2022 中国数学奥林匹克（CMO）

首先分析“好三角形对”的几何结构。由条件 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{X}\htmlData{tutor-start=10,tutor-end=11}{Y}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{E}\htmlData{tutor-start=14,tutor-end=15}{F} \htmlData{tutor-start=16,tutor-end=22}{\perp }\htmlData{tutor-start=22,tutor-end=23}{Y}\htmlData{tutor-start=23,tutor-end=24}{Z}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{F}\htmlData{tutor-start=27,tutor-end=28}{D} \htmlData{tutor-start=29,tutor-end=35}{\perp }\htmlData{tutor-start=35,tutor-end=36}{Z}\htmlData{tutor-start=36,tutor-end=37}{X}$，我们可以将 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$ 绕平面上任意一点旋转 $90^\circ$（顺时针或逆时针均可，这里取逆时针）。设旋转后的像为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}'\htmlData{tutor-start=12,tutor-end=13}{Y}'\htmlData{tutor-start=14,tutor-end=15}{Z}'$。由于旋转保持角度不变且改变方向 $90^\circ$，原垂直关系转化为平行关系：$\htmlData{tutor-start=0,tutor-end=1}{X}'\htmlData{tutor-start=2,tutor-end=3}{Y}' \htmlData{tutor-start=5,tutor-end=15}{\parallel }\htmlData{tutor-start=15,tutor-end=16}{D}\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{Y}'\htmlData{tutor-start=21,tutor-end=22}{Z}' \htmlData{tutor-start=24,tutor-end=34}{\parallel }\htmlData{tutor-start=34,tutor-end=35}{E}\htmlData{tutor-start=35,tutor-end=36}{F}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{Z}'\htmlData{tutor-start=40,tutor-end=41}{X}' \htmlData{tutor-start=43,tutor-end=53}{\parallel }\htmlData{tutor-start=53,tutor-end=54}{F}\htmlData{tutor-start=54,tutor-end=55}{D}$。这意味着 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}'\htmlData{tutor-start=12,tutor-end=13}{Y}'\htmlData{tutor-start=14,tutor-end=15}{Z}'$ 与 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 对应边平行，故两者直接相似（directly similar）。因此，对于任意好三角形对，$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$ 与 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 的形状是固定的（仅大小和位置不同），且面积比 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{S}_{\htmlData{tutor-start=13,tutor-end=23}{\triangle }\htmlData{tutor-start=23,tutor-end=24}{X}\htmlData{tutor-start=24,tutor-end=25}{Y}\htmlData{tutor-start=25,tutor-end=26}{Z}}}{\htmlData{tutor-start=29,tutor-end=30}{S}_{\htmlData{tutor-start=32,tutor-end=42}{\triangle }\htmlData{tutor-start=42,tutor-end=43}{D}\htmlData{tutor-start=43,tutor-end=44}{E}\htmlData{tutor-start=44,tutor-end=45}{F}}}$ 是一个常数。我们的目标转化为求定值 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=22}{\triangle }\htmlData{tutor-start=22,tutor-end=23}{D}\htmlData{tutor-start=23,tutor-end=24}{E}\htmlData{tutor-start=24,tutor-end=25}{F}}}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{1} \htmlData{tutor-start=30,tutor-end=31}{+} \frac{\htmlData{tutor-start=38,tutor-end=39}{1}}{\htmlData{tutor-start=41,tutor-end=42}{k}}\htmlData{tutor-start=43,tutor-end=44}{)}$。

为了计算面积比，我们将整个图形（包括 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 和 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$）进行上述 $90^\circ$ 旋转、缩放和平移，使得 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$ 的像恰好重合于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$。在此变换下，原正三角形 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 变为一个新的正三角形 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$。根据变换性质： 1. $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 是正三角形； 2. 点 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{F}$ 分别位于直线 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{B}_{\htmlData{tutor-start=32,tutor-end=33}{1}}$ 上（因为原 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{Y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{Z}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B}$ 上，且变换把 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{Y}\htmlData{tutor-start=2,tutor-end=3}{Z}$ 映射为 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F}$）； 3. 对应边互相垂直：$\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{B}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=17}{\perp }\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B}$ 等。

