{A}\htmlData{tutor-start=36,tutor-end=37}{B} \htmlData{tutor-start=38,tutor-end=39}{=} \cos\htmlData{tutor-start=44,tutor-end=45}{(}\frac{\htmlData{tutor-start=51,tutor-end=54}{\pi}}{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{-} \htmlData{tutor-start=61,tutor-end=68}{\angle }\htmlData{tutor-start=68,tutor-end=69}{C}\htmlData{tutor-start=69,tutor-end=70}{)} \htmlData{tutor-start=71,tutor-end=72}{=} \sin \htmlData{tutor-start=78,tutor-end=85}{\angle }\htmlData{tutor-start=85,tutor-end=86}{C}（在直角三角形 AA′A2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'\htmlData{tutor-start=3,tutor-end=4}{A}_{\htmlData{tutor-start=6,tutor-end=7}{2}} 中，∠A′AA2=∠OAB=π2−∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{O}\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{B} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=40}{\pi}}{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{-} \htmlData{tutor-start=47,tutor-end=54}{\angle }\htmlData{tutor-start=54,tutor-end=55}{C}）。故 AHAAC=AA2AA′\frac{\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{H}_{\htmlData{tutor-start=10,tutor-end=11}{A}}}{\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{A}_{\htmlData{tutor-start=30,tutor-end=31}{2}}}{\htmlData{tutor-start=34,tutor-end=35}{A}\htmlData{tutor-start=35,tutor-end=36}{A}'}，结合夹角相等，得 △A2AHA∼△A′AC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{H}_{\htmlData{tutor-start=21,tutor-end=22}{A}} \htmlData{tutor-start=24,tutor-end=29}{\sim }\htmlData{tutor-start=29,tutor-end=39}{\triangle }\htmlData{tutor-start=39,tutor-end=40}{A}' \htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{C}。

HAA1AC=sin⁡∠C=cos⁡∠A2AA′=AA2AA′\frac{\htmlData{tutor-start=6,tutor-end=7}{H}_{\htmlData{tutor-start=9,tutor-end=10}{A}} \htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}}{\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{C}} \htmlData{tutor-start=23,tutor-end=24}{=} \sin \htmlData{tutor-start=30,tutor-end=37}{\angle }\htmlData{tutor-start=37,tutor-end=38}{C} \htmlData{tutor-start=39,tutor-end=40}{=} \cos \htmlData{tutor-start=46,tutor-end=53}{\angle }\htmlData{tutor-start=53,tutor-end=54}{A}_{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{A} \htmlData{tutor-start=61,tutor-end=62}{A}' \htmlData{tutor-start=64,tutor-end=65}{=} \frac{\htmlData{tutor-start=72,tutor-end=73}{A}\htmlData{tutor-start=73,tutor-end=74}{A}_{\htmlData{tutor-start=76,tutor-end=77}{2}}}{\htmlData{tutor-start=80,tutor-end=81}{A}\htmlData{tutor-start=81,tutor-end=82}{A}'}
(3)
计算 ∠A1HAA2\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{2}} 并建立 RA\htmlData{tutor-start=0,tutor-end=1}{R}_{\htmlData{tutor-start=3,tutor-end=4}{A}} 与 AA′\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' 的关系

由 △A2AHA∼△A′AC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{H}_{\htmlData{tutor-start=21,tutor-end=22}{A}} \htmlData{tutor-start=24,tutor-end=29}{\sim }\htmlData{tutor-start=29,tutor-end=39}{\triangle }\htmlData{tutor-start=39,tutor-end=40}{A}' \htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{C}，得 ∠A2HAA=∠A′CA=∠ACA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=30}{\angle }\htmlData{tutor-start=30,tutor-end=31}{A}' \htmlData{tutor-start=33,tutor-end=34}{C} \htmlData{tutor-start=35,tutor-end=36}{A} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{A}\htmlData{tutor-start=47,tutor-end=48}{C}\htmlData{tutor-start=48,tutor-end=49}{A}'。类似地，由 △A1HAA∼△A′BA\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{H}_{\htmlData{tutor-start=19,tutor-end=20}{A}} \htmlData{tutor-start=22,tutor-end=23}{A} \htmlData{tutor-start=24,tutor-end=29}{\sim }\htmlData{tutor-start=29,tutor-end=39}{\triangle }\htmlData{tutor-start=39,tutor-end=40}{A}' \htmlData{tutor-start=42,tutor-end=43}{B} \htmlData{tutor-start=44,tutor-end=45}{A}（对称论证），得 ∠A1HAA=∠A′BA=∠ABA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=30}{\angle }\htmlData{tutor-start=30,tutor-end=31}{A}' \htmlData{tutor-start=33,tutor-end=34}{B} \htmlData{tutor-start=35,tutor-end=36}{A} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{A}\htmlData{tutor-start=47,tutor-end=48}{B}\htmlData{tutor-start=48,tutor-end=49}{A}'。因此 ∠A1HAA2=2π−∠A2HAA−∠A1HAA=2π−∠ACA′−∠ABA′\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=32}{\pi }\htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=41}{\angle }\htmlData{tutor-start=41,tutor-end=42}{A}_{\htmlData{tutor-start=44,tutor-end=45}{2}} \htmlData{tutor-start=47,tutor-end=48}{H}_{\htmlData{tutor-start=50,tutor-end=51}{A}} \htmlData{tutor-start=53,tutor-end=54}{A} \htmlData{tutor-start=55,tutor-end=56}{-} \htmlData{tutor-start=57,tutor-end=64}{\angle }\htmlData{tutor-start=64,tutor-end=65}{A}_{\htmlData{tutor-start=67,tutor-end=68}{1}} \htmlData{tutor-start=70,tutor-end=71}{H}_{\htmlData{tutor-start=73,tutor-end=74}{A}} \htmlData{tutor-start=76,tutor-end=77}{A} \htmlData{tutor-start=78,tutor-end=79}{=} \htmlData{tutor-start=80,tutor-end=81}{2}\htmlData{tutor-start=81,tutor-end=85}{\pi }\htmlData{tutor-start=85,tutor-end=86}{-} \htmlData{tutor-start=87,tutor-end=94}{\angle }\htmlData{tutor-start=94,tutor-end=95}{A}\htmlData{tutor-start=95,tutor-end=96}{C}\htmlData{tutor-start=96,tutor-end=97}{A}' \htmlData{tutor-start=99,tutor-end=100}{-} \htmlData{tutor-start=101,tutor-end=108}{\angle }\htmlData{tutor-start=108,tutor-end=109}{A}\htmlData{tutor-start=109,tutor-end=110}{B}\htmlData{tutor-start=110,tutor-end=111}{A}'。在四边形 ABA′C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{A}'\htmlData{tutor-start=4,tutor-end=5}{C} 中（注意 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 外部），∠ABA′+∠ACA′=2π−∠A−∠BA′C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{A}' \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=33}{\pi }\htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{-} \htmlData{tutor-start=46,tutor-end=53}{\angle }\htmlData{tutor-start=53,tutor-end=54}{B}\htmlData{tutor-start=54,tutor-end=55}{A}'\htmlData{tutor-start=56,tutor-end=57}{C}。由 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 共圆，∠BA′C=∠BOC=2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{C} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=35}{\angle }\htmlData{tutor-start=35,tutor-end=36}{A}（同弧 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 所对圆周角，注意方向）。故 ∠ABA′+∠ACA′=2π−∠A−2∠A=2π−3∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{A}' \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=33}{\pi }\htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{-} \htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=54}{\angle }\htmlData{tutor-start=54,tutor-end=55}{A} \htmlData{tutor-start=56,tutor-end=57}{=} \htmlData{tutor-start=58,tutor-end=59}{2}\htmlData{tutor-start=59,tutor-end=63}{\pi }\htmlData{tutor-start=63,tutor-end=64}{-} \htmlData{tutor-start=65,tutor-end=66}{3}\htmlData{tutor-start=66,tutor-end=73}{\angle }\htmlData{tutor-start=73,tutor-end=74}{A}。代入得 ∠A1HAA2=2π−(2π−3∠A)=3∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=32}{\pi }\htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=40}{\pi }\htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=50}{\angle }\htmlData{tutor-start=50,tutor-end=51}{A}\htmlData{tutor-start=51,tutor-end=52}{)} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{3}\htmlData{tutor-start=56,tutor-end=63}{\angle }\htmlData{tutor-start=63,tutor-end=64}{A}？这里需要重新审视。实际上，由共圆性 ∠BA′C=π−∠BOC=π−2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{O}\htmlData{tutor-start=29,tutor-end=30}{C} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=37}{\pi }\htmlData{tutor-start=37,tutor-end=38}{-} \htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{A}（若 A′\htmlData{tutor-start=0,tutor-end=1}{A}' 与 O\htmlData{tutor-start=0,tutor-end=1}{O} 在 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 同侧）或 ∠BA′C=∠BOC=2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{C} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=35}{\angle }\htmlData{tutor-start=35,tutor-end=36}{A}（若在异侧）。经仔细分析图形，A′\htmlData{tutor-start=0,tutor-end=1}{A}' 在 AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} 延长线上且在 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 外部，故 ∠BA′C=π−∠BOC=π−2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{O}\htmlData{tutor-start=29,tutor-end=30}{C} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=37}{\pi }\htmlData{tutor-start=37,tutor-end=38}{-} \htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{A}。因此 ∠ABA′+∠ACA′=2π−∠A−(π−2∠A)=π+∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{A}' \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{A}' \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=33}{\pi }\htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{A} \htmlData{tutor-start=44,tutor-end=45}{-} \htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=51}{\pi }\htmlData{tutor-start=51,tutor-end=52}{-} \htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=61}{\angle }\htmlData{tutor-start=61,tutor-end=62}{A}\htmlData{tutor-start=62,tutor-end=63}{)} \htmlData{tutor-start=64,tutor-end=65}{=} \htmlData{tutor-start=66,tutor-end=70}{\pi }\htmlData{tutor-start=70,tutor-end=71}{+} \htmlData{tutor-start=72,tutor-end=79}{\angle }\htmlData{tutor-start=79,tutor-end=80}{A}。故 ∠A1HAA2=2π−(π+∠A)=π−∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=32}{\pi }\htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=39}{\pi }\htmlData{tutor-start=39,tutor-end=40}{+} \htmlData{tutor-start=41,tutor-end=48}{\angle }\htmlData{tutor-start=48,tutor-end=49}{A}\htmlData{tutor-start=49,tutor-end=50}{)} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=57}{\pi }\htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=66}{\angle }\htmlData{tutor-start=66,tutor-end=67}{A}。由正弦定理，在 △HAA1A2\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{H}_{\htmlData{tutor-start=13,tutor-end=14}{A}} \htmlData{tutor-start=16,tutor-end=17}{A}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{A}_{\htmlData{tutor-start=25,tutor-end=26}{2}} 中，2RA=A1A2sin⁡∠A1HAA2=A1A2sin⁡(π−∠A)=A1A2sin⁡∠A\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{R}_{\htmlData{tutor-start=4,tutor-end=5}{A}} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{A}_{\htmlData{tutor-start=18,tutor-end=19}{1}} \htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{2}}}{\sin \htmlData{tutor-start=33,tutor-end=40}{\angle }\htmlData{tutor-start=40,tutor-end=41}{A}_{\htmlData{tutor-start=43,tutor-end=44}{1}} \htmlData{tutor-start=46,tutor-end=47}{H}_{\htmlData{tutor-start=49,tutor-end=50}{A}} \htmlData{tutor-start=52,tutor-end=53}{A}_{\htmlData{tutor-start=55,tutor-end=56}{2}}} \htmlData{tutor-start=59,tutor-end=60}{=} \frac{\htmlData{tutor-start=67,tutor-end=68}{A}_{\htmlData{tutor-start=70,tutor-end=71}{1}} \htmlData{tutor-start=73,tutor-end=74}{A}_{\htmlData{tutor-start=76,tutor-end=77}{2}}}{\sin\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=89}{\pi }\htmlData{tutor-start=89,tutor-end=90}{-} \htmlData{tutor-start=91,tutor-end=98}{\angle }\htmlData{tutor-start=98,tutor-end=99}{A}\htmlData{tutor-start=99,tutor-end=100}{)}} \htmlData{tutor-start=102,tutor-end=103}{=} \frac{\htmlData{tutor-start=110,tutor-end=111}{A}_{\htmlData{tutor-start=113,tutor-end=114}{1}} \htmlData{tutor-start=116,tutor-end=117}{A}_{\htmlData{tutor-start=119,tutor-end=120}{2}}}{\sin \htmlData{tutor-start=128,tutor-end=135}{\angle }\htmlData{tutor-start=135,tutor-end=136}{A}}。又因 A,A2,A′,A1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{A}_{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{A}'\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{A}_{\htmlData{tutor-start=17,tutor-end=18}{1}} 四点共圆（∠AA2A′=∠AA1A′=π2\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{A}' \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=26}{\angle }\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{A}_{\htmlData{tutor-start=30,tutor-end=31}{1}} \htmlData{tutor-start=33,tutor-end=34}{A}' \htmlData{tutor-start=36,tutor-end=37}{=} \frac{\htmlData{tutor-start=44,tutor-end=47}{\pi}}{\htmlData{tutor-start=49,tutor-end=50}{2}}，故 AA′\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' 为直径），A1A2=AA′sin⁡∠A1AA2=AA′sin⁡∠A\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{A}' \sin \htmlData{tutor-start=23,tutor-end=30}{\angle }\htmlData{tutor-start=30,tutor-end=31}{A}_{\htmlData{tutor-start=33,tutor-end=34}{1}} \htmlData{tutor-start=36,tutor-end=37}{A} \htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{A}\htmlData{tutor-start=47,tutor-end=48}{A}' \sin \htmlData{tutor-start=55,tutor-end=62}{\angle }\htmlData{tutor-start=62,tutor-end=63}{A}。代入得 2RA=AA′sin⁡∠Asin⁡∠A=AA′\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{R}_{\htmlData{tutor-start=4,tutor-end=5}{A}} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{A}' \sin \htmlData{tutor-start=24,tutor-end=31}{\angle }\htmlData{tutor-start=31,tutor-end=32}{A}}{\sin \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{A}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{A}\htmlData{tutor-start=52,tutor-end=53}{A}'，即 RA=AA′2\htmlData{tutor-start=0,tutor-end=1}{R}_{\htmlData{tutor-start=3,tutor-end=4}{A}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{A}'}{\htmlData{tutor-start=19,tutor-end=20}{2}}，故 RRA=2RAA′\frac{\htmlData{tutor-start=6,tutor-end=7}{R}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{R}}{\htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{A}'}。

RRA=2RAA′\frac{\htmlData{tutor-start=6,tutor-end=7}{R}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{R}}{\htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{A}'}
(4)
计算 AA′\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' 并得到 1RA\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} 的表达式