考察四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{A}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{F}$。由于 $\angle EAF = 60^\circ$（原三角形角）且 $\angle EA_{1}F = 60^\circ$（新三角形角，注意方向），且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{E}$（因为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$ 即 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{F}$ 所在直线？不，是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{A}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$，即 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{E}$ 不一定成立，需仔细推导）。 更准确的论证：由 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{A}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$ 知 $\angle (AC, C_{1}A_{1}) = 90^\circ$。又 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{A}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{A}_{\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{B}_{\htmlData{tutor-start=32,tutor-end=33}{1}}$ 上。考虑点 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 对 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 的位置。事实上，可以证明 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 外接圆上 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的对径点。这是因为 $\angle A_{1} E A = 90^\circ$（由 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=17}{\perp }\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{A}$ 及 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{C}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{A}_{\htmlData{tutor-start=12,tutor-end=13}{1}}$ 交点附近推导，严格来说是利用直角投影性质）。同理 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{1}}$ 分别是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{F}\htmlData{tutor-start=12,tutor-end=13}{D}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=25}{\triangle }\htmlData{tutor-start=25,tutor-end=26}{C}\htmlData{tutor-start=26,tutor-end=27}{D}\htmlData{tutor-start=27,tutor-end=28}{E}$ 外接圆上 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{C}$ 的对径点。

利用有向面积公式，我们有恒等式： $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=33}{\triangle }\htmlData{tutor-start=33,tutor-end=34}{A}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{B}_{\htmlData{tutor-start=41,tutor-end=42}{1}}\htmlData{tutor-start=43,tutor-end=44}{C}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{(}\htmlData{tutor-start=54,tutor-end=55}{S}_{\htmlData{tutor-start=57,tutor-end=67}{\triangle }\htmlData{tutor-start=67,tutor-end=68}{D}\htmlData{tutor-start=68,tutor-end=69}{E}\htmlData{tutor-start=69,tutor-end=70}{F}} \htmlData{tutor-start=72,tutor-end=73}{+} \htmlData{tutor-start=74,tutor-end=75}{S}_{\htmlData{tutor-start=77,tutor-end=87}{\triangle }\htmlData{tutor-start=87,tutor-end=88}{E}\htmlData{tutor-start=88,tutor-end=89}{O}_{\htmlData{tutor-start=91,tutor-end=92}{1}}\htmlData{tutor-start=93,tutor-end=94}{F}} \htmlData{tutor-start=96,tutor-end=97}{+} \htmlData{tutor-start=98,tutor-end=99}{S}_{\htmlData{tutor-start=101,tutor-end=111}{\triangle }\htmlData{tutor-start=111,tutor-end=112}{F}\htmlData{tutor-start=112,tutor-end=113}{O}_{\htmlData{tutor-start=115,tutor-end=116}{2}}\htmlData{tutor-start=117,tutor-end=118}{D}} \htmlData{tutor-start=120,tutor-end=121}{+} \htmlData{tutor-start=122,tutor-end=123}{S}_{\htmlData{tutor-start=125,tutor-end=135}{\triangle }\htmlData{tutor-start=135,tutor-end=136}{D}\htmlData{tutor-start=136,tutor-end=137}{O}_{\htmlData{tutor-start=139,tutor-end=140}{3}}\htmlData{tutor-start=141,tutor-end=142}{E}}\htmlData{tutor-start=143,tutor-end=144}{)}$ 其中 $\htmlData{tutor-start=0,tutor-end=1}{O}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{O}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{O}_{\htmlData{tutor-start=17,tutor-end=18}{3}}$ 分别是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=25}{\triangle }\htmlData{tutor-start=25,tutor-end=26}{B}\htmlData{tutor-start=26,tutor-end=27}{F}\htmlData{tutor-start=27,tutor-end=28}{D}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=40}{\triangle }\htmlData{tutor-start=40,tutor-end=41}{C}\htmlData{tutor-start=41,tutor-end=42}{D}\htmlData{tutor-start=42,tutor-end=43}{E}$ 的外心。这个公式的本质是将两个大正三角形的面积分解为中心小三角形 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F}$ 加上周围三个“外心三角形”的面积之和的两倍。

现在问题归结为计算 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 及其外心三角形的面积。已知 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 三边比例为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{:}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{8}$。设比例系数为 $\htmlData{tutor-start=0,tutor-end=1}{t}$，则边长为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{t}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{8}\htmlData{tutor-start=12,tutor-end=13}{t}$。 半周长 $\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{8}}{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{t} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{4}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{t}$。 由海伦公式，$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{E}\htmlData{tutor-start=15,tutor-end=16}{F}} \htmlData{tutor-start=18,tutor-end=19}{=} \sqrt{\htmlData{tutor-start=26,tutor-end=27}{4}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{t}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{t}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{8}\htmlData{tutor-start=37,tutor-end=38}{t}\htmlData{tutor-start=38,tutor-end=39}{)}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{t}\htmlData{tutor-start=42,tutor-end=43}{)}} \htmlData{tutor-start=45,tutor-end=46}{=} \sqrt{\htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{4}\htmlData{tutor-start=55,tutor-end=56}{4}\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{0} \htmlData{tutor-start=59,tutor-end=60}{t}^{\htmlData{tutor-start=62,tutor-end=63}{4}}} \htmlData{tutor-start=66,tutor-end=67}{=} \htmlData{tutor-start=68,tutor-end=69}{1}\htmlData{tutor-start=69,tutor-end=70}{2}\htmlData{tutor-start=70,tutor-end=71}{0}\sqrt{\htmlData{tutor-start=77,tutor-end=78}{2}} \htmlData{tutor-start=80,tutor-end=81}{t}^{\htmlData{tutor-start=83,tutor-end=84}{2}}$。

接下来计算 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{E}\htmlData{tutor-start=14,tutor-end=15}{O}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{F}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{S}_{\htmlData{tutor-start=27,tutor-end=37}{\triangle }\htmlData{tutor-start=37,tutor-end=38}{F}\htmlData{tutor-start=38,tutor-end=39}{O}_{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{D}} \htmlData{tutor-start=46,tutor-end=47}{+} \htmlData{tutor-start=48,tutor-end=49}{S}_{\htmlData{tutor-start=51,tutor-end=61}{\triangle }\htmlData{tutor-start=61,tutor-end=62}{D}\htmlData{tutor-start=62,tutor-end=63}{O}_{\htmlData{tutor-start=65,tutor-end=66}{3}}\htmlData{tutor-start=67,tutor-end=68}{E}}$。注意到 $\htmlData{tutor-start=0,tutor-end=1}{O}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 外心，且 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 是对径点，故 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{E}\htmlData{tutor-start=11,tutor-end=12}{O}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{F}$ 是以 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 为底，高为 $\htmlData{tutor-start=0,tutor-end=1}{R}_{\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{E}\htmlData{tutor-start=5,tutor-end=6}{F}}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}$ 的三角形？