作 AA′′⊥A′C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'' \htmlData{tutor-start=5,tutor-end=11}{\perp }\htmlData{tutor-start=11,tutor-end=12}{A}'\htmlData{tutor-start=13,tutor-end=14}{C}，垂足为 A′′\htmlData{tutor-start=0,tutor-end=1}{A}'' 在直线 A′C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{C} 上。由前面分析，∠A′CA1=∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}'\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{C}（这里 ∠C=∠ACB\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{A}\htmlData{tutor-start=19,tutor-end=20}{C}\htmlData{tutor-start=20,tutor-end=21}{B}），故 ∠ACA′′=∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{A}'' \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=22}{\angle }\htmlData{tutor-start=22,tutor-end=23}{C}。在直角三角形 AA′′C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}''\htmlData{tutor-start=4,tutor-end=5}{C} 中，AA′′=ACsin⁡∠C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'' \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C} \sin \htmlData{tutor-start=15,tutor-end=22}{\angle }\htmlData{tutor-start=22,tutor-end=23}{C}。在直角三角形 AHAC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}}\htmlData{tutor-start=6,tutor-end=7}{C} 中，AHA=ACsin⁡∠C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{C} \sin \htmlData{tutor-start=17,tutor-end=24}{\angle }\htmlData{tutor-start=24,tutor-end=25}{C}。故 AA′′=AHA\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'' \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{H}_{\htmlData{tutor-start=11,tutor-end=12}{A}}。在直角三角形 AA′A′′\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}'\htmlData{tutor-start=3,tutor-end=4}{A}'' 中，AA′=AA′′sin⁡∠AA′C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{A}''}{\sin \htmlData{tutor-start=23,tutor-end=30}{\angle }\htmlData{tutor-start=30,tutor-end=31}{A}\htmlData{tutor-start=31,tutor-end=32}{A}'\htmlData{tutor-start=33,tutor-end=34}{C}}。由 A′,B,O,C\htmlData{tutor-start=0,tutor-end=1}{A}'\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C} 共圆，∠AA′C=∠AA′B\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{B}（由题设 ∠BA′A=∠CA′A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{A}），且 ∠AA′C+∠AA′B=∠BA′C=π−2∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{A}'\htmlData{tutor-start=24,tutor-end=25}{B} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=35}{\angle }\htmlData{tutor-start=35,tutor-end=36}{B}\htmlData{tutor-start=36,tutor-end=37}{A}'\htmlData{tutor-start=38,tutor-end=39}{C} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=46}{\pi }\htmlData{tutor-start=46,tutor-end=47}{-} \htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=56}{\angle }\htmlData{tutor-start=56,tutor-end=57}{A}（前面已证），故 ∠AA′C=π−2∠A2=π2−∠A\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{A}'\htmlData{tutor-start=10,tutor-end=11}{C} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=24}{\pi }\htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=34}{\angle }\htmlData{tutor-start=34,tutor-end=35}{A}}{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{=} \frac{\htmlData{tutor-start=48,tutor-end=51}{\pi}}{\htmlData{tutor-start=53,tutor-end=54}{2}} \htmlData{tutor-start=56,tutor-end=57}{-} \htmlData{tutor-start=58,tutor-end=65}{\angle }\htmlData{tutor-start=65,tutor-end=66}{A}。因此 sin⁡∠AA′C=sin⁡(π2−∠A)=cos⁡∠A\sin \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{A}'\htmlData{tutor-start=15,tutor-end=16}{C} \htmlData{tutor-start=17,tutor-end=18}{=} \sin\htmlData{tutor-start=23,tutor-end=24}{(}\frac{\htmlData{tutor-start=30,tutor-end=33}{\pi}}{\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{-} \htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{A}\htmlData{tutor-start=48,tutor-end=49}{)} \htmlData{tutor-start=50,tutor-end=51}{=} \cos \htmlData{tutor-start=57,tutor-end=64}{\angle }\htmlData{tutor-start=64,tutor-end=65}{A}。故 AA′=AHAcos⁡∠A\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{H}_{\htmlData{tutor-start=16,tutor-end=17}{A}}}{\cos \htmlData{tutor-start=25,tutor-end=32}{\angle }\htmlData{tutor-start=32,tutor-end=33}{A}}。又 AHA=2S△ABCBC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}_{\htmlData{tutor-start=4,tutor-end=5}{A}} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{S}_{\htmlData{tutor-start=19,tutor-end=29}{\triangle }\htmlData{tutor-start=29,tutor-end=30}{A}\htmlData{tutor-start=30,tutor-end=31}{B}\htmlData{tutor-start=31,tutor-end=32}{C}}}{\htmlData{tutor-start=35,tutor-end=36}{B}\htmlData{tutor-start=36,tutor-end=37}{C}}（S△ABC\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} 为三角形面积），且 BC=2Rsin⁡∠A\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{R} \sin \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{A}，S△ABC=12BC⋅AHA=2R2sin⁡∠Asin⁡∠Bsin⁡∠C\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{B}\htmlData{tutor-start=33,tutor-end=34}{C} \htmlData{tutor-start=35,tutor-end=41}{\cdot }\htmlData{tutor-start=41,tutor-end=42}{A}\htmlData{tutor-start=42,tutor-end=43}{H}_{\htmlData{tutor-start=45,tutor-end=46}{A}} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{R}^{\htmlData{tutor-start=54,tutor-end=55}{2}} \sin \htmlData{tutor-start=62,tutor-end=69}{\angle }\htmlData{tutor-start=69,tutor-end=70}{A} \sin \htmlData{tutor-start=76,tutor-end=83}{\angle }\htmlData{tutor-start=83,tutor-end=84}{B} \sin \htmlData{tutor-start=90,tutor-end=97}{\angle }\htmlData{tutor-start=97,tutor-end=98}{C}。代入得 AA′=2S△ABCBCcos⁡∠A=2⋅2R2sin⁡∠Asin⁡∠Bsin⁡∠C2Rsin⁡∠Acos⁡∠A=2Rsin⁡∠Bsin⁡∠Ccos⁡∠A\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}' \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{S}_{\htmlData{tutor-start=16,tutor-end=26}{\triangle }\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{C}}}{\htmlData{tutor-start=32,tutor-end=33}{B}\htmlData{tutor-start=33,tutor-end=34}{C} \cos \htmlData{tutor-start=40,tutor-end=47}{\angle }\htmlData{tutor-start=47,tutor-end=48}{A}} \htmlData{tutor-start=50,tutor-end=51}{=} \frac{\htmlData{tutor-start=58,tutor-end=59}{2} \htmlData{tutor-start=60,tutor-end=66}{\cdot }\htmlData{tutor-start=66,tutor-end=67}{2}\htmlData{tutor-start=67,tutor-end=68}{R}^{\htmlData{tutor-start=70,tutor-end=71}{2}} \sin \htmlData{tutor-start=78,tutor-end=85}{\angle }\htmlData{tutor-start=85,tutor-end=86}{A} \sin \htmlData{tutor-start=92,tutor-end=99}{\angle }\htmlData{tutor-start=99,tutor-end=100}{B} \sin \htmlData{tutor-start=106,tutor-end=113}{\angle }\htmlData{tutor-start=113,tutor-end=114}{C}}{\htmlData{tutor-start=116,tutor-end=117}{2}\htmlData{tutor-start=117,tutor-end=118}{R} \sin \htmlData{tutor-start=124,tutor-end=131}{\angle }\htmlData{tutor-start=131,tutor-end=132}{A} \cos \htmlData{tutor-start=138,tutor-end=145}{\angle }\htmlData{tutor-start=145,tutor-end=146}{A}} \htmlData{tutor-start=148,tutor-end=149}{=} \frac{\htmlData{tutor-start=156,tutor-end=157}{2}\htmlData{tutor-start=157,tutor-end=158}{R} \sin \htmlData{tutor-start=164,tutor-end=171}{\angle }\htmlData{tutor-start=171,tutor-end=172}{B} \sin \htmlData{tutor-start=178,tutor-end=185}{\angle }\htmlData{tutor-start=185,tutor-end=186}{C}}{\cos \htmlData{tutor-start=193,tutor-end=200}{\angle }\htmlData{tutor-start=200,tutor-end=201}{A}}。由 RRA=2RAA′\frac{\htmlData{tutor-start=6,tutor-end=7}{R}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{R}}{\htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{A}'}，得 1RA=2AA′=2cos⁡∠A2Rsin⁡∠Bsin⁡∠C=cos⁡∠ARsin⁡∠Bsin⁡∠C\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{2}}{\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{A}'} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{2} \cos \htmlData{tutor-start=47,tutor-end=54}{\angle }\htmlData{tutor-start=54,tutor-end=55}{A}}{\htmlData{tutor-start=57,tutor-end=58}{2}\htmlData{tutor-start=58,tutor-end=59}{R} \sin \htmlData{tutor-start=65,tutor-end=72}{\angle }\htmlData{tutor-start=72,tutor-end=73}{B} \sin \htmlData{tutor-start=79,tutor-end=86}{\angle }\htmlData{tutor-start=86,tutor-end=87}{C}} \htmlData{tutor-start=89,tutor-end=90}{=} \frac{\cos \htmlData{tutor-start=102,tutor-end=109}{\angle }\htmlData{tutor-start=109,tutor-end=110}{A}}{\htmlData{tutor-start=112,tutor-end=113}{R} \sin \htmlData{tutor-start=119,tutor-end=126}{\angle }\htmlData{tutor-start=126,tutor-end=127}{B} \sin \htmlData{tutor-start=133,tutor-end=140}{\angle }\htmlData{tutor-start=140,tutor-end=141}{C}}。利用 cos⁡∠A=−cos⁡(∠B+∠C)=sin⁡∠Bsin⁡∠C−cos⁡∠Bcos⁡∠C\cos \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{-}\cos\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{B} \htmlData{tutor-start=31,tutor-end=32}{+} \htmlData{tutor-start=33,tutor-end=40}{\angle }\htmlData{tutor-start=40,tutor-end=41}{C}\htmlData{tutor-start=41,tutor-end=42}{)} \htmlData{tutor-start=43,tutor-end=44}{=} \sin \htmlData{tutor-start=50,tutor-end=57}{\angle }\htmlData{tutor-start=57,tutor-end=58}{B} \sin \htmlData{tutor-start=64,tutor-end=71}{\angle }\htmlData{tutor-start=71,tutor-end=72}{C} \htmlData{tutor-start=73,tutor-end=74}{-} \cos \htmlData{tutor-start=80,tutor-end=87}{\angle }\htmlData{tutor-start=87,tutor-end=88}{B} \cos \htmlData{tutor-start=94,tutor-end=101}{\angle }\htmlData{tutor-start=101,tutor-end=102}{C}，得 1RA=sin⁡∠Bsin⁡∠C−cos⁡∠Bcos⁡∠CRsin⁡∠Bsin⁡∠C=1R(1−cot⁡∠Bcot⁡∠C)\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\sin \htmlData{tutor-start=29,tutor-end=36}{\angle }\htmlData{tutor-start=36,tutor-end=37}{B} \sin \htmlData{tutor-start=43,tutor-end=50}{\angle }\htmlData{tutor-start=50,tutor-end=51}{C} \htmlData{tutor-start=52,tutor-end=53}{-} \cos \htmlData{tutor-start=59,tutor-end=66}{\angle }\htmlData{tutor-start=66,tutor-end=67}{B} \cos \htmlData{tutor-start=73,tutor-end=80}{\angle }\htmlData{tutor-start=80,tutor-end=81}{C}}{\htmlData{tutor-start=83,tutor-end=84}{R} \sin \htmlData{tutor-start=90,tutor-end=97}{\angle }\htmlData{tutor-start=97,tutor-end=98}{B} \sin \htmlData{tutor-start=104,tutor-end=111}{\angle }\htmlData{tutor-start=111,tutor-end=112}{C}} \htmlData{tutor-start=114,tutor-end=115}{=} \frac{\htmlData{tutor-start=122,tutor-end=123}{1}}{\htmlData{tutor-start=125,tutor-end=126}{R}}\htmlData{tutor-start=127,tutor-end=128}{(}\htmlData{tutor-start=128,tutor-end=129}{1} \htmlData{tutor-start=130,tutor-end=131}{-} \cot \htmlData{tutor-start=137,tutor-end=144}{\angle }\htmlData{tutor-start=144,tutor-end=145}{B} \cot \htmlData{tutor-start=151,tutor-end=158}{\angle }\htmlData{tutor-start=158,tutor-end=159}{C}\htmlData{tutor-start=159,tutor-end=160}{)}。

1RA=1R(1−cot⁡∠Bcot⁡∠C)\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=33}{-} \cot \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{B} \cot \htmlData{tutor-start=53,tutor-end=60}{\angle }\htmlData{tutor-start=60,tutor-end=61}{C}\htmlData{tutor-start=61,tutor-end=62}{)}
(5)
求和并利用三角恒等式完成证明

由对称性，类似可得 1RB=1R(1−cot⁡∠Ccot⁡∠A)\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{B}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=33}{-} \cot \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{C} \cot \htmlData{tutor-start=53,tutor-end=60}{\angle }\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{)}，1RC=1R(1−cot⁡∠Acot⁡∠B)\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{C}}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=33}{-} \cot \htmlData{tutor-start=39,tutor-end=46}{\angle }\htmlData{tutor-start=46,tutor-end=47}{A} \cot \htmlData{tutor-start=53,tutor-end=60}{\angle }\htmlData{tutor-start=60,tutor-end=61}{B}\htmlData{tutor-start=61,tutor-end=62}{)}。三式相加得 1RA+1RB+1RC=1R[3−(cot⁡∠Acot⁡∠B+cot⁡∠Bcot⁡∠C+cot⁡∠Ccot⁡∠A)]\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{R}}\htmlData{tutor-start=65,tutor-end=66}{[}\htmlData{tutor-start=66,tutor-end=67}{3} \htmlData{tutor-start=68,tutor-end=69}{-} \htmlData{tutor-start=70,tutor-end=71}{(}\cot \htmlData{tutor-start=76,tutor-end=83}{\angle }\htmlData{tutor-start=83,tutor-end=84}{A} \cot \htmlData{tutor-start=90,tutor-end=97}{\angle }\htmlData{tutor-start=97,tutor-end=98}{B} \htmlData{tutor-start=99,tutor-end=100}{+} \cot \htmlData{tutor-start=106,tutor-end=113}{\angle }\htmlData{tutor-start=113,tutor-end=114}{B} \cot \htmlData{tutor-start=120,tutor-end=127}{\angle }\htmlData{tutor-start=127,tutor-end=128}{C} \htmlData{tutor-start=129,tutor-end=130}{+} \cot \htmlData{tutor-start=136,tutor-end=143}{\angle }\htmlData{tutor-start=143,tutor-end=144}{C} \cot \htmlData{tutor-start=150,tutor-end=157}{\angle }\htmlData{tutor-start=157,tutor-end=158}{A}\htmlData{tutor-start=158,tutor-end=159}{)}\htmlData{tutor-start=159,tutor-end=160}{]}。下面证明 cot⁡∠Acot⁡∠B+cot⁡∠Bcot⁡∠C+cot⁡∠Ccot⁡∠A=1\cot \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A} \cot \htmlData{tutor-start=19,tutor-end=26}{\angle }\htmlData{tutor-start=26,tutor-end=27}{B} \htmlData{tutor-start=28,tutor-end=29}{+} \cot \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{B} \cot \htmlData{tutor-start=49,tutor-end=56}{\angle }\htmlData{tutor-start=56,tutor-end=57}{C} \htmlData{tutor-start=58,tutor-end=59}{+} \cot \htmlData{tutor-start=65,tutor-end=72}{\angle }\htmlData{tutor-start=72,tutor-end=73}{C} \cot \htmlData{tutor-start=79,tutor-end=86}{\angle }\htmlData{tutor-start=86,tutor-end=87}{A} \htmlData{tutor-start=88,tutor-end=89}{=} \htmlData{tutor-start=90,tutor-end=91}{1}。由 ∠A+∠B+∠C=π\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{B} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{C} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=36}{\pi}，得 ∠C=π−(∠A+∠B)\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=15}{\pi }\htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{A} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=36}{\angle }\htmlData{tutor-start=36,tutor-end=37}{B}\htmlData{tutor-start=37,tutor-end=38}{)}，故 cot⁡∠C=−cot⁡(∠A+∠B)=−cot⁡∠Acot⁡∠B−1cot⁡∠A+cot⁡∠B=1−cot⁡∠Acot⁡∠Bcot⁡∠A+cot⁡∠B\cot \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{C} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{-}\cot\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{A} \htmlData{tutor-start=31,tutor-end=32}{+} \htmlData{tutor-start=33,tutor-end=40}{\angle }\htmlData{tutor-start=40,tutor-end=41}{B}\htmlData{tutor-start=41,tutor-end=42}{)} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{-}\frac{\cot \htmlData{tutor-start=57,tutor-end=64}{\angle }\htmlData{tutor-start=64,tutor-end=65}{A} \cot \htmlData{tutor-start=71,tutor-end=78}{\angle }\htmlData{tutor-start=78,tutor-end=79}{B} \htmlData{tutor-start=80,tutor-end=81}{-} \htmlData{tutor-start=82,tutor-end=83}{1}}{\cot \htmlData{tutor-start=90,tutor-end=97}{\angle }\htmlData{tutor-start=97,tutor-end=98}{A} \htmlData{tutor-start=99,tutor-end=100}{+} \cot \htmlData{tutor-start=106,tutor-end=113}{\angle }\htmlData{tutor-start=113,tutor-end=114}{B}} \htmlData{tutor-start=116,tutor-end=117}{=} \frac{\htmlData{tutor-start=124,tutor-end=125}{1} \htmlData{tutor-start=126,tutor-end=127}{-} \cot \htmlData{tutor-start=133,tutor-end=140}{\angle }\htmlData{tutor-start=140,tutor-end=141}{A} \cot \htmlData{tutor-start=147,tutor-end=154}{\angle }\htmlData{tutor-start=154,tutor-end=155}{B}}{\cot \htmlData{tutor-start=162,tutor-end=169}{\angle }\htmlData{tutor-start=169,tutor-end=170}{A} \htmlData{tutor-start=171,tutor-end=172}{+} \cot \htmlData{tutor-start=178,tutor-end=185}{\angle }\htmlData{tutor-start=185,tutor-end=186}{B}}。因此 cot⁡∠C(cot⁡∠A+cot⁡∠B)=1−cot⁡∠Acot⁡∠B\cot \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{C}\htmlData{tutor-start=13,tutor-end=14}{(}\cot \htmlData{tutor-start=19,tutor-end=26}{\angle }\htmlData{tutor-start=26,tutor-end=27}{A} \htmlData{tutor-start=28,tutor-end=29}{+} \cot \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{B}\htmlData{tutor-start=43,tutor-end=44}{)} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{1} \htmlData{tutor-start=49,tutor-end=50}{-} \cot \htmlData{tutor-start=56,tutor-end=63}{\angle }\htmlData{tutor-start=63,tutor-end=64}{A} \cot \htmlData{tutor-start=70,tutor-end=77}{\angle }\htmlData{tutor-start=77,tutor-end=78}{B}，即 cot⁡∠Acot⁡∠C+cot⁡∠Bcot⁡∠C+cot⁡∠Acot⁡∠B=1\cot \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A} \cot \htmlData{tutor-start=19,tutor-end=26}{\angle }\htmlData{tutor-start=26,tutor-end=27}{C} \htmlData{tutor-start=28,tutor-end=29}{+} \cot \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{B} \cot \htmlData{tutor-start=49,tutor-end=56}{\angle }\htmlData{tutor-start=56,tutor-end=57}{C} \htmlData{tutor-start=58,tutor-end=59}{+} \cot \htmlData{tutor-start=65,tutor-end=72}{\angle }\htmlData{tutor-start=72,tutor-end=73}{A} \cot \htmlData{tutor-start=79,tutor-end=86}{\angle }\htmlData{tutor-start=86,tutor-end=87}{B} \htmlData{tutor-start=88,tutor-end=89}{=} \htmlData{tutor-start=90,tutor-end=91}{1}。代入求和式得 1RA+1RB+1RC=1R(3−1)=2R\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{R}_{\htmlData{tutor-start=12,tutor-end=13}{A}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{R}_{\htmlData{tutor-start=30,tutor-end=31}{B}}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{R}_{\htmlData{tutor-start=48,tutor-end=49}{C}}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{R}}\htmlData{tutor-start=65,tutor-end=66}{(}\htmlData{tutor-start=66,tutor-end=67}{3} \htmlData{tutor-start=68,tutor-end=69}{-} \htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{)} \htmlData{tutor-start=73,tutor-end=74}{=} \frac{\htmlData{tutor-start=81,tutor-end=82}{2}}{\htmlData{tutor-start=84,tutor-end=85}{R}}。命题得证。

cot⁡∠Acot⁡∠B+cot⁡∠Bcot⁡∠C+cot⁡∠Ccot⁡∠A=1\cot \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A} \cot \htmlData{tutor-start=19,tutor-end=26}{\angle }\htmlData{tutor-start=26,tutor-end=27}{B} \htmlData{tutor-start=28,tutor-end=29}{+} \cot \htmlData{tutor-start=35,tutor-end=42}{\angle }\htmlData{tutor-start=42,tutor-end=43}{B} \cot \htmlData{tutor-start=49,tutor-end=56}{\angle }\htmlData{tutor-start=56,tutor-end=57}{C} \htmlData{tutor-start=58,tutor-end=59}{+} \cot \htmlData{tutor-start=65,tutor-end=72}{\angle }\htmlData{tutor-start=72,tutor-end=73}{C} \cot \htmlData{tutor-start=79,tutor-end=86}{\angle }\htmlData{tutor-start=86,tutor-end=87}{A} \htmlData{tutor-start=88,tutor-end=89}{=} \htmlData{tutor-start=90,tutor-end=91}{1}
2

Day 1 · 组合数学

Given an integer n≥3\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{3}, prove that the set X={1,2,3,…,n2−n}\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{,} \dots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{n}^{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{n}\htmlData{tutor-start=31,tutor-end=33}{\}} can be divided into two non-intersecting subsets such that neither of them contains n\htmlData{tutor-start=0,tutor-end=1}{n} elements a1,a2,…,an\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}} with a1<a2<…<an\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{<} \dots \htmlData{tutor-start=22,tutor-end=23}{<} \htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{n}} and ak≤ak−1+ak+12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=10}{\le }\frac{\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{k}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{a}_{\htmlData{tutor-start=29,tutor-end=30}{k}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{1}}}{\htmlData{tutor-start=35,tutor-end=36}{2}} for all k=2,…,n−1\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \dots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}.