不完全是。实际上，有一个关于外心三角形面积的结论：若 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 边长为 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{e}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{f}$，则这三个外心三角形面积之和为 $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{d}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{e}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{f}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{)}$？让我们重新核对参考解答中的公式。 参考解答给出：$\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{m} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}{\htmlData{tutor-start=22,tutor-end=23}{3}} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{1}^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{+} \htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{0}^{\htmlData{tutor-start=39,tutor-end=40}{2}} \htmlData{tutor-start=42,tutor-end=43}{+} \htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{9}^{\htmlData{tutor-start=48,tutor-end=49}{2}}\htmlData{tutor-start=50,tutor-end=51}{)} \htmlData{tutor-start=52,tutor-end=53}{t}^{\htmlData{tutor-start=55,tutor-end=56}{2}}$。注意这里用的是半边长 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{9}$。即 $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{(}\frac{\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{0}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{)}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{+} \htmlData{tutor-start=41,tutor-end=42}{(}\frac{\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{2}}{\htmlData{tutor-start=52,tutor-end=53}{2}}\htmlData{tutor-start=54,tutor-end=55}{)}^{\htmlData{tutor-start=57,tutor-end=58}{2}} \htmlData{tutor-start=60,tutor-end=61}{+} \htmlData{tutor-start=62,tutor-end=63}{(}\frac{\htmlData{tutor-start=69,tutor-end=70}{3}\htmlData{tutor-start=70,tutor-end=71}{8}}{\htmlData{tutor-start=73,tutor-end=74}{2}}\htmlData{tutor-start=75,tutor-end=76}{)}^{\htmlData{tutor-start=78,tutor-end=79}{2}}\htmlData{tutor-start=80,tutor-end=81}{)} \htmlData{tutor-start=82,tutor-end=83}{t}^{\htmlData{tutor-start=85,tutor-end=86}{2}} \htmlData{tutor-start=88,tutor-end=89}{=} \frac{\sqrt{\htmlData{tutor-start=102,tutor-end=103}{3}}}{\htmlData{tutor-start=106,tutor-end=107}{1}\htmlData{tutor-start=107,tutor-end=108}{2}} \htmlData{tutor-start=110,tutor-end=111}{(}\htmlData{tutor-start=111,tutor-end=112}{2}\htmlData{tutor-start=112,tutor-end=113}{0}^{\htmlData{tutor-start=115,tutor-end=116}{2}} \htmlData{tutor-start=118,tutor-end=119}{+} \htmlData{tutor-start=120,tutor-end=121}{2}\htmlData{tutor-start=121,tutor-end=122}{2}^{\htmlData{tutor-start=124,tutor-end=125}{2}} \htmlData{tutor-start=127,tutor-end=128}{+} \htmlData{tutor-start=129,tutor-end=130}{3}\htmlData{tutor-start=130,tutor-end=131}{8}^{\htmlData{tutor-start=133,tutor-end=134}{2}}\htmlData{tutor-start=135,tutor-end=136}{)} \htmlData{tutor-start=137,tutor-end=138}{t}^{\htmlData{tutor-start=140,tutor-end=141}{2}}$。 计算括号内：$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0} \htmlData{tutor-start=4,tutor-end=5}{+} \htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{8}\htmlData{tutor-start=8,tutor-end=9}{4} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{4} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{8}$。 所以外心三角形面积和 $\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=29}{\times }\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{3}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{8} \htmlData{tutor-start=34,tutor-end=35}{t}^{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{9}\htmlData{tutor-start=44,tutor-end=45}{4}\sqrt{\htmlData{tutor-start=51,tutor-end=52}{3}} \htmlData{tutor-start=54,tutor-end=55}{t}^{\htmlData{tutor-start=57,tutor-end=58}{2}}$。