答案：命题得证。将 X 分划为 S 与 T 即可。

题目标签：2008 CMO 第2题：集合的凸性分划

解题过程

（1）构造分划并验证 S 不含满足条件的 n 元子列

构造 X = S \cup T 的分划，并证明 S 中不存在满足 a_1 < \dots < a_n 且 a_k \le \frac{a_{k-1}+a_{k+1}}{2} 的 n 元子列

(1)
构造分划 S 与 T

对每个 k = 1, 2, \dots, n-1，定义两个相邻的整数块： S_k = \{k^2 - k + 1, k^2 - k + 2, \dots, k^2\}，共 k 个元素； T_k = \{k^2 + 1, k^2 + 2, \dots, k^2 + k\}，共 k 个元素。 令 S = \bigcup_{k=1}^{n-1} S_k，T = \bigcup_{k=1}^{n-1} T_k。

验证 S \cup T = X：S_k 与 T_k 合起来正好是 \{k^2 - k + 1, \dots, k^2 + k\} = \{k(k-1)+1, \dots, k(k+1)\}。当 k 从 1 取到 n-1 时，这些块首尾相接，覆盖 \{1, 2, \dots, (n-1)n\} = X。 验证 S \cap T = \emptyset：每个 S_k 与 T_k 不相交，且不同 k 对应的块也不相交。

计算 |S| = \sum_{k=1}^{n-1} k = \frac{n(n-1)}{2}，|T| = \frac{n(n-1)}{2}，两者大小相等。

Sk={k2−k+1,…,k2},Tk={k2+1,…,k2+k},S=⋃k=1n−1Sk,T=⋃k=1n−1Tk\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{k}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{k} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{,} \dots\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{k}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=39}{\}}\htmlData{tutor-start=39,tutor-end=40}{,}\quad \htmlData{tutor-start=46,tutor-end=47}{T}_{\htmlData{tutor-start=49,tutor-end=50}{k}} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=56}{\{}\htmlData{tutor-start=56,tutor-end=57}{k}^{\htmlData{tutor-start=59,tutor-end=60}{2}} \htmlData{tutor-start=62,tutor-end=63}{+} \htmlData{tutor-start=64,tutor-end=65}{1}\htmlData{tutor-start=65,tutor-end=66}{,} \dots\htmlData{tutor-start=72,tutor-end=73}{,} \htmlData{tutor-start=74,tutor-end=75}{k}^{\htmlData{tutor-start=77,tutor-end=78}{2}} \htmlData{tutor-start=80,tutor-end=81}{+} \htmlData{tutor-start=82,tutor-end=83}{k}\htmlData{tutor-start=83,tutor-end=85}{\}}\htmlData{tutor-start=85,tutor-end=86}{,}\quad \htmlData{tutor-start=92,tutor-end=93}{S} \htmlData{tutor-start=94,tutor-end=95}{=} \bigcup_{\htmlData{tutor-start=105,tutor-end=106}{k}\htmlData{tutor-start=106,tutor-end=107}{=}\htmlData{tutor-start=107,tutor-end=108}{1}}^{\htmlData{tutor-start=111,tutor-end=112}{n}\htmlData{tutor-start=112,tutor-end=113}{-}\htmlData{tutor-start=113,tutor-end=114}{1}} \htmlData{tutor-start=116,tutor-end=117}{S}_{\htmlData{tutor-start=119,tutor-end=120}{k}}\htmlData{tutor-start=121,tutor-end=122}{,}\quad \htmlData{tutor-start=128,tutor-end=129}{T} \htmlData{tutor-start=130,tutor-end=131}{=} \bigcup_{\htmlData{tutor-start=141,tutor-end=142}{k}\htmlData{tutor-start=142,tutor-end=143}{=}\htmlData{tutor-start=143,tutor-end=144}{1}}^{\htmlData{tutor-start=147,tutor-end=148}{n}\htmlData{tutor-start=148,tutor-end=149}{-}\htmlData{tutor-start=149,tutor-end=150}{1}} \htmlData{tutor-start=152,tutor-end=153}{T}_{\htmlData{tutor-start=155,tutor-end=156}{k}}
(2)
将凸性条件转化为差分单调不减

设 S 中存在 a_1 < a_2 < \dots < a_n 满足 a_k \le \frac{a_{k-1}+a_{k+1}}{2}（k = 2, \dots, n-1）。 两边乘以 2 并移项：2a_k \le a_{k-1} + a_{k+1}，即 a_k - a_{k-1} \le a_{k+1} - a_k。 令 d_k = a_{k+1} - a_k（k = 1, \dots, n-1），则条件等价于 d_1 \le d_2 \le \dots \le d_{n-1}，即差分序列单调不减。

ak≤ak−1+ak+12  ⟺  ak+1−ak≥ak−ak−1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=10}{\le }\frac{\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{k}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}}}{\htmlData{tutor-start=33,tutor-end=34}{2}} \iff \htmlData{tutor-start=41,tutor-end=42}{a}_{\htmlData{tutor-start=44,tutor-end=45}{k}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{1}} \htmlData{tutor-start=49,tutor-end=50}{-} \htmlData{tutor-start=51,tutor-end=52}{a}_{\htmlData{tutor-start=54,tutor-end=55}{k}} \htmlData{tutor-start=57,tutor-end=61}{\ge }\htmlData{tutor-start=61,tutor-end=62}{a}_{\htmlData{tutor-start=64,tutor-end=65}{k}} \htmlData{tutor-start=67,tutor-end=68}{-} \htmlData{tutor-start=69,tutor-end=70}{a}_{\htmlData{tutor-start=72,tutor-end=73}{k}\htmlData{tutor-start=73,tutor-end=74}{-}\htmlData{tutor-start=74,tutor-end=75}{1}}
(3)
用鸽巢原理定位两个相邻元素落入同一块 S_j

设 a_1 \in S_i。由于 |S_{n-1}| = n-1 < n，a_1 不可能在 S_{n-1} 中（否则 S 中最多只有 n-1 个元素 \ge a_1，不够取 n 个），故 i \le n-2。

在 \{a_1, \dots, a_n\} 中，属于 S_1 \cup \dots \cup S_i 的元素至多 |S_i| = i 个（因为 a_1 是其中最小的，且 S_i 中比 a_1 大的元素至多 i-1 个，加上 a_1 本身至多 i 个）。因此属于 S_{i+1} \cup \dots \cup S_{n-1} 的元素至少有 n - i 个。

这些元素分布在 n - 1 - i 个块 S_{i+1}, \dots, S_{n-1} 中。由鸽巢原理，至少有一个块 S_j（i < j \le n-1）包含其中至少 \lceil \frac{n-i}{n-1-i} \rceil \ge 2 个元素。

取最小的这样的 j，并设 a_k, a_{k+1} 是落入 S_j 的两个相邻元素（在子列中相邻），则 a_{k-1} \in S_1 \cup \dots \cup S_{j-1}（由 j 的最小性，a_{k-1} 不在 S_j 中，且 a_{k-1} < a_k，故 a_{k-1} 在更前面的块中）。

若a1∈Si, 则 i≤n−2.∃ j∈{i+1,…,n−1}, ∃ k: ak,ak+1∈Sj, ak−1∈S1∪…∪Sj−1\htmlData{tutor-start=0,tutor-end=6}{\text{}\htmlData{tutor-start=6,tutor-end=7}{若} } \htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=20}{\in }\htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{,}\text{ \htmlData{tutor-start=33,tutor-end=34}{则} } \htmlData{tutor-start=37,tutor-end=38}{i} \htmlData{tutor-start=39,tutor-end=43}{\le }\htmlData{tutor-start=43,tutor-end=44}{n}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{2}\htmlData{tutor-start=46,tutor-end=47}{.}\quad \htmlData{tutor-start=53,tutor-end=60}{\exists}\, \htmlData{tutor-start=63,tutor-end=64}{j} \htmlData{tutor-start=65,tutor-end=69}{\in }\htmlData{tutor-start=69,tutor-end=71}{\{}\htmlData{tutor-start=71,tutor-end=72}{i}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{1}\htmlData{tutor-start=74,tutor-end=75}{,} \dots\htmlData{tutor-start=81,tutor-end=82}{,} \htmlData{tutor-start=83,tutor-end=84}{n}\htmlData{tutor-start=84,tutor-end=85}{-}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=88}{\}}\htmlData{tutor-start=88,tutor-end=89}{,}\htmlData{tutor-start=89,tutor-end=91}{\ }\htmlData{tutor-start=91,tutor-end=98}{\exists}\, \htmlData{tutor-start=101,tutor-end=102}{k}\htmlData{tutor-start=102,tutor-end=103}{:}\htmlData{tutor-start=103,tutor-end=105}{\ }\htmlData{tutor-start=105,tutor-end=106}{a}_{\htmlData{tutor-start=108,tutor-end=109}{k}}\htmlData{tutor-start=110,tutor-end=111}{,} \htmlData{tutor-start=112,tutor-end=113}{a}_{\htmlData{tutor-start=115,tutor-end=116}{k}\htmlData{tutor-start=116,tutor-end=117}{+}\htmlData{tutor-start=117,tutor-end=118}{1}} \htmlData{tutor-start=120,tutor-end=124}{\in }\htmlData{tutor-start=124,tutor-end=125}{S}_{\htmlData{tutor-start=127,tutor-end=128}{j}}\htmlData{tutor-start=129,tutor-end=130}{,}\htmlData{tutor-start=130,tutor-end=132}{\ }\htmlData{tutor-start=132,tutor-end=133}{a}_{\htmlData{tutor-start=135,tutor-end=136}{k}\htmlData{tutor-start=136,tutor-end=137}{-}\htmlData{tutor-start=137,tutor-end=138}{1}} \htmlData{tutor-start=140,tutor-end=144}{\in }\htmlData{tutor-start=144,tutor-end=145}{S}_{\htmlData{tutor-start=147,tutor-end=148}{1}} \htmlData{tutor-start=150,tutor-end=155}{\cup }\dots \htmlData{tutor-start=161,tutor-end=166}{\cup }\htmlData{tutor-start=166,tutor-end=167}{S}_{\htmlData{tutor-start=169,tutor-end=170}{j}\htmlData{tutor-start=170,tutor-end=171}{-}\htmlData{tutor-start=171,tutor-end=172}{1}}
(4)
比较跨块差分与块内差分，导出矛盾

由 a_k, a_{k+1} \in S_j 且 a_k < a_{k+1}，两者都在 S_j = \{j^2 - j + 1, \dots, j^2\} 中，故 a_{k+1} - a_k \le |S_j| - 1 = j - 1。

由 a_{k-1} \in S_1 \cup \dots \cup S_{j-1} 且 a_k \in S_j，a_{k-1} 最大为 S_{j-1} 的最后一个元素 (j-1)^2，a_k 最小为 S_j 的第一个元素 j^2 - j + 1。但 a_{k-1} 与 a_k 之间还隔着整个 T_{j-1} = \{(j-1)^2 + 1, \dots, (j-1)^2 + (j-1)\}，共 j-1 个元素都不在 S 中。 因此 a_k - a_{k-1} \ge (j^2 - j + 1) - (j-1)^2 = j^2 - j + 1 - j^2 + 2j - 1 = j。

于是 a_{k+1} - a_k \le j - 1 < j \le a_k - a_{k-1}，即 d_k < d_{k-1}，与差分单调不减矛盾。

故 S 中不存在满足条件的 n 元子列。

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（2）对 T 的对称论证

证明 T 中也不存在满足条件的 n 元子列

(1)
T 的块结构与 S 完全对称

T = \bigcup_{k=1}^{n-1} T_k，其中 T_k = \{k^2 + 1, \dots, k^2 + k\}，|T_k| = k。 T 的块结构与 S 完全相同：块长同样为 1, 2, \dots, n-1，且相邻 T 块之间隔着 S 块（|S_k| = k）。

若 T 中存在 a_1 < \dots < a_n 满足差分单调不减，设 a_1 \in T_i（i \le n-2，理由同 S 的情形）。 同理，由鸽巢原理存在最小的 j > i 使得 T_j 包含子列中两个相邻元素 a_k, a_{k+1}，且 a_{k-1} \in T_1 \cup \dots \cup T_{j-1}。

此时 a_{k+1} - a_k \le |T_j| - 1 = j - 1。 而 a_k - a_{k-1} \ge (j^2 + 1) - ((j-1)^2 + (j-1)) = j^2 + 1 - j^2 + 2j - 1 - j + 1 = j + 1 > j - 1。 （这里 a_k 最小为 T_j 的首元素 j^2 + 1，a_{k-1} 最大为 T_{j-1} 的末元素 (j-1)^2 + (j-1) = j^2 - j，中间隔了 S_j 的 j 个元素。）

故 a_{k+1} - a_k < a_k - a_{k-1}，与差分单调不减矛盾。

因此 T 中也不存在满足条件的 n 元子列。结合 S 与 T 构成 X 的分划，命题得证。

ak+1−ak≤j−1,ak−ak−1≥j+1⟹ak+1−ak<ak−ak−1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{k}} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{j} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\quad \htmlData{tutor-start=32,tutor-end=33}{a}_{\htmlData{tutor-start=35,tutor-end=36}{k}} \htmlData{tutor-start=38,tutor-end=39}{-} \htmlData{tutor-start=40,tutor-end=41}{a}_{\htmlData{tutor-start=43,tutor-end=44}{k}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{1}} \htmlData{tutor-start=48,tutor-end=52}{\ge }\htmlData{tutor-start=52,tutor-end=53}{j} \htmlData{tutor-start=54,tutor-end=55}{+} \htmlData{tutor-start=56,tutor-end=57}{1} \quad \htmlData{tutor-start=64,tutor-end=80}{\Longrightarrow }\quad \htmlData{tutor-start=86,tutor-end=87}{a}_{\htmlData{tutor-start=89,tutor-end=90}{k}\htmlData{tutor-start=90,tutor-end=91}{+}\htmlData{tutor-start=91,tutor-end=92}{1}} \htmlData{tutor-start=94,tutor-end=95}{-} \htmlData{tutor-start=96,tutor-end=97}{a}_{\htmlData{tutor-start=99,tutor-end=100}{k}} \htmlData{tutor-start=102,tutor-end=103}{<} \htmlData{tutor-start=104,tutor-end=105}{a}_{\htmlData{tutor-start=107,tutor-end=108}{k}} \htmlData{tutor-start=110,tutor-end=111}{-} \htmlData{tutor-start=112,tutor-end=113}{a}_{\htmlData{tutor-start=115,tutor-end=116}{k}\htmlData{tutor-start=116,tutor-end=117}{-}\htmlData{tutor-start=117,tutor-end=118}{1}}
(2)
汇总结论

综上，我们构造了 X 的一个分划 X = S \cup T，S \cap T = \emptyset，且 S 与 T 中都不存在 n 个元素 a_1 < a_2 < \dots < a_n 满足 a_k \le \frac{a_{k-1}+a_{k+1}}{2}（k = 2, \dots, n-1）。

具体地： - S = \bigcup_{k=1}^{n-1} \{k^2 - k + 1, \dots, k^2\}； - T = \bigcup_{k=1}^{n-1} \{k^2 + 1, \dots, k^2 + k\}。

对 S 的论证：若存在满足条件的子列，由鸽巢原理必有某块 S_j 包含子列中两个相邻元素 a_k, a_{k+1}，且前一个元素 a_{k-1} 在更前的块中。此时块内差分 a_{k+1} - a_k \le j - 1，而跨块差分 a_k - a_{k-1} \ge j，与差分单调不减矛盾。

对 T 的论证完全对称：块内差分 \le j - 1，跨块差分 \ge j + 1，同样矛盾。

因此这样的分划 S, T 满足题目要求，命题得证。

X=S∪T,S∩T=∅,S,T 均不含满足条件的 n 元子列\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{S} \htmlData{tutor-start=6,tutor-end=11}{\cup }\htmlData{tutor-start=11,tutor-end=12}{T}\htmlData{tutor-start=12,tutor-end=13}{,}\quad \htmlData{tutor-start=19,tutor-end=20}{S} \htmlData{tutor-start=21,tutor-end=26}{\cap }\htmlData{tutor-start=26,tutor-end=27}{T} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=39}{\emptyset}\htmlData{tutor-start=39,tutor-end=40}{,}\quad \htmlData{tutor-start=46,tutor-end=47}{S}\htmlData{tutor-start=47,tutor-end=48}{,} \htmlData{tutor-start=49,tutor-end=50}{T} \text{ \htmlData{tutor-start=58,tutor-end=59}{均}\htmlData{tutor-start=59,tutor-end=60}{不}\htmlData{tutor-start=60,tutor-end=61}{含}\htmlData{tutor-start=61,tutor-end=62}{满}\htmlData{tutor-start=62,tutor-end=63}{足}\htmlData{tutor-start=63,tutor-end=64}{条}\htmlData{tutor-start=64,tutor-end=65}{件}\htmlData{tutor-start=65,tutor-end=66}{的} } \htmlData{tutor-start=69,tutor-end=70}{n} \text{ \htmlData{tutor-start=78,tutor-end=79}{元}\htmlData{tutor-start=79,tutor-end=80}{子}\htmlData{tutor-start=80,tutor-end=81}{列}}
3

Day 1 · 代数

Given an integer n>0\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0} and real numbers x1≤x2≤…≤xn\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=10}{\le }\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\le }\dots \htmlData{tutor-start=26,tutor-end=30}{\le }\htmlData{tutor-start=30,tutor-end=31}{x}_{\htmlData{tutor-start=33,tutor-end=34}{n}}, y1≥y2≥…≥yn\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=10}{\ge }\htmlData{tutor-start=10,tutor-end=11}{y}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\ge }\dots \htmlData{tutor-start=26,tutor-end=30}{\ge }\htmlData{tutor-start=30,tutor-end=31}{y}_{\htmlData{tutor-start=33,tutor-end=34}{n}}, satisfying ∑i=1nixi=∑i=1niyi\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{i}\htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{i}} \htmlData{tutor-start=22,tutor-end=23}{=} \sum_{\htmlData{tutor-start=30,tutor-end=31}{i}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{1}}^{\htmlData{tutor-start=36,tutor-end=37}{n}} \htmlData{tutor-start=39,tutor-end=40}{i}\htmlData{tutor-start=40,tutor-end=41}{y}_{\htmlData{tutor-start=43,tutor-end=44}{i}}. Prove that for any real number α\htmlData{tutor-start=0,tutor-end=6}{\alpha}, ∑i=1nxi[iα]≥∑i=1nyi[iα]\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{x}_{\htmlData{tutor-start=18,tutor-end=19}{i}}\htmlData{tutor-start=20,tutor-end=21}{[}\htmlData{tutor-start=21,tutor-end=22}{i}\htmlData{tutor-start=22,tutor-end=28}{\alpha}\htmlData{tutor-start=28,tutor-end=29}{]} \htmlData{tutor-start=30,tutor-end=34}{\ge }\sum_{\htmlData{tutor-start=40,tutor-end=41}{i}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{1}}^{\htmlData{tutor-start=46,tutor-end=47}{n}} \htmlData{tutor-start=49,tutor-end=50}{y}_{\htmlData{tutor-start=52,tutor-end=53}{i}}\htmlData{tutor-start=54,tutor-end=55}{[}\htmlData{tutor-start=55,tutor-end=56}{i}\htmlData{tutor-start=56,tutor-end=62}{\alpha}\htmlData{tutor-start=62,tutor-end=63}{]}, where [β]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=6}{\beta}\htmlData{tutor-start=6,tutor-end=7}{]} is defined as the greatest integer less than or equal to β\htmlData{tutor-start=0,tutor-end=5}{\beta}.

答案：命题得证。

题目标签：2008 CMO 第 3 题：带取整权重的排序不等式

解题过程

主问题

证明 ∑i=1nxi[iα]≥∑i=1nyi[iα]\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{x}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=35}{\ge }\sum_{\htmlData{tutor-start=41,tutor-end=42}{i}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1}}^{\htmlData{tutor-start=47,tutor-end=48}{n}} \htmlData{tutor-start=50,tutor-end=51}{y}_{\htmlData{tutor-start=53,tutor-end=54}{i}} \htmlData{tutor-start=56,tutor-end=57}{[}\htmlData{tutor-start=57,tutor-end=58}{i}\htmlData{tutor-start=58,tutor-end=64}{\alpha}\htmlData{tutor-start=64,tutor-end=65}{]}。

(1)
差分替换，将问题转化为关于 zi=xi−yi\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{i}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{i}} 的不等式

令 zi=xi−yi\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{i}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{i}}（i=1,2,…,n\htmlData{tutor-start=0,tutor-end=1}{i} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \dots\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{n}）。由 x1≤x2≤…≤xn\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=10}{\le }\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\le }\dots \htmlData{tutor-start=26,tutor-end=30}{\le }\htmlData{tutor-start=30,tutor-end=31}{x}_{\htmlData{tutor-start=33,tutor-end=34}{n}} 与 y1≥y2≥…≥yn\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=10}{\ge }\htmlData{tutor-start=10,tutor-end=11}{y}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\ge }\dots \htmlData{tutor-start=26,tutor-end=30}{\ge }\htmlData{tutor-start=30,tutor-end=31}{y}_{\htmlData{tutor-start=33,tutor-end=34}{n}} 可知 z1≤z2≤…≤zn\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=10}{\le }\htmlData{tutor-start=10,tutor-end=11}{z}_{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\le }\dots \htmlData{tutor-start=26,tutor-end=30}{\le }\htmlData{tutor-start=30,tutor-end=31}{z}_{\htmlData{tutor-start=33,tutor-end=34}{n}}；由题设条件得 ∑i=1nizi=0\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{i} \htmlData{tutor-start=17,tutor-end=18}{z}_{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{0}。原不等式等价于证明 ∑i=1nzi[iα]≥0.\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=35}{\ge }\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{.}

∑i=1nzi[iα]≥0\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=35}{\ge }\htmlData{tutor-start=35,tutor-end=36}{0}
(2)
二阶差分展开，利用约束消去 Δ1\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{1}}

令 Δ1=z1\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{z}_{\htmlData{tutor-start=16,tutor-end=17}{1}}，Δj=zj−zj−1\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{j}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{z}_{\htmlData{tutor-start=16,tutor-end=17}{j}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{z}_{\htmlData{tutor-start=24,tutor-end=25}{j}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}}（j=2,…,n\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \dots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{n}）。由 zi\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 单调递增知 Δj≥0\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{j}} \htmlData{tutor-start=11,tutor-end=15}{\ge }\htmlData{tutor-start=15,tutor-end=16}{0}（j≥2\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{2}），且 zi=∑j=1iΔj\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \sum_{\htmlData{tutor-start=14,tutor-end=15}{j}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}}^{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=29}{\Delta}_{\htmlData{tutor-start=31,tutor-end=32}{j}}。交换求和顺序： ∑i=1nizi=∑j=1nΔj∑i=jni=0.\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{i} \htmlData{tutor-start=17,tutor-end=18}{z}_{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=24}{=} \sum_{\htmlData{tutor-start=31,tutor-end=32}{j}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{1}}^{\htmlData{tutor-start=37,tutor-end=38}{n}} \htmlData{tutor-start=40,tutor-end=46}{\Delta}_{\htmlData{tutor-start=48,tutor-end=49}{j}} \sum_{\htmlData{tutor-start=57,tutor-end=58}{i}\htmlData{tutor-start=58,tutor-end=59}{=}\htmlData{tutor-start=59,tutor-end=60}{j}}^{\htmlData{tutor-start=63,tutor-end=64}{n}} \htmlData{tutor-start=66,tutor-end=67}{i} \htmlData{tutor-start=68,tutor-end=69}{=} \htmlData{tutor-start=70,tutor-end=71}{0}\htmlData{tutor-start=71,tutor-end=72}{.} 由此解出 Δ1=−∑j=2nΔj∑i=jni∑i=1ni.\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\frac{\sum_{\htmlData{tutor-start=26,tutor-end=27}{j}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{2}}^{\htmlData{tutor-start=32,tutor-end=33}{n}} \htmlData{tutor-start=35,tutor-end=41}{\Delta}_{\htmlData{tutor-start=43,tutor-end=44}{j}} \sum_{\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=54}{=}\htmlData{tutor-start=54,tutor-end=55}{j}}^{\htmlData{tutor-start=58,tutor-end=59}{n}} \htmlData{tutor-start=61,tutor-end=62}{i}}{\sum_{\htmlData{tutor-start=70,tutor-end=71}{i}\htmlData{tutor-start=71,tutor-end=72}{=}\htmlData{tutor-start=72,tutor-end=73}{1}}^{\htmlData{tutor-start=76,tutor-end=77}{n}} \htmlData{tutor-start=79,tutor-end=80}{i}}\htmlData{tutor-start=81,tutor-end=82}{.} 再对目标式同样交换求和顺序： ∑i=1nzi[iα]=∑j=1nΔj∑i=jn[iα].\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=32}{=} \sum_{\htmlData{tutor-start=39,tutor-end=40}{j}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{1}}^{\htmlData{tutor-start=45,tutor-end=46}{n}} \htmlData{tutor-start=48,tutor-end=54}{\Delta}_{\htmlData{tutor-start=56,tutor-end=57}{j}} \sum_{\htmlData{tutor-start=65,tutor-end=66}{i}\htmlData{tutor-start=66,tutor-end=67}{=}\htmlData{tutor-start=67,tutor-end=68}{j}}^{\htmlData{tutor-start=71,tutor-end=72}{n}} \htmlData{tutor-start=74,tutor-end=75}{[}\htmlData{tutor-start=75,tutor-end=76}{i}\htmlData{tutor-start=76,tutor-end=82}{\alpha}\htmlData{tutor-start=82,tutor-end=83}{]}\htmlData{tutor-start=83,tutor-end=84}{.} 将 Δ1\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{1}} 的表达式代入，整理得 ∑i=1nzi[iα]=∑j=2nΔj(∑i=jn[iα]−∑i=jni∑i=1ni∑i=1n[iα]).\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=32}{=} \sum_{\htmlData{tutor-start=39,tutor-end=40}{j}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{2}}^{\htmlData{tutor-start=45,tutor-end=46}{n}} \htmlData{tutor-start=48,tutor-end=54}{\Delta}_{\htmlData{tutor-start=56,tutor-end=57}{j}} \left( \sum_{\htmlData{tutor-start=72,tutor-end=73}{i}\htmlData{tutor-start=73,tutor-end=74}{=}\htmlData{tutor-start=74,tutor-end=75}{j}}^{\htmlData{tutor-start=78,tutor-end=79}{n}} \htmlData{tutor-start=81,tutor-end=82}{[}\htmlData{tutor-start=82,tutor-end=83}{i}\htmlData{tutor-start=83,tutor-end=89}{\alpha}\htmlData{tutor-start=89,tutor-end=90}{]} \htmlData{tutor-start=91,tutor-end=92}{-} \frac{\sum_{\htmlData{tutor-start=105,tutor-end=106}{i}\htmlData{tutor-start=106,tutor-end=107}{=}\htmlData{tutor-start=107,tutor-end=108}{j}}^{\htmlData{tutor-start=111,tutor-end=112}{n}} \htmlData{tutor-start=114,tutor-end=115}{i}}{\sum_{\htmlData{tutor-start=123,tutor-end=124}{i}\htmlData{tutor-start=124,tutor-end=125}{=}\htmlData{tutor-start=125,tutor-end=126}{1}}^{\htmlData{tutor-start=129,tutor-end=130}{n}} \htmlData{tutor-start=132,tutor-end=133}{i}} \sum_{\htmlData{tutor-start=141,tutor-end=142}{i}\htmlData{tutor-start=142,tutor-end=143}{=}\htmlData{tutor-start=143,tutor-end=144}{1}}^{\htmlData{tutor-start=147,tutor-end=148}{n}} \htmlData{tutor-start=150,tutor-end=151}{[}\htmlData{tutor-start=151,tutor-end=152}{i}\htmlData{tutor-start=152,tutor-end=158}{\alpha}\htmlData{tutor-start=158,tutor-end=159}{]} \right)\htmlData{tutor-start=167,tutor-end=168}{.}

∑i=1nzi[iα]=∑j=2nΔj(∑i=jn[iα]−∑i=jni∑i=1ni∑i=1n[iα])\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=32}{=} \sum_{\htmlData{tutor-start=39,tutor-end=40}{j}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{2}}^{\htmlData{tutor-start=45,tutor-end=46}{n}} \htmlData{tutor-start=48,tutor-end=54}{\Delta}_{\htmlData{tutor-start=56,tutor-end=57}{j}} \left( \sum_{\htmlData{tutor-start=72,tutor-end=73}{i}\htmlData{tutor-start=73,tutor-end=74}{=}\htmlData{tutor-start=74,tutor-end=75}{j}}^{\htmlData{tutor-start=78,tutor-end=79}{n}} \htmlData{tutor-start=81,tutor-end=82}{[}\htmlData{tutor-start=82,tutor-end=83}{i}\htmlData{tutor-start=83,tutor-end=89}{\alpha}\htmlData{tutor-start=89,tutor-end=90}{]} \htmlData{tutor-start=91,tutor-end=92}{-} \frac{\sum_{\htmlData{tutor-start=105,tutor-end=106}{i}\htmlData{tutor-start=106,tutor-end=107}{=}\htmlData{tutor-start=107,tutor-end=108}{j}}^{\htmlData{tutor-start=111,tutor-end=112}{n}} \htmlData{tutor-start=114,tutor-end=115}{i}}{\sum_{\htmlData{tutor-start=123,tutor-end=124}{i}\htmlData{tutor-start=124,tutor-end=125}{=}\htmlData{tutor-start=125,tutor-end=126}{1}}^{\htmlData{tutor-start=129,tutor-end=130}{n}} \htmlData{tutor-start=132,tutor-end=133}{i}} \sum_{\htmlData{tutor-start=141,tutor-end=142}{i}\htmlData{tutor-start=142,tutor-end=143}{=}\htmlData{tutor-start=143,tutor-end=144}{1}}^{\htmlData{tutor-start=147,tutor-end=148}{n}} \htmlData{tutor-start=150,tutor-end=151}{[}\htmlData{tutor-start=151,tutor-end=152}{i}\htmlData{tutor-start=152,tutor-end=158}{\alpha}\htmlData{tutor-start=158,tutor-end=159}{]} \right)
(3)
化归为平均值不等式：证明 ∑i=jn[iα]∑i=jni≥∑i=1n[iα]∑i=1ni\frac{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{j}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]}}{\sum_{\htmlData{tutor-start=38,tutor-end=39}{i}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{j}}^{\htmlData{tutor-start=44,tutor-end=45}{n}} \htmlData{tutor-start=47,tutor-end=48}{i}} \htmlData{tutor-start=50,tutor-end=54}{\ge }\frac{\sum_{\htmlData{tutor-start=66,tutor-end=67}{i}\htmlData{tutor-start=67,tutor-end=68}{=}\htmlData{tutor-start=68,tutor-end=69}{1}}^{\htmlData{tutor-start=72,tutor-end=73}{n}} \htmlData{tutor-start=75,tutor-end=76}{[}\htmlData{tutor-start=76,tutor-end=77}{i}\htmlData{tutor-start=77,tutor-end=83}{\alpha}\htmlData{tutor-start=83,tutor-end=84}{]}}{\sum_{\htmlData{tutor-start=92,tutor-end=93}{i}\htmlData{tutor-start=93,tutor-end=94}{=}\htmlData{tutor-start=94,tutor-end=95}{1}}^{\htmlData{tutor-start=98,tutor-end=99}{n}} \htmlData{tutor-start=101,tutor-end=102}{i}}