代入总面积公式： $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=33}{\triangle }\htmlData{tutor-start=33,tutor-end=34}{A}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{B}_{\htmlData{tutor-start=41,tutor-end=42}{1}}\htmlData{tutor-start=43,tutor-end=44}{C}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{(}\htmlData{tutor-start=54,tutor-end=55}{1}\htmlData{tutor-start=55,tutor-end=56}{2}\htmlData{tutor-start=56,tutor-end=57}{0}\sqrt{\htmlData{tutor-start=63,tutor-end=64}{2}} \htmlData{tutor-start=66,tutor-end=67}{t}^{\htmlData{tutor-start=69,tutor-end=70}{2}} \htmlData{tutor-start=72,tutor-end=73}{+} \htmlData{tutor-start=74,tutor-end=75}{1}\htmlData{tutor-start=75,tutor-end=76}{9}\htmlData{tutor-start=76,tutor-end=77}{4}\sqrt{\htmlData{tutor-start=83,tutor-end=84}{3}} \htmlData{tutor-start=86,tutor-end=87}{t}^{\htmlData{tutor-start=89,tutor-end=90}{2}}\htmlData{tutor-start=91,tutor-end=92}{)}$。 另一方面，由相似比可知 $\frac{\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=19}{\triangle }\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{C}}}{\htmlData{tutor-start=25,tutor-end=26}{S}_{\htmlData{tutor-start=28,tutor-end=38}{\triangle }\htmlData{tutor-start=38,tutor-end=39}{D}\htmlData{tutor-start=39,tutor-end=40}{E}\htmlData{tutor-start=40,tutor-end=41}{F}}} \htmlData{tutor-start=44,tutor-end=45}{=} \frac{\htmlData{tutor-start=52,tutor-end=53}{S}_{\htmlData{tutor-start=55,tutor-end=65}{\triangle }\htmlData{tutor-start=65,tutor-end=66}{A}_{\htmlData{tutor-start=68,tutor-end=69}{1}}\htmlData{tutor-start=70,tutor-end=71}{B}_{\htmlData{tutor-start=73,tutor-end=74}{1}}\htmlData{tutor-start=75,tutor-end=76}{C}_{\htmlData{tutor-start=78,tutor-end=79}{1}}}}{\htmlData{tutor-start=83,tutor-end=84}{S}_{\htmlData{tutor-start=86,tutor-end=96}{\triangle }\htmlData{tutor-start=96,tutor-end=97}{X}\htmlData{tutor-start=97,tutor-end=98}{Y}\htmlData{tutor-start=98,tutor-end=99}{Z}}}$。但这不够直接。我们需要的是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=22}{\triangle }\htmlData{tutor-start=22,tutor-end=23}{D}\htmlData{tutor-start=23,tutor-end=24}{E}\htmlData{tutor-start=24,tutor-end=25}{F}}} \htmlData{tutor-start=28,tutor-end=29}{+} \frac{\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{S}_{\htmlData{tutor-start=42,tutor-end=52}{\triangle }\htmlData{tutor-start=52,tutor-end=53}{X}\htmlData{tutor-start=53,tutor-end=54}{Y}\htmlData{tutor-start=54,tutor-end=55}{Z}}}$。 利用前面的相似性推导，实际上有恒等式： $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=22}{\triangle }\htmlData{tutor-start=22,tutor-end=23}{D}\htmlData{tutor-start=23,tutor-end=24}{E}\htmlData{tutor-start=24,tutor-end=25}{F}}} \htmlData{tutor-start=28,tutor-end=29}{+} \frac{\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{S}_{\htmlData{tutor-start=42,tutor-end=52}{\triangle }\htmlData{tutor-start=52,tutor-end=53}{X}\htmlData{tutor-start=53,tutor-end=54}{Y}\htmlData{tutor-start=54,tutor-end=55}{Z}}} \htmlData{tutor-start=58,tutor-end=59}{=} \frac{\htmlData{tutor-start=66,tutor-end=67}{S}_{\htmlData{tutor-start=69,tutor-end=79}{\triangle }\htmlData{tutor-start=79,tutor-end=80}{A}\htmlData{tutor-start=80,tutor-end=81}{B}\htmlData{tutor-start=81,tutor-end=82}{C}} \htmlData{tutor-start=84,tutor-end=85}{+} \htmlData{tutor-start=86,tutor-end=87}{S}_{\htmlData{tutor-start=89,tutor-end=99}{\triangle }\htmlData{tutor-start=99,tutor-end=100}{A}_{\htmlData{tutor-start=102,tutor-end=103}{1}}\htmlData{tutor-start=104,tutor-end=105}{B}_{\htmlData{tutor-start=107,tutor-end=108}{1}}\htmlData{tutor-start=109,tutor-end=110}{C}_{\htmlData{tutor-start=112,tutor-end=113}{1}}}}{\htmlData{tutor-start=117,tutor-end=118}{S}_{\htmlData{tutor-start=120,tutor-end=130}{\triangle }