由于 Δj≥0\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{j}} \htmlData{tutor-start=11,tutor-end=15}{\ge }\htmlData{tutor-start=15,tutor-end=16}{0}（j≥2\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{2}），只需对每个 j∈{2,…,n}\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=8}{\{}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,} \dots\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=21}{\}} 证明 ∑i=jn[iα]∑i=jni≥∑i=1n[iα]∑i=1ni.\frac{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{j}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]}}{\sum_{\htmlData{tutor-start=38,tutor-end=39}{i}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{j}}^{\htmlData{tutor-start=44,tutor-end=45}{n}} \htmlData{tutor-start=47,tutor-end=48}{i}} \htmlData{tutor-start=50,tutor-end=54}{\ge }\frac{\sum_{\htmlData{tutor-start=66,tutor-end=67}{i}\htmlData{tutor-start=67,tutor-end=68}{=}\htmlData{tutor-start=68,tutor-end=69}{1}}^{\htmlData{tutor-start=72,tutor-end=73}{n}} \htmlData{tutor-start=75,tutor-end=76}{[}\htmlData{tutor-start=76,tutor-end=77}{i}\htmlData{tutor-start=77,tutor-end=83}{\alpha}\htmlData{tutor-start=83,tutor-end=84}{]}}{\sum_{\htmlData{tutor-start=92,tutor-end=93}{i}\htmlData{tutor-start=93,tutor-end=94}{=}\htmlData{tutor-start=94,tutor-end=95}{1}}^{\htmlData{tutor-start=98,tutor-end=99}{n}} \htmlData{tutor-start=101,tutor-end=102}{i}}\htmlData{tutor-start=103,tutor-end=104}{.} 利用分比性质，这等价于 ∑i=jn[iα]∑i=jni≥∑i=1j−1[iα]∑i=1j−1i,\frac{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{j}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]}}{\sum_{\htmlData{tutor-start=38,tutor-end=39}{i}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{j}}^{\htmlData{tutor-start=44,tutor-end=45}{n}} \htmlData{tutor-start=47,tutor-end=48}{i}} \htmlData{tutor-start=50,tutor-end=54}{\ge }\frac{\sum_{\htmlData{tutor-start=66,tutor-end=67}{i}\htmlData{tutor-start=67,tutor-end=68}{=}\htmlData{tutor-start=68,tutor-end=69}{1}}^{\htmlData{tutor-start=72,tutor-end=73}{j}\htmlData{tutor-start=73,tutor-end=74}{-}\htmlData{tutor-start=74,tutor-end=75}{1}} \htmlData{tutor-start=77,tutor-end=78}{[}\htmlData{tutor-start=78,tutor-end=79}{i}\htmlData{tutor-start=79,tutor-end=85}{\alpha}\htmlData{tutor-start=85,tutor-end=86}{]}}{\sum_{\htmlData{tutor-start=94,tutor-end=95}{i}\htmlData{tutor-start=95,tutor-end=96}{=}\htmlData{tutor-start=96,tutor-end=97}{1}}^{\htmlData{tutor-start=100,tutor-end=101}{j}\htmlData{tutor-start=101,tutor-end=102}{-}\htmlData{tutor-start=102,tutor-end=103}{1}} \htmlData{tutor-start=105,tutor-end=106}{i}}\htmlData{tutor-start=107,tutor-end=108}{,} 也等价于对任意 1≤k<n\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{k} \htmlData{tutor-start=8,tutor-end=9}{<} \htmlData{tutor-start=10,tutor-end=11}{n}， ∑i=1k+1[iα]∑i=1k+1i≥∑i=1k[iα]∑i=1ki.\frac{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{1}} \htmlData{tutor-start=23,tutor-end=24}{[}\htmlData{tutor-start=24,tutor-end=25}{i}\htmlData{tutor-start=25,tutor-end=31}{\alpha}\htmlData{tutor-start=31,tutor-end=32}{]}}{\sum_{\htmlData{tutor-start=40,tutor-end=41}{i}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{1}}^{\htmlData{tutor-start=46,tutor-end=47}{k}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{1}} \htmlData{tutor-start=51,tutor-end=52}{i}} \htmlData{tutor-start=54,tutor-end=58}{\ge }\frac{\sum_{\htmlData{tutor-start=70,tutor-end=71}{i}\htmlData{tutor-start=71,tutor-end=72}{=}\htmlData{tutor-start=72,tutor-end=73}{1}}^{\htmlData{tutor-start=76,tutor-end=77}{k}} \htmlData{tutor-start=79,tutor-end=80}{[}\htmlData{tutor-start=80,tutor-end=81}{i}\htmlData{tutor-start=81,tutor-end=87}{\alpha}\htmlData{tutor-start=87,tutor-end=88}{]}}{\sum_{\htmlData{tutor-start=96,tutor-end=97}{i}\htmlData{tutor-start=97,tutor-end=98}{=}\htmlData{tutor-start=98,tutor-end=99}{1}}^{\htmlData{tutor-start=102,tutor-end=103}{k}} \htmlData{tutor-start=105,tutor-end=106}{i}}\htmlData{tutor-start=107,tutor-end=108}{.} 通分后（注意 ∑i=1ki=k(k+1)2\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{k}} \htmlData{tutor-start=15,tutor-end=16}{i} \htmlData{tutor-start=17,tutor-end=18}{=} \frac{\htmlData{tutor-start=25,tutor-end=26}{k}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}}{\htmlData{tutor-start=33,tutor-end=34}{2}}，∑i=1k+1i=(k+1)(k+2)2\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}} \htmlData{tutor-start=17,tutor-end=18}{i} \htmlData{tutor-start=19,tutor-end=20}{=} \frac{\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{k}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{k}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{)}}{\htmlData{tutor-start=39,tutor-end=40}{2}}），上式等价于 [(k+1)α]⋅k2≥∑i=1k[iα].\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\frac{\htmlData{tutor-start=26,tutor-end=27}{k}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=36}{\ge }\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}} \htmlData{tutor-start=51,tutor-end=52}{[}\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=59}{\alpha}\htmlData{tutor-start=59,tutor-end=60}{]}\htmlData{tutor-start=60,tutor-end=61}{.}

[(k+1)α]⋅k2≥∑i=1k[iα]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\frac{\htmlData{tutor-start=26,tutor-end=27}{k}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=36}{\ge }\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}} \htmlData{tutor-start=51,tutor-end=52}{[}\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=59}{\alpha}\htmlData{tutor-start=59,tutor-end=60}{]}
(4)
利用取整函数的超可加性完成证明

对任意实数 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}，取整函数满足 [x+y]≥[x]+[y]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{]} \htmlData{tutor-start=6,tutor-end=10}{\ge }\htmlData{tutor-start=10,tutor-end=11}{[}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{[}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{]}（超可加性）。因此对每个 i∈{1,…,k}\htmlData{tutor-start=0,tutor-end=1}{i} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=8}{\{}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \dots\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=21}{\}}， [(k+1)α]=[iα+(k+1−i)α]≥[iα]+[(k+1−i)α].\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{[}\htmlData{tutor-start=17,tutor-end=18}{i}\htmlData{tutor-start=18,tutor-end=25}{\alpha }\htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{k}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{i}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=40}{\alpha}\htmlData{tutor-start=40,tutor-end=41}{]} \htmlData{tutor-start=42,tutor-end=46}{\ge }\htmlData{tutor-start=46,tutor-end=47}{[}\htmlData{tutor-start=47,tutor-end=48}{i}\htmlData{tutor-start=48,tutor-end=54}{\alpha}\htmlData{tutor-start=54,tutor-end=55}{]} \htmlData{tutor-start=56,tutor-end=57}{+} \htmlData{tutor-start=58,tutor-end=59}{[}\htmlData{tutor-start=59,tutor-end=60}{(}\htmlData{tutor-start=60,tutor-end=61}{k}\htmlData{tutor-start=61,tutor-end=62}{+}\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{i}\htmlData{tutor-start=65,tutor-end=66}{)}\htmlData{tutor-start=66,tutor-end=72}{\alpha}\htmlData{tutor-start=72,tutor-end=73}{]}\htmlData{tutor-start=73,tutor-end=74}{.} 对 i=1,2,…,k\htmlData{tutor-start=0,tutor-end=1}{i} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \dots\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{k} 求和： k⋅[(k+1)α]≥∑i=1k[iα]+∑i=1k[(k+1−i)α]=2∑i=1k[iα].\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{[}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=20}{\alpha}\htmlData{tutor-start=20,tutor-end=21}{]} \htmlData{tutor-start=22,tutor-end=26}{\ge }\sum_{\htmlData{tutor-start=32,tutor-end=33}{i}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{1}}^{\htmlData{tutor-start=38,tutor-end=39}{k}} \htmlData{tutor-start=41,tutor-end=42}{[}\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=49}{\alpha}\htmlData{tutor-start=49,tutor-end=50}{]} \htmlData{tutor-start=51,tutor-end=52}{+} \sum_{\htmlData{tutor-start=59,tutor-end=60}{i}\htmlData{tutor-start=60,tutor-end=61}{=}\htmlData{tutor-start=61,tutor-end=62}{1}}^{\htmlData{tutor-start=65,tutor-end=66}{k}} \htmlData{tutor-start=68,tutor-end=69}{[}\htmlData{tutor-start=69,tutor-end=70}{(}\htmlData{tutor-start=70,tutor-end=71}{k}\htmlData{tutor-start=71,tutor-end=72}{+}\htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{-}\htmlData{tutor-start=74,tutor-end=75}{i}\htmlData{tutor-start=75,tutor-end=76}{)}\htmlData{tutor-start=76,tutor-end=82}{\alpha}\htmlData{tutor-start=82,tutor-end=83}{]} \htmlData{tutor-start=84,tutor-end=85}{=} \htmlData{tutor-start=86,tutor-end=87}{2} \sum_{\htmlData{tutor-start=94,tutor-end=95}{i}\htmlData{tutor-start=95,tutor-end=96}{=}\htmlData{tutor-start=96,tutor-end=97}{1}}^{\htmlData{tutor-start=100,tutor-end=101}{k}} \htmlData{tutor-start=103,tutor-end=104}{[}\htmlData{tutor-start=104,tutor-end=105}{i}\htmlData{tutor-start=105,tutor-end=111}{\alpha}\htmlData{tutor-start=111,tutor-end=112}{]}\htmlData{tutor-start=112,tutor-end=113}{.} 两边除以 2 即得 [(k+1)α]⋅k2≥∑i=1k[iα]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\frac{\htmlData{tutor-start=26,tutor-end=27}{k}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=36}{\ge }\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}} \htmlData{tutor-start=51,tutor-end=52}{[}\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=59}{\alpha}\htmlData{tutor-start=59,tutor-end=60}{]}，这正是上一步所需的不等式。由此所有 Δj\htmlData{tutor-start=0,tutor-end=6}{\Delta}_{\htmlData{tutor-start=8,tutor-end=9}{j}}（j≥2\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{2}）前的系数均非负，故 ∑i=1nzi[iα]≥0\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{z}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=35}{\ge }\htmlData{tutor-start=35,tutor-end=36}{0}，原不等式得证。

[(k+1)α]⋅k2≥∑i=1k[iα]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{]} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\frac{\htmlData{tutor-start=26,tutor-end=27}{k}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=36}{\ge }\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}} \htmlData{tutor-start=51,tutor-end=52}{[}\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=59}{\alpha}\htmlData{tutor-start=59,tutor-end=60}{]}
4

Day 2 · 数论

Let n>1\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1} be a given integer and A\htmlData{tutor-start=0,tutor-end=1}{A} be an infinite set of positive integers satisfying: for any prime p∤n\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=8}{\nmid }\htmlData{tutor-start=8,tutor-end=9}{n}, there exist infinitely many elements of A\htmlData{tutor-start=0,tutor-end=1}{A} not divisible by p\htmlData{tutor-start=0,tutor-end=1}{p}. Prove that for any integer m>1\htmlData{tutor-start=0,tutor-end=1}{m} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}, (m,n)=1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{1}, there exists a finite subset of A\htmlData{tutor-start=0,tutor-end=1}{A} whose sum of elements, say S\htmlData{tutor-start=0,tutor-end=1}{S}, satisfies S≡1(modm)S \equiv 1 \pmod m and S≡0(modn)S \equiv 0 \pmod n.

答案：命题得证。对 m\htmlData{tutor-start=0,tutor-end=1}{m} 的每个素因子幂 piai\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{i}}^{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{i}}}，利用抽屉原理与孙子定理构造有限子集 Bi\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}，使其元素和模 piai\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{i}}^{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{i}}} 余 1\htmlData{tutor-start=0,tutor-end=1}{1}、模 mn/piai\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{p}_{\htmlData{tutor-start=6,tutor-end=7}{i}}^{\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{i}}} 余 0\htmlData{tutor-start=0,tutor-end=1}{0}；再取这些子集的并即得所求。

题目标签：2008 CMO 第4题：无穷集上的模同余子集和

解题过程

主问题：构造满足双模条件的有限子集

证明存在 A\htmlData{tutor-start=0,tutor-end=1}{A} 的有限子集 B\htmlData{tutor-start=0,tutor-end=1}{B}，其元素之和 S\htmlData{tutor-start=0,tutor-end=1}{S} 满足 S≡1htmlDatatutor−start=10,tutor−end=11pmodm\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{1}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{m} 且 S≡0htmlDatatutor−start=10,tutor−end=11pmodn\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{0}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{n}。

(1)
对 m\htmlData{tutor-start=0,tutor-end=1}{m} 的每个素因子幂 pa\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{a}} 构造局部子集

设 pa∣m\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{a}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{m}。由题设 p∤n\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{n}，故 A\htmlData{tutor-start=0,tutor-end=1}{A} 中有无穷多个元素不被 p\htmlData{tutor-start=0,tutor-end=1}{p} 整除，记此无穷子集为 A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}。对 A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 中元素模 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} 分类，由抽屉原理，存在无穷子集 A2⊆A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=15}{\subseteq }\htmlData{tutor-start=15,tutor-end=16}{A}_{\htmlData{tutor-start=18,tutor-end=19}{1}} 及正整数 a\htmlData{tutor-start=0,tutor-end=1}{a}（p∤a\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{a}），使得 A2\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 中每个元素 x\htmlData{tutor-start=0,tutor-end=1}{x} 都满足 x≡ahtmlDatatutor−start=10,tutor−end=11pmodmn\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{a}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d{\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{n}}。

由于 (m,n)=1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}，故 (pa, mn/pa)=1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}^{\htmlData{tutor-start=4,tutor-end=5}{a}}\htmlData{tutor-start=6,tutor-end=7}{,}\,\htmlData{tutor-start=9,tutor-end=10}{m}\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{p}^{\htmlData{tutor-start=15,tutor-end=16}{a}}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{1}。由孙子定理，同余方程组 {x≡a−1(modpa),x≡0(modmn/pa)\begin{cases} x\equiv a^{-1}\pmod{p^{a}},\\ x\equiv 0\pmod{mn/p^{a}} \end{cases} 有无穷多解。取其中一个正整数解 x\htmlData{tutor-start=0,tutor-end=1}{x}，令 Bp\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{p}} 为 A2\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的前 x\htmlData{tutor-start=0,tutor-end=1}{x} 个元素构成的子集，其元素和记为 Sp\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{p}}。因 A2\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 中每个元素模 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} 均余 a\htmlData{tutor-start=0,tutor-end=1}{a}，故 Sp≡ax(modmn)\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{p}}\htmlData{tutor-start=5,tutor-end=12}{\equiv }\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{x}\pmod{\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{n}}。由方程组得 Sp≡ax≡1(modpa),Sp≡ax≡0(modmn/pa).S_{p}\equiv ax\equiv 1\pmod{p^{a}},\quad S_{p}\equiv ax\equiv 0\pmod{mn/p^{a}}.

{x≡a−1(modpa),x≡0(modmn/pa)  ⟹  Sp≡1(modpa), Sp≡0(modmn/pa)\begin{cases}x\equiv a^{-1}\pmod{p^{a}},\\x\equiv 0\pmod{mn/p^{a}}\end{cases}\implies S_{p}\equiv 1\pmod{p^{a}},\ S_{p}\equiv 0\pmod{mn/p^{a}}
(2)
合并各素因子幂对应的子集

设 m=p1a1⋯pkak\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{p}_{\htmlData{tutor-start=5,tutor-end=6}{1}}^{\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}}\cdots \htmlData{tutor-start=22,tutor-end=23}{p}_{\htmlData{tutor-start=25,tutor-end=26}{k}}^{\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{k}}}。对每个 pi\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{i}}（1≤i≤k\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{i}\htmlData{tutor-start=6,tutor-end=10}{\le }\htmlData{tutor-start=10,tutor-end=11}{k}），按上述方法选取有限子集 Bi⊆A\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=15}{\subseteq }\htmlData{tutor-start=15,tutor-end=16}{A}，且要求 Bi⊆A∖(B1∪⋯∪Bi−1)\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=15}{\subseteq }\htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=25}{\setminus}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{B}_{\htmlData{tutor-start=29,tutor-end=30}{1}}\htmlData{tutor-start=31,tutor-end=35}{\cup}\cdots\htmlData{tutor-start=41,tutor-end=46}{\cup }\htmlData{tutor-start=46,tutor-end=47}{B}_{\htmlData{tutor-start=49,tutor-end=50}{i}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{1}}\htmlData{tutor-start=53,tutor-end=54}{)}（因 A\htmlData{tutor-start=0,tutor-end=1}{A} 无穷，每次只取有限个，故总能做到）。设 Bi\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 的元素和为 Si\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{i}}，则 Si≡1(modpiai),Si≡0(modmn/piai).S_{i}\equiv 1\pmod{p_{i}^{a_{i}}},\quad S_{i}\equiv 0\pmod{mn/p_{i}^{a_{i}}}. 令 B=⋃i=1kBi\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\bigcup_{\htmlData{tutor-start=11,tutor-end=12}{i}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}}^{\htmlData{tutor-start=17,tutor-end=18}{k}} \htmlData{tutor-start=20,tutor-end=21}{B}_{\htmlData{tutor-start=23,tutor-end=24}{i}}，其元素和 S=∑i=1kSi\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\sum_{\htmlData{tutor-start=8,tutor-end=9}{i}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}}^{\htmlData{tutor-start=14,tutor-end=15}{k}} \htmlData{tutor-start=17,tutor-end=18}{S}_{\htmlData{tutor-start=20,tutor-end=21}{i}}。对每个 i\htmlData{tutor-start=0,tutor-end=1}{i}，因 Sj≡0(modpiai)\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{j}}\htmlData{tutor-start=5,tutor-end=12}{\equiv }\htmlData{tutor-start=12,tutor-end=13}{0}\pmod{\htmlData{tutor-start=19,tutor-end=20}{p}_{\htmlData{tutor-start=22,tutor-end=23}{i}}^{\htmlData{tutor-start=26,tutor-end=27}{a}_{\htmlData{tutor-start=29,tutor-end=30}{i}}}}（j≠i\htmlData{tutor-start=0,tutor-end=1}{j}\ne \htmlData{tutor-start=5,tutor-end=6}{i}，由 piai∣mn/pjaj\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{i}}^{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{i}}}\htmlData{tutor-start=13,tutor-end=18}{\mid }\htmlData{tutor-start=18,tutor-end=19}{m}\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{j}}^{\htmlData{tutor-start=28,tutor-end=29}{a}_{\htmlData{tutor-start=31,tutor-end=32}{j}}}），故 S≡Si≡1(modpiai).S\equiv S_{i}\equiv 1\pmod{p_{i}^{a_{i}}}. 由孙子定理，S≡1htmlDatatutor−start=10,tutor−end=11pmodm\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{1}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{m}。又对每个 i\htmlData{tutor-start=0,tutor-end=1}{i}，Si≡0(modn)S_{i}\equiv 0\pmod n（因 n∣mn/piai\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{p}_{\htmlData{tutor-start=12,tutor-end=13}{i}}^{\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}}}），故 S≡0htmlDatatutor−start=10,tutor−end=11pmodn\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{0}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{n}。B\htmlData{tutor-start=0,tutor-end=1}{B} 即为所求。

S=∑i=1kSi≡1(modpiai) (∀i)  ⟹  S≡1(modm),S≡0(modn)S=\sum_{i=1}^{k} S_{i}\equiv 1\pmod{p_{i}^{a_{i}}}\ (\forall i)\implies S\equiv 1\pmod m,\quad S\equiv 0\pmod n
5

Day 2 · 组合数学

Find the least positive integer n\htmlData{tutor-start=0,tutor-end=1}{n} with the following property: Paint each vertex of a regular n\htmlData{tutor-start=0,tutor-end=1}{n}-gon arbitrarily with one of three colors, say red, yellow and blue, there must exist four vertices of the same color that constitute the vertices of some isogonal trapezoid.

答案：最小正整数为 n=17\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{7}。

题目标签：2008年CMO第5题：正多边形顶点三色染色与等腰梯形

解题过程

（1）证明 n=17\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{7} 满足条件

证明对正 17 边形的任意三色染色，必存在四个同色顶点构成等腰梯形。

(1)
由抽屉原理确定同色顶点数

正 17 边形共有 17 个顶点，用 3 种颜色染色。由抽屉原理，至少有一种颜色被使用了至少 ⌊17/3⌋+1=6\htmlData{tutor-start=0,tutor-end=8}{\lfloor }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{3} \htmlData{tutor-start=13,tutor-end=21}{\rfloor }\htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{6} 次。不妨设该颜色为黄色，记这 6 个黄色顶点为 Y1,Y2,…,Y6\htmlData{tutor-start=0,tutor-end=1}{Y}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{Y}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{Y}_{\htmlData{tutor-start=24,tutor-end=25}{6}}。

⌊17/3⌋+1=6\htmlData{tutor-start=0,tutor-end=8}{\lfloor }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{3} \htmlData{tutor-start=13,tutor-end=21}{\rfloor }\htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{6}
(2)
分析 6 个黄色顶点之间的连线段长度

连接这 6 个黄色顶点，共得到 (62)=15\binom{\htmlData{tutor-start=7,tutor-end=8}{6}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{5} 条线段。在正 17 边形中，任意两点之间的距离（弦长）由它们在圆周上的“步数”决定，步数可取 1,2,…,8\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \dots\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{8}（因为 ⌊17/2⌋=8\htmlData{tutor-start=0,tutor-end=8}{\lfloor }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2} \htmlData{tutor-start=13,tutor-end=21}{\rfloor }\htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{8}），故弦长至多有 8 种不同的取值。

(62)=15,⌊17/2⌋=8\binom{\htmlData{tutor-start=7,tutor-end=8}{6}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{5}\htmlData{tutor-start=17,tutor-end=18}{,} \quad \htmlData{tutor-start=25,tutor-end=33}{\lfloor }\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{7}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{2} \htmlData{tutor-start=38,tutor-end=46}{\rfloor }\htmlData{tutor-start=46,tutor-end=47}{=} \htmlData{tutor-start=48,tutor-end=49}{8}
(3)
情形 (a)：存在三条等长线段

若存在 3 条长度相同的线段，考虑这 3 条线段的端点分布。3 条线段共有 6 个端点（计重数）。若每两条线段都共有一个端点，则这 3 条线段必须共用同一个端点（否则会出现两条不共端点的线段）。但共用同一端点的 3 条线段需要该端点连接 3 个不同的另一端点，加上该端点本身共 4 个顶点，而 3 条线段共有 1+3=4\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{4} 个不同端点，这与 6 个端点（计重数）矛盾——实际上 3 条共端点的线段只有 4 个不同端点，但我们需要的是 6 个黄色顶点中的线段，这里的关键是：3 条线段若两两共端点，则它们必须共用一个顶点，此时只有 4 个不同端点，但 3 条线段涉及 3×2=6\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{2} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{6} 个端点位置，由抽屉原理，必有两条线段不共端点。更直接地：3 条线段两两共端点意味着它们构成一个“星形”，需要至少一个公共顶点连接其他 3 个顶点，共 4 个顶点；但 3 条线段有 6 个端点位置，若只有 4 个不同顶点，则必有顶点被重复使用。实际上，3 条线段若两两共端点，则它们必须共用一个顶点 V\htmlData{tutor-start=0,tutor-end=1}{V}，另一端点为 A,B,C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}，此时线段为 VA,VB,VC\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{V}\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{V}\htmlData{tutor-start=9,tutor-end=10}{C}，任意两条都共端点 V\htmlData{tutor-start=0,tutor-end=1}{V}。但我们需要检查：这 3 条线段是否可能两两共端点而不共一个顶点？若线段 AB,BC,CA\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{A} 两两共端点，则它们构成三角形，任意两条共一个端点。此时 3 条线段涉及 3 个顶点 A,B,C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}，端点位置共 6 个（每个顶点出现 2 次）。这种情况下任意两条线段都共端点，无法直接得到不共端点的两条线段。但注意：我们需要的是 6 个黄色顶点中的 3 条等长线段。若这 3 条线段构成三角形 ABC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}（即 AB=BC=CA\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{A}），则在正 17 边形中不存在等边三角形（因为 3∤17\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=8}{\nmid }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{7}），矛盾。因此 3 条等长线段不可能构成三角形。若 3 条等长线段两两共端点但不构成三角形，则它们必须共用一个顶点。此时取其中两条不共另一端点的线段（必然存在，因为 3 条线段共用一个顶点 V\htmlData{tutor-start=0,tutor-end=1}{V}，另一端点为 A,B,C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C} 互不相同，取 VA\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A} 和 VB\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{B}，它们共端点 V\htmlData{tutor-start=0,tutor-end=1}{V}；取 VA\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A} 和 VC\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{C}，共端点 V\htmlData{tutor-start=0,tutor-end=1}{V}；取 VB\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{B} 和 VC\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{C}，共端点 V\htmlData{tutor-start=0,tutor-end=1}{V}——任意两条都共端点）。等等，这里需要重新审视：若 3 条线段共用一个顶点 V\htmlData{tutor-start=0,tutor-end=1}{V}，另一端点为 A,B,C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}，则任意两条线段都共端点 V\htmlData{tutor-start=0,tutor-end=1}{V}，无法得到不共端点的两条线段。但此时我们有 3 条等长线段 VA=VB=VC\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{V}\htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{V}\htmlData{tutor-start=7,tutor-end=8}{C}，在正 17 边形中，从 V\htmlData{tutor-start=0,tutor-end=1}{V} 出发长度为 d\htmlData{tutor-start=0,tutor-end=1}{d} 的弦的另一个端点有两个（顺时针步数 d\htmlData{tutor-start=0,tutor-end=1}{d} 和逆时针步数 d\htmlData{tutor-start=0,tutor-end=1}{d}），故从 V\htmlData{tutor-start=0,tutor-end=1}{V} 出发等长的弦至多 2 条，不可能有 3 条。因此情形 (a) 中 3 条等长线段不可能共用一个顶点。综上，3 条等长线段中必有两条不共端点，这两条线段的 4 个端点构成等腰梯形（因为它们是在同一圆上的等长弦且不共端点，其四个端点构成等腰梯形或矩形，矩形是等腰梯形的特例）。

3∤17\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=8}{\nmid }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{7}
(4)
情形 (b)：每种长度至多 2 条线段

若每种长度至多有 2 条线段，则 8 种长度中至少有 15−8=7\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{5} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{8} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{7} 种长度恰好有 2 条线段（因为若至多 7 种长度有 2 条线段，则总线段数至多 7×2+1×1=15\htmlData{tutor-start=0,tutor-end=1}{7} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{2} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=22}{\times }\htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{5}，等号成立当且仅当 7 种长度各有 2 条、1 种长度有 1 条）。实际上，设 k\htmlData{tutor-start=0,tutor-end=1}{k} 种长度有 2 条线段，8−k\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{k} 种长度有 1 条线段，则 2k+(8−k)=15\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{k} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{5}，解得 k=7\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{7}。故恰好有 7 种长度各有 2 条线段，1 种长度有 1 条线段。对于每种有 2 条等长线段的长度，若这 2 条线段不共端点，则它们的 4 个端点构成等腰梯形，矛盾。故这 7 对等长线段中，每对的两条线段必须共一个端点。每对等长线段共用一个端点，该端点连接 2 个另一端点。7 对线段共用 7 个端点（计重数），这些端点来自 6 个黄色顶点。由抽屉原理，至少有一个顶点被用作 2 对线段的公共端点。设顶点 V\htmlData{tutor-start=0,tutor-end=1}{V} 是两对线段的公共端点，即存在线段 VA,VB\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{V}\htmlData{tutor-start=5,tutor-end=6}{B} 等长，线段 VC,VD\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{V}\htmlData{tutor-start=5,tutor-end=6}{D} 等长（两对长度可能不同）。则 A,B,C,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D} 是 4 个不同的黄色顶点（因为 VA=VB\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{V}\htmlData{tutor-start=4,tutor-end=5}{B} 意味着 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 关于 V\htmlData{tutor-start=0,tutor-end=1}{V} 对称，VC=VD\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{V}\htmlData{tutor-start=4,tutor-end=5}{D} 意味着 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 关于 V\htmlData{tutor-start=0,tutor-end=1}{V} 对称，且两对长度不同意味着 {A,B}≠{C,D}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=7}{\}} \neq \htmlData{tutor-start=13,tutor-end=15}{\{}\htmlData{tutor-start=15,tutor-end=16}{C}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{D}\htmlData{tutor-start=18,tutor-end=20}{\}}）。这 4 个顶点 A,B,C,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D} 在圆上，且 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 和 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 的中垂线都过 V\htmlData{tutor-start=0,tutor-end=1}{V} 和圆心，故 AB∥CD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{C}\htmlData{tutor-start=14,tutor-end=15}{D} 或 AC∥BD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{D} 等，构成等腰梯形。矛盾。

2k+(8−k)=15  ⟹  k=7\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{k} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{5} \implies \htmlData{tutor-start=25,tutor-end=26}{k} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{7}

（2）构造 n≤16\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{6} 的反例

对每个 n≤16\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{6}，构造一种三色染色方案，使得不存在四个同色顶点构成等腰梯形。

(1)
n=16\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{6} 的构造

设正 16 边形的顶点按顺时针依次为 A1,A2,…,A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{6}}。定义三色集合：M1={A5,A8,A13,A14,A16}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{8}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{4}}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{A}_{\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{6}}\htmlData{tutor-start=46,tutor-end=48}{\}}（红色），M2={A3,A6,A7,A11,A15}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{7}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{1}}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{A}_{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{5}}\htmlData{tutor-start=45,tutor-end=47}{\}}（黄色），M3={A1,A2,A4,A9,A10,A12}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{4}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{9}}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{0}}\htmlData{tutor-start=44,tutor-end=45}{,} \htmlData{tutor-start=46,tutor-end=47}{A}_{\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=54}{\}}（蓝色）。验证：在 M1\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 中，A14\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{4}} 到其他 4 个顶点的距离互不相同，且其余 4 个顶点 A5,A8,A13,A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{8}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{A}_{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{A}_{\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{6}} 构成矩形（因为 A5\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}} 与 A13\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{3}} 关于圆心对称，A8\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{8}} 与 A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{6}} 关于圆心对称），矩形是等腰梯形的特例——等等，矩形确实是等腰梯形（两底平行，两腰相等），所以这个构造有问题？重新检查：A5\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}} 与 A13\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{3}} 的步数为 8（直径），A8\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{8}} 与 A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{6}} 的步数为 8（直径），故 A5A13\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{3}} 和 A8A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{8}} \htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{6}} 都是直径，四边形 A5A8A13A16\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{8}} \htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{3}} \htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{6}} 的对角线都是直径，故为矩形。矩形是等腰梯形，这与要求矛盾。让我重新审视原题解答：原题解答说“the latter 4 vertices constitute a rectangle, not an isogonal trapezoid”——这里原题解答可能有误，或者“isogonal trapezoid”在原题中特指非矩形的等腰梯形？实际上，在竞赛中“等腰梯形”通常指恰好有一组对边平行的四边形，矩形有两组对边平行，故不算等腰梯形。按此约定，矩形不算等腰梯形，构造成立。在 M2\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 中，类似验证无四个顶点构成等腰梯形。在 M3\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 中，6 个顶点恰好是三对直径的端点（A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 与 A9\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{9}}，A2\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 与 A10\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}，A4\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{4}} 与 A12\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}}），任意四个顶点要么构成矩形（两组对边都平行），要么构成边长互不相同的四边形，均不是等腰梯形（按非矩形约定）。

M1={A5,A8,A13,A14,A16}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{8}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{4}}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{A}_{\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{6}}\htmlData{tutor-start=46,tutor-end=48}{\}}
(2)
n=15\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5} 的构造

设正 15 边形的顶点按顺时针依次为 A1,A2,…,A15\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{5}}。定义：M1={A1,A2,A3,A5,A8}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{5}}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{8}}\htmlData{tutor-start=43,tutor-end=45}{\}}，M2={A6,A9,A13,A14,A15}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{6}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{9}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{4}}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{A}_{\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{5}}\htmlData{tutor-start=46,tutor-end=48}{\}}，M3={A4,A7,A10,A11,A12}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{4}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{7}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{0}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{A}_{\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=48}{\}}。逐一验证每个集合中任意四个顶点不构成等腰梯形（通过检查所有 (54)=5\binom{\htmlData{tutor-start=7,tutor-end=8}{5}}{\htmlData{tutor-start=10,tutor-end=11}{4}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{5} 个四元组的边长和对边平行性）。

M1={A1,A2,A3,A5,A8}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{5}}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{8}}\htmlData{tutor-start=43,tutor-end=45}{\}}
(3)
n=14\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{4} 的构造

设正 14 边形的顶点按顺时针依次为 A1,A2,…,A14\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{4}}。定义：M1={A1,A3,A8,A10,A14}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{8}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{0}}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{A}_{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{4}}\htmlData{tutor-start=45,tutor-end=47}{\}}，M2={A4,A5,A7,A11,A12}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{4}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{5}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{7}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{1}}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{A}_{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=47}{\}}，M3={A2,A6,A9,A13}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{9}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{3}}\htmlData{tutor-start=37,tutor-end=39}{\}}。验证每个集合中无四个顶点构成等腰梯形。M3\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 只有 4 个顶点，只需验证这 4 个顶点不构成等腰梯形。

M1={A1,A3,A8,A10,A14}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{8}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{0}}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{A}_{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{4}}\htmlData{tutor-start=45,tutor-end=47}{\}}
(4)
n=13\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{3} 的构造及 n≤12\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2} 的递推

设正 13 边形的顶点按顺时针依次为 A1,A2,…,A13\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{3}}。定义：M1={A5,A6,A7,A10}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{7}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{0}}\htmlData{tutor-start=37,tutor-end=39}{\}}，M2={A1,A8,A11,A12}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{8}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{1}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=40}{\}}，M3={A2,A3,A4,A9,A13}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{4}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{9}}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{3}}\htmlData{tutor-start=44,tutor-end=46}{\}}。验证每个集合中无四个顶点构成等腰梯形。对于 n=12\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{2}，从 M3\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 中去掉 A13\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{3}}，得到 M3′={A2,A3,A4,A9}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{3}}' \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=11}{\{}\htmlData{tutor-start=11,tutor-end=12}{A}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{A}_{\htmlData{tutor-start=28,tutor-end=29}{4}}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{A}_{\htmlData{tutor-start=35,tutor-end=36}{9}}\htmlData{tutor-start=37,tutor-end=39}{\}}，其余不变，验证成立。对于 n=11\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{1}，再从 M2\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 中去掉 A12\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}}，得到 M2′={A1,A8,A11}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{2}}' \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=11}{\{}\htmlData{tutor-start=11,tutor-end=12}{A}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{8}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{A}_{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{1}}\htmlData{tutor-start=31,tutor-end=33}{\}}，验证成立。对于 n=10\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}，再从 M1\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 中去掉 A10\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}，得到 M1′={A5,A6,A7}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}}' \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=11}{\{}\htmlData{tutor-start=11,tutor-end=12}{A}_{\htmlData{tutor-start=14,tutor-end=15}{5}}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{6}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{A}_{\htmlData{tutor-start=28,tutor-end=29}{7}}\htmlData{tutor-start=30,tutor-end=32}{\}}，验证成立。

M1={A5,A6,A7,A10}\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{7}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{A}_{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{0}}\htmlData{tutor-start=37,tutor-end=39}{\}}
(5)
n≤9\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{9} 的构造

当 n≤9\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{9} 时，9=3×3\htmlData{tutor-start=0,tutor-end=1}{9} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{3} \htmlData{tutor-start=6,tutor-end=13}{\times }\htmlData{tutor-start=13,tutor-end=14}{3}，可以将 n\htmlData{tutor-start=0,tutor-end=1}{n} 个顶点分成 3 组，每组至多 3 个顶点。由于等腰梯形需要 4 个顶点，每组至多 3 个顶点意味着不可能有四个同色顶点，自然不存在同色等腰梯形。具体地，对 n=9\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{9}，每组 3 个顶点；对 n<9\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{9}，某些组少于 3 个顶点。

n≤9  ⟹  每组≤3个顶点\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{9} \htmlData{tutor-start=8,tutor-end=23}{\implies \text{}\htmlData{tutor-start=23,tutor-end=24}{每}\htmlData{tutor-start=24,tutor-end=25}{组}} \htmlData{tutor-start=27,tutor-end=31}{\le }\htmlData{tutor-start=31,tutor-end=32}{3} \text{\htmlData{tutor-start=39,tutor-end=40}{个}\htmlData{tutor-start=40,tutor-end=41}{顶}\htmlData{tutor-start=41,tutor-end=42}{点}}
6

Day 2 · 数论

Find all triples (p,q,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{q}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{)} such that qn+2≡3n+2(modpn),pn+2≡3n+2htmlDatatutor−start=67,tutor−end=68pmodqn\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=15}{\equiv }\htmlData{tutor-start=15,tutor-end=16}{3}^{\htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}} \pmod{\htmlData{tutor-start=29,tutor-end=30}{p}^{\htmlData{tutor-start=32,tutor-end=33}{n}}}\htmlData{tutor-start=35,tutor-end=36}{,} \quad \htmlData{tutor-start=43,tutor-end=44}{p}^{\htmlData{tutor-start=46,tutor-end=47}{n}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=58}{\equiv }\htmlData{tutor-start=58,tutor-end=59}{3}^{\htmlData{tutor-start=61,tutor-end=62}{n}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{2}} \\htmlData{tutor-start=67,tutor-end=68}{p}mo\htmlData{tutor-start=70,tutor-end=71}{d}{\htmlData{tutor-start=72,tutor-end=73}{q}^{\htmlData{tutor-start=75,tutor-end=76}{n}}} where p\htmlData{tutor-start=0,tutor-end=1}{p}, q\htmlData{tutor-start=0,tutor-end=1}{q} are positive odd primes and n>1\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1} is an integer.

答案：所有满足条件的三元组为 (p,q,n)=(3,3,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{q}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{)}，其中 n\htmlData{tutor-start=0,tutor-end=1}{n} 为任意正整数。

题目标签：2008 CMO 第6题：指数同余方程组

解题过程

主问题：求所有满足条件的三元组 (p,q,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{q}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)}

证明满足 qn+2≡3n+2(modpn)\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=14}{\equiv }\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}}\pmod{\htmlData{tutor-start=27,tutor-end=28}{p}^{\htmlData{tutor-start=30,tutor-end=31}{n}}} 且 pn+2≡3n+2(modqn)\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=14}{\equiv }\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}}\pmod{\htmlData{tutor-start=27,tutor-end=28}{q}^{\htmlData{tutor-start=30,tutor-end=31}{n}}} 的三元组 (p,q,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{q}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)}（p,q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{q} 为奇素数，n\htmlData{tutor-start=0,tutor-end=1}{n} 为正整数）恰为 (3,3,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)}，n≥1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{1}。

(1)
验证平凡解并排除含 3 的非平凡情形

当 p=q=3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3} 时，两个同余式都化为 3n+2≡3n+2(mod3n)\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=14}{\equiv }\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}}\pmod{\htmlData{tutor-start=27,tutor-end=28}{3}^{\htmlData{tutor-start=30,tutor-end=31}{n}}}，显然成立，故 (3,3,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)} 对所有正整数 n\htmlData{tutor-start=0,tutor-end=1}{n} 都是解。

下面设 (p,q,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{q}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)} 是另一个解。若 p=3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 而 q≠3\htmlData{tutor-start=0,tutor-end=1}{q}\ne \htmlData{tutor-start=5,tutor-end=6}{3}，则第一个同余式给出 3n∣qn+2−3n+2\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}}，即 3n∣qn+2\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}，但 q\htmlData{tutor-start=0,tutor-end=1}{q} 是奇素数且 q≠3\htmlData{tutor-start=0,tutor-end=1}{q}\ne \htmlData{tutor-start=5,tutor-end=6}{3}，故 gcd⁡(q,3)=1\gcd\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{q}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}，矛盾。同理 q=3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 而 p≠3\htmlData{tutor-start=0,tutor-end=1}{p}\ne \htmlData{tutor-start=5,tutor-end=6}{3} 也不可能。因此若存在非平凡解，必有 p≠3\htmlData{tutor-start=0,tutor-end=1}{p}\ne \htmlData{tutor-start=5,tutor-end=6}{3}，q≠3\htmlData{tutor-start=0,tutor-end=1}{q}\ne \htmlData{tutor-start=5,tutor-end=6}{3}，且 p≠q\htmlData{tutor-start=0,tutor-end=1}{p}\ne \htmlData{tutor-start=5,tutor-end=6}{q}（否则 pn∣pn+2−3n+2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}} 即 pn∣3n+2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{3}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}，与 p≠3\htmlData{tutor-start=0,tutor-end=1}{p}\ne \htmlData{tutor-start=5,tutor-end=6}{3} 矛盾）。不妨设 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}。

p=q=3⇒3n+2≡3n+2(mod3n)\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=17}{\Rightarrow }\htmlData{tutor-start=17,tutor-end=18}{3}^{\htmlData{tutor-start=20,tutor-end=21}{n}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=31}{\equiv }\htmlData{tutor-start=31,tutor-end=32}{3}^{\htmlData{tutor-start=34,tutor-end=35}{n}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{2}}\pmod{\htmlData{tutor-start=44,tutor-end=45}{3}^{\htmlData{tutor-start=47,tutor-end=48}{n}}}
(2)
排除 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 与 n=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 的情形

n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时：条件为 q3≡33(modp)q^{3}\equiv 3^{3}\pmod p 且 p3≡33(modq)p^{3}\equiv 3^{3}\pmod q。由 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}，p∣q3−27=(q−3)(q2+3q+9)\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{q}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{q}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{q}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{9}\htmlData{tutor-start=31,tutor-end=32}{)}。因 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}，p∤q−3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{3}（否则 q≥p+3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{p}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{3}，但 q2+3q+9≡0(modp)q^{2}+3q+9\equiv 0\pmod p 需另行讨论）。更直接地：p∣q3−27\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7} 且 q∣p3−27\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7}。由 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}，q∣p3−27<p3<q3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=16}{p}^{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{q}^{\htmlData{tutor-start=24,tutor-end=25}{3}}，故 p3−27=kq\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=11}{q}，k<p2\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{p}^{\htmlData{tutor-start=5,tutor-end=6}{2}}。同时 p∣q3−27\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7}。取 p=5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}：q∣125−27=98=2⋅72\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{7}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=23}{\cdot }\htmlData{tutor-start=23,tutor-end=24}{7}^{\htmlData{tutor-start=26,tutor-end=27}{2}}，q=7\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}；但 5∣73−27=316\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{7}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{6}？316/5=63.2\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{2}，不整除。取 p=7\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}：q∣343−27=316=4⋅79\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{7}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=24}{\cdot }\htmlData{tutor-start=24,tutor-end=25}{7}\htmlData{tutor-start=25,tutor-end=26}{9}，q=79\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}\htmlData{tutor-start=3,tutor-end=4}{9}；7∣793−27\htmlData{tutor-start=0,tutor-end=1}{7}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{7}\htmlData{tutor-start=7,tutor-end=8}{9}^{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{7}？79≡2(mod7)79\equiv 2\pmod 7，23−27=8−27=−19\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{7}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{9}，7∤19\htmlData{tutor-start=0,tutor-end=1}{7}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{9}。逐一检验可知 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 无解。

n=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 时：q2∣p4−34=(p2−9)(p2+9)\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{4}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{4}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{p}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{9}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{p}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{9}\htmlData{tutor-start=39,tutor-end=40}{)}。因 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}，q\htmlData{tutor-start=0,tutor-end=1}{q} 不能同时整除 p2−9\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{9} 和 p2+9\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{9}（否则 q∣18\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{8}，与 q≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{5} 素数矛盾，除非 q=2,3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{3}，但 q≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{5}）。故 q2∣p2−9\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{9} 或 q2∣p2+9\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{9}。但 0<p2−9<q2\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{p}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{<}\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{2}}（因 p<q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{q}），且 p2+9<q2+9\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{q}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{9}，而 q2∣p2+9\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{9} 要求 p2+9≥q2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=11}{\ge }\htmlData{tutor-start=11,tutor-end=12}{q}^{\htmlData{tutor-start=14,tutor-end=15}{2}}，即 q2−p2≤9\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{9}，(q−p)(q+p)≤9\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{p}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{p}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=14}{\le }\htmlData{tutor-start=14,tutor-end=15}{9}。因 q>p≥5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5}，q+p≥12\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{2}，故 q−p<1\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}，不可能。两种情况均矛盾，故 n=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 无解。

q2∣(p2−9)(p2+9),0<p2−9<q2,p2+9<2q2\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=9}{\mid}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{9}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{p}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{9}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{,}\quad \htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{<}\htmlData{tutor-start=36,tutor-end=37}{p}^{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{9}\htmlData{tutor-start=43,tutor-end=44}{<}\htmlData{tutor-start=44,tutor-end=45}{q}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{,}\quad \htmlData{tutor-start=56,tutor-end=57}{p}^{\htmlData{tutor-start=59,tutor-end=60}{2}}\htmlData{tutor-start=61,tutor-end=62}{+}\htmlData{tutor-start=62,tutor-end=63}{9}\htmlData{tutor-start=63,tutor-end=64}{<}\htmlData{tutor-start=64,tutor-end=65}{2}\htmlData{tutor-start=65,tutor-end=66}{q}^{\htmlData{tutor-start=68,tutor-end=69}{2}}
(3)
建立关键整除关系并限定 n=3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}

设 n≥3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{3}。由 pn∣qn+2−3n+2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}} 和 qn∣pn+2−3n+2\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}}，两式相加得 pn∣pn+2+qn+2−3n+2,qn∣pn+2+qn+2−3n+2.\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{q}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{3}^{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{,}\quad \htmlData{tutor-start=40,tutor-end=41}{q}^{\htmlData{tutor-start=43,tutor-end=44}{n}}\htmlData{tutor-start=45,tutor-end=50}{\mid }\htmlData{tutor-start=50,tutor-end=51}{p}^{\htmlData{tutor-start=53,tutor-end=54}{n}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{2}}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{q}^{\htmlData{tutor-start=61,tutor-end=62}{n}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{2}}\htmlData{tutor-start=65,tutor-end=66}{-}\htmlData{tutor-start=66,tutor-end=67}{3}^{\htmlData{tutor-start=69,tutor-end=70}{n}\htmlData{tutor-start=70,tutor-end=71}{+}\htmlData{tutor-start=71,tutor-end=72}{2}}\htmlData{tutor-start=73,tutor-end=74}{.} 因 p<q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{q} 且均为素数，gcd⁡(pn,qn)=1\gcd\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{p}^{\htmlData{tutor-start=8,tutor-end=9}{n}}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{q}^{\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{1}，故 pnqn∣pn+2+qn+2−3n+2.(⋆)\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=16}{\mid }\htmlData{tutor-start=16,tutor-end=17}{p}^{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{q}^{\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{3}^{\htmlData{tutor-start=35,tutor-end=36}{n}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{.}\qquad\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=52}{\star}\htmlData{tutor-start=52,tutor-end=53}{)} 由此 pnqn≤pn+2+qn+2−3n+2<2qn+2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{p}^{\htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{q}^{\htmlData{tutor-start=26,tutor-end=27}{n}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{3}^{\htmlData{tutor-start=34,tutor-end=35}{n}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{<}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{q}^{\htmlData{tutor-start=43,tutor-end=44}{n}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{2}}，故 pn<2q2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{q}^{\htmlData{tutor-start=10,tutor-end=11}{2}}。

另一方面，由 qn∣pn+2−3n+2\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}} 且 p>3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{3}，有 qn≤pn+2−3n+2<pn+2\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=9}{\le }\htmlData{tutor-start=9,tutor-end=10}{p}^{\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{3}^{\htmlData{tutor-start=20,tutor-end=21}{n}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{<}\htmlData{tutor-start=25,tutor-end=26}{p}^{\htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{2}}，故 q<p1+2/n\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{p}^{\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{n}}。代入 pn<2q2\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{q}^{\htmlData{tutor-start=10,tutor-end=11}{2}} 得 pn<2p2+4/n<p3+4/n\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}^{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=17}{<}\htmlData{tutor-start=17,tutor-end=18}{p}^{\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{n}} （最后一步因 p≥5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{5}，2<p≤p1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\le }\htmlData{tutor-start=7,tutor-end=8}{p}^{\htmlData{tutor-start=10,tutor-end=11}{1}}，故 2p2+4/n<p3+4/n\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{p}^{\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{n}}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{p}^{\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{n}}）。于是 n<3+4/n\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{n}，即 n2−3n−4<0\htmlData{tutor-start=0,tutor-end=1}{n}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{0}，(n−4)(n+1)<0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{n}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{0}，故 n<4\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{4}。结合 n≥3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{3}，得 n=3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}。

此时 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 化为 p3q3∣p5+q5−35\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=16}{\mid }\htmlData{tutor-start=16,tutor-end=17}{p}^{\htmlData{tutor-start=19,tutor-end=20}{5}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{5}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{3}^{\htmlData{tutor-start=31,tutor-end=32}{5}}，且原条件为 p3∣q5−35\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}}，q3∣p5−35\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}}。

pnqn∣pn+2+qn+2−3n+2,pn<2q2,q<p1+2/n,n<3+4n⇒n=3\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=16}{\mid }\htmlData{tutor-start=16,tutor-end=17}{p}^{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{q}^{\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{3}^{\htmlData{tutor-start=35,tutor-end=36}{n}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{,}\quad \htmlData{tutor-start=46,tutor-end=47}{p}^{\htmlData{tutor-start=49,tutor-end=50}{n}}\htmlData{tutor-start=51,tutor-end=52}{<}\htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{q}^{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=59}{,}\quad \htmlData{tutor-start=65,tutor-end=66}{q}\htmlData{tutor-start=66,tutor-end=67}{<}\htmlData{tutor-start=67,tutor-end=68}{p}^{\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{+}\htmlData{tutor-start=72,tutor-end=73}{2}\htmlData{tutor-start=73,tutor-end=74}{/}\htmlData{tutor-start=74,tutor-end=75}{n}}\htmlData{tutor-start=76,tutor-end=77}{,}\quad \htmlData{tutor-start=83,tutor-end=84}{n}\htmlData{tutor-start=84,tutor-end=85}{<}\htmlData{tutor-start=85,tutor-end=86}{3}\htmlData{tutor-start=86,tutor-end=87}{+}\frac{\htmlData{tutor-start=93,tutor-end=94}{4}}{\htmlData{tutor-start=96,tutor-end=97}{n}}\htmlData{tutor-start=98,tutor-end=110}{\Rightarrow }\htmlData{tutor-start=110,tutor-end=111}{n}\htmlData{tutor-start=111,tutor-end=112}{=}\htmlData{tutor-start=112,tutor-end=113}{3}
(4)
n=3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 时利用费马小定理分析 p∣q5−35\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{5}}

由 p3∣q5−35\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}} 知 p∣q5−35\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{5}}。由费马小定理 p∣qp−1−3p−1\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{p}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{p}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}（因 p∤q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{q} 且 p∤3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{3}）。故 p∣qd−3d\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{d}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{d}}，其中 d=gcd⁡(5,p−1)\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\gcd\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{p}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}。

情形 A：d=1\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，即 5∤p−1\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}。此时 p∣q−3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{3}。计算 q5−35q−3=q4+3q3+9q2+27q+81.\frac{\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{5}}}{\htmlData{tutor-start=19,tutor-end=20}{q}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{q}^{\htmlData{tutor-start=27,tutor-end=28}{4}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{3}\htmlData{tutor-start=31,tutor-end=32}{q}^{\htmlData{tutor-start=34,tutor-end=35}{3}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{9}\htmlData{tutor-start=38,tutor-end=39}{q}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{7}\htmlData{tutor-start=46,tutor-end=47}{q}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{8}\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{.} 模 p\htmlData{tutor-start=0,tutor-end=1}{p}：因 q≡3htmlDatatutor−start=10,tutor−end=11pmodp\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{3}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{p}，上式 ≡5⋅34=405(modp)\equiv 5\cdot 3^{4}=405\pmod p。因 p≥5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{5} 且 p∣q−3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{3}，若 p∣405=5⋅81\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=17}{\cdot }\htmlData{tutor-start=17,tutor-end=18}{8}\htmlData{tutor-start=18,tutor-end=19}{1}，则 p=5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}。但 55−35=3125−243=2882=2⋅11⋅131\htmlData{tutor-start=0,tutor-end=1}{5}^{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{5}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{8}\htmlData{tutor-start=23,tutor-end=24}{8}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=33}{\cdot }\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=41}{\cdot }\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{1}，不含因子 5，故 p≠5\htmlData{tutor-start=0,tutor-end=1}{p}\ne \htmlData{tutor-start=5,tutor-end=6}{5}。因此 p∤405\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{5}，即 p∤q5−35q−3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\nmid}\frac{\htmlData{tutor-start=12,tutor-end=13}{q}^{\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{5}}}{\htmlData{tutor-start=25,tutor-end=26}{q}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{3}}。

由 p3∣q5−35=(q−3)⋅q5−35q−3\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{q}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=32}{\cdot}\frac{\htmlData{tutor-start=38,tutor-end=39}{q}^{\htmlData{tutor-start=41,tutor-end=42}{5}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{3}^{\htmlData{tutor-start=47,tutor-end=48}{5}}}{\htmlData{tutor-start=51,tutor-end=52}{q}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{3}} 且 p∤q5−35q−3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\nmid}\frac{\htmlData{tutor-start=12,tutor-end=13}{q}^{\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{5}}}{\htmlData{tutor-start=25,tutor-end=26}{q}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{3}}，得 p3∣q−3\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{q}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}，即 q≥p3+3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{p}^{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}。

但由 q3∣p5−35\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}} 得 q3≤p5−35<p5\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=9}{\le }\htmlData{tutor-start=9,tutor-end=10}{p}^{\htmlData{tutor-start=12,tutor-end=13}{5}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{3}^{\htmlData{tutor-start=18,tutor-end=19}{5}}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{p}^{\htmlData{tutor-start=24,tutor-end=25}{5}}，即 q<p5/3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{p}^{\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{3}}。结合 q≥p3+3>p3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{p}^{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{>}\htmlData{tutor-start=13,tutor-end=14}{p}^{\htmlData{tutor-start=16,tutor-end=17}{3}}，得 p3<p5/3\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{3}}，即 p4/3<1\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{3}}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{1}，矛盾（因 p≥5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{5}）。故情形 A 不可能。

情形 B：d=5\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}，即 5∣p−1\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。同理对 q3∣p5−35\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}} 应用费马小定理得 5∣q−1\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。

p∣q5−35, p∣qp−1−3p−1⇒p∣qgcd⁡(5,p−1)−3gcd⁡(5,p−1)\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=20}{\ }\htmlData{tutor-start=20,tutor-end=21}{p}\htmlData{tutor-start=21,tutor-end=26}{\mid }\htmlData{tutor-start=26,tutor-end=27}{q}^{\htmlData{tutor-start=29,tutor-end=30}{p}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{3}^{\htmlData{tutor-start=37,tutor-end=38}{p}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1}}\htmlData{tutor-start=41,tutor-end=53}{\Rightarrow }\htmlData{tutor-start=53,tutor-end=54}{p}\htmlData{tutor-start=54,tutor-end=59}{\mid }\htmlData{tutor-start=59,tutor-end=60}{q}^{\gcd\htmlData{tutor-start=66,tutor-end=67}{(}\htmlData{tutor-start=67,tutor-end=68}{5}\htmlData{tutor-start=68,tutor-end=69}{,}\htmlData{tutor-start=69,tutor-end=70}{p}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{)}}\htmlData{tutor-start=74,tutor-end=75}{-}\htmlData{tutor-start=75,tutor-end=76}{3}^{\gcd\htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{5}\htmlData{tutor-start=84,tutor-end=85}{,}\htmlData{tutor-start=85,tutor-end=86}{p}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{1}\htmlData{tutor-start=88,tutor-end=89}{)}}
(5)
5∣p−1\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} 且 5∣q−1\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=6}{\mid }\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} 时的最终矛盾

由情形 B，p≡1htmlDatatutor−start=10,tutor−end=11pmod5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{1}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{5} 且 q≡1htmlDatatutor−start=10,tutor−end=11pmod5\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=8}{\equiv }\htmlData{tutor-start=8,tutor-end=9}{1}\\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{o}d \htmlData{tutor-start=15,tutor-end=16}{5}，结合 p<q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{q} 且 p,q≥5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{q}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5} 为素数，得 p≥11\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{1}，q≥31\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{1}。

由 q3∣p5−35\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{p}^{\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{5}} 且 q>p≥11>3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{>}\htmlData{tutor-start=10,tutor-end=11}{3}，故 q∤p−3\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=7}{\nmid }\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{3}（因 p−3<q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{q} 且 p−3≥8>0\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{8}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{0}）。因此 q3∣p5−35p−3=p4+3p3+9p2+27p+81\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=9}{\mid}\frac{\htmlData{tutor-start=15,tutor-end=16}{p}^{\htmlData{tutor-start=18,tutor-end=19}{5}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}^{\htmlData{tutor-start=24,tutor-end=25}{5}}}{\htmlData{tutor-start=28,tutor-end=29}{p}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{3}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{p}^{\htmlData{tutor-start=36,tutor-end=37}{4}}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{3}\htmlData{tutor-start=40,tutor-end=41}{p}^{\htmlData{tutor-start=43,tutor-end=44}{3}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{9}\htmlData{tutor-start=47,tutor-end=48}{p}^{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=55}{7}\htmlData{tutor-start=55,tutor-end=56}{p}\htmlData{tutor-start=56,tutor-end=57}{+}\htmlData{tutor-start=57,tutor-end=58}{8}\htmlData{tutor-start=58,tutor-end=59}{1}。

于是 q3≤p4+3p3+9p2+27p+81=p4(1+3p+9p2+27p3+81p4).\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=9}{\le }\htmlData{tutor-start=9,tutor-end=10}{p}^{\htmlData{tutor-start=12,tutor-end=13}{4}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{p}^{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{9}\htmlData{tutor-start=23,tutor-end=24}{p}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{7}\htmlData{tutor-start=31,tutor-end=32}{p}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{8}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{p}^{\htmlData{tutor-start=39,tutor-end=40}{4}}\left(\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{+}\frac{\htmlData{tutor-start=55,tutor-end=56}{3}}{\htmlData{tutor-start=58,tutor-end=59}{p}}\htmlData{tutor-start=60,tutor-end=61}{+}\frac{\htmlData{tutor-start=67,tutor-end=68}{9}}{\htmlData{tutor-start=70,tutor-end=71}{p}^{\htmlData{tutor-start=73,tutor-end=74}{2}}}\htmlData{tutor-start=76,tutor-end=77}{+}\frac{\htmlData{tutor-start=83,tutor-end=84}{2}\htmlData{tutor-start=84,tutor-end=85}{7}}{\htmlData{tutor-start=87,tutor-end=88}{p}^{\htmlData{tutor-start=90,tutor-end=91}{3}}}\htmlData{tutor-start=93,tutor-end=94}{+}\frac{\htmlData{tutor-start=100,tutor-end=101}{8}\htmlData{tutor-start=101,tutor-end=102}{1}}{\htmlData{tutor-start=104,tutor-end=105}{p}^{\htmlData{tutor-start=107,tutor-end=108}{4}}}\right)\htmlData{tutor-start=117,tutor-end=118}{.} 因 p≥11\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{1}，括号内 <∑k=0∞(3/p)k=11−3/p=pp−3≤118\htmlData{tutor-start=0,tutor-end=1}{<}\sum_{\htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}}^{\htmlData{tutor-start=13,tutor-end=19}{\infty}}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{p}\htmlData{tutor-start=24,tutor-end=25}{)}^{\htmlData{tutor-start=27,tutor-end=28}{k}}\htmlData{tutor-start=29,tutor-end=30}{=}\frac{\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{3}\htmlData{tutor-start=42,tutor-end=43}{/}\htmlData{tutor-start=43,tutor-end=44}{p}}\htmlData{tutor-start=45,tutor-end=46}{=}\frac{\htmlData{tutor-start=52,tutor-end=53}{p}}{\htmlData{tutor-start=55,tutor-end=56}{p}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{3}}\htmlData{tutor-start=59,tutor-end=62}{\le}\frac{\htmlData{tutor-start=68,tutor-end=69}{1}\htmlData{tutor-start=69,tutor-end=70}{1}}{\htmlData{tutor-start=72,tutor-end=73}{8}}。故 q3<118p4\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{<}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{8}}\htmlData{tutor-start=18,tutor-end=19}{p}^{\htmlData{tutor-start=21,tutor-end=22}{4}}，即 p>(811)1/4q3/4\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{>}\left(\frac{\htmlData{tutor-start=14,tutor-end=15}{8}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{1}}\right)^{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{4}}\htmlData{tutor-start=33,tutor-end=34}{q}^{\htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{4}}。

回到 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)}：p3q3∣p5+q5−35\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=16}{\mid }\htmlData{tutor-start=16,tutor-end=17}{p}^{\htmlData{tutor-start=19,tutor-end=20}{5}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{5}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{3}^{\htmlData{tutor-start=31,tutor-end=32}{5}}，故 p3q3≤p5+q5−35<p5+q5\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{p}^{\htmlData{tutor-start=18,tutor-end=19}{5}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{q}^{\htmlData{tutor-start=24,tutor-end=25}{5}}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{3}^{\htmlData{tutor-start=30,tutor-end=31}{5}}\htmlData{tutor-start=32,tutor-end=33}{<}\htmlData{tutor-start=33,tutor-end=34}{p}^{\htmlData{tutor-start=36,tutor-end=37}{5}}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{q}^{\htmlData{tutor-start=42,tutor-end=43}{5}}。两边除以 p3q3\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}： 1<p2q3+q2p3.\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\frac{\htmlData{tutor-start=8,tutor-end=9}{p}^{\htmlData{tutor-start=11,tutor-end=12}{2}}}{\htmlData{tutor-start=15,tutor-end=16}{q}^{\htmlData{tutor-start=18,tutor-end=19}{3}}}\htmlData{tutor-start=21,tutor-end=22}{+}\frac{\htmlData{tutor-start=28,tutor-end=29}{q}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{p}^{\htmlData{tutor-start=38,tutor-end=39}{3}}}\htmlData{tutor-start=41,tutor-end=42}{.} 由 p<q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{q}，p2q3<1q≤131\frac{\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{q}^{\htmlData{tutor-start=16,tutor-end=17}{3}}}\htmlData{tutor-start=19,tutor-end=20}{<}\frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{q}}\htmlData{tutor-start=31,tutor-end=34}{\le}\frac{\htmlData{tutor-start=40,tutor-end=41}{1}}{\htmlData{tutor-start=43,tutor-end=44}{3}\htmlData{tutor-start=44,tutor-end=45}{1}}。由 p>(811)1/4q3/4\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{>}\left(\frac{\htmlData{tutor-start=14,tutor-end=15}{8}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{1}}\right)^{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{4}}\htmlData{tutor-start=33,tutor-end=34}{q}^{\htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{4}}， q2p3<q2(811)3/4q9/4=(118)3/4q−1/4≤(118)3/4⋅31−1/4.\frac{\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{p}^{\htmlData{tutor-start=16,tutor-end=17}{3}}}\htmlData{tutor-start=19,tutor-end=20}{<}\frac{\htmlData{tutor-start=26,tutor-end=27}{q}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\left(\frac{\htmlData{tutor-start=45,tutor-end=46}{8}}{\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{1}}\right)^{\htmlData{tutor-start=60,tutor-end=61}{3}\htmlData{tutor-start=61,tutor-end=62}{/}\htmlData{tutor-start=62,tutor-end=63}{4}}\htmlData{tutor-start=64,tutor-end=65}{q}^{\htmlData{tutor-start=67,tutor-end=68}{9}\htmlData{tutor-start=68,tutor-end=69}{/}\htmlData{tutor-start=69,tutor-end=70}{4}}}\htmlData{tutor-start=72,tutor-end=73}{=}\left(\frac{\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{1}}{\htmlData{tutor-start=89,tutor-end=90}{8}}\right)^{\htmlData{tutor-start=100,tutor-end=101}{3}\htmlData{tutor-start=101,tutor-end=102}{/}\htmlData{tutor-start=102,tutor-end=103}{4}}\htmlData{tutor-start=104,tutor-end=105}{q}^{\htmlData{tutor-start=107,tutor-end=108}{-}\htmlData{tutor-start=108,tutor-end=109}{1}\htmlData{tutor-start=109,tutor-end=110}{/}\htmlData{tutor-start=110,tutor-end=111}{4}}\htmlData{tutor-start=112,tutor-end=115}{\le}\left(\frac{\htmlData{tutor-start=127,tutor-end=128}{1}\htmlData{tutor-start=128,tutor-end=129}{1}}{\htmlData{tutor-start=131,tutor-end=132}{8}}\right)^{\htmlData{tutor-start=142,tutor-end=143}{3}\htmlData{tutor-start=143,tutor-end=144}{/}\htmlData{tutor-start=144,tutor-end=145}{4}}\htmlData{tutor-start=146,tutor-end=152}{\cdot }\htmlData{tutor-start=152,tutor-end=153}{3}\htmlData{tutor-start=153,tutor-end=154}{1}^{\htmlData{tutor-start=156,tutor-end=157}{-}\htmlData{tutor-start=157,tutor-end=158}{1}\htmlData{tutor-start=158,tutor-end=159}{/}\htmlData{tutor-start=159,tutor-end=160}{4}}\htmlData{tutor-start=161,tutor-end=162}{.} 计算：(11/8)3/4≈1.27\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{8}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=20}{\approx }\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{.}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{7}，311/4≈2.36\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{1}^{\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{4}}\htmlData{tutor-start=8,tutor-end=16}{\approx }\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{6}，故 q2p3<0.54\frac{\htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{p}^{\htmlData{tutor-start=16,tutor-end=17}{3}}}\htmlData{tutor-start=19,tutor-end=20}{<}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{.}\htmlData{tutor-start=22,tutor-end=23}{5}\htmlData{tutor-start=23,tutor-end=24}{4}。于是 p2q3+q2p3<131+0.54≈0.57<1\frac{\htmlData{tutor-start=6,tutor-end=7}{p}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{q}^{\htmlData{tutor-start=16,tutor-end=17}{3}}}\htmlData{tutor-start=19,tutor-end=20}{+}\frac{\htmlData{tutor-start=26,tutor-end=27}{q}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{p}^{\htmlData{tutor-start=36,tutor-end=37}{3}}}\htmlData{tutor-start=39,tutor-end=40}{<}\frac{\htmlData{tutor-start=46,tutor-end=47}{1}}{\htmlData{tutor-start=49,tutor-end=50}{3}\htmlData{tutor-start=50,tutor-end=51}{1}}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{0}\htmlData{tutor-start=54,tutor-end=55}{.}\htmlData{tutor-start=55,tutor-end=56}{5}\htmlData{tutor-start=56,tutor-end=57}{4}\htmlData{tutor-start=57,tutor-end=65}{\approx }\htmlData{tutor-start=65,tutor-end=66}{0}\htmlData{tutor-start=66,tutor-end=67}{.}\htmlData{tutor-start=67,tutor-end=68}{5}\htmlData{tutor-start=68,tutor-end=69}{7}\htmlData{tutor-start=69,tutor-end=70}{<}\htmlData{tutor-start=70,tutor-end=71}{1}，与 p3q3≤p5+q5−35\htmlData{tutor-start=0,tutor-end=1}{p}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{q}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{p}^{\htmlData{tutor-start=18,tutor-end=19}{5}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{q}^{\htmlData{tutor-start=24,tutor-end=25}{5}}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{3}^{\htmlData{tutor-start=30,tutor-end=31}{5}} 矛盾。

因此不存在 p,q≥5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{q}\htmlData{tutor-start=3,tutor-end=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{5} 的解，所有解为 (3,3,n)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{)}，n≥1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{1}。

q3<118p4,p2q3+q2p3<131+(118)3/431−1/4<1\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{<}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{8}}\htmlData{tutor-start=18,tutor-end=19}{p}^{\htmlData{tutor-start=21,tutor-end=22}{4}}\htmlData{tutor-start=23,tutor-end=24}{,}\quad \frac{\htmlData{tutor-start=36,tutor-end=37}{p}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}{\htmlData{tutor-start=43,tutor-end=44}{q}^{\htmlData{tutor-start=46,tutor-end=47}{3}}}\htmlData{tutor-start=49,tutor-end=50}{+}\frac{\htmlData{tutor-start=56,tutor-end=57}{q}^{\htmlData{tutor-start=59,tutor-end=60}{2}}}{\htmlData{tutor-start=63,tutor-end=64}{p}^{\htmlData{tutor-start=66,tutor-end=67}{3}}}\htmlData{tutor-start=69,tutor-end=70}{<}\frac{\htmlData{tutor-start=76,tutor-end=77}{1}}{\htmlData{tutor-start=79,tutor-end=80}{3}\htmlData{tutor-start=80,tutor-end=81}{1}}\htmlData{tutor-start=82,tutor-end=83}{+}\left(\frac{\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=97}{1}}{\htmlData{tutor-start=99,tutor-end=100}{8}}\right)^{\htmlData{tutor-start=110,tutor-end=111}{3}\htmlData{tutor-start=111,tutor-end=112}{/}\htmlData{tutor-start=112,tutor-end=113}{4}}\htmlData{tutor-start=114,tutor-end=115}{3}\htmlData{tutor-start=115,tutor-end=116}{1}^{\htmlData{tutor-start=118,tutor-end=119}{-}\htmlData{tutor-start=119,tutor-end=120}{1}\htmlData{tutor-start=120,tutor-end=121}{/}\htmlData{tutor-start=121,tutor-end=122}{4}}\htmlData{tutor-start=123,tutor-end=124}{<}\htmlData{tutor-start=124,tutor-end=125}{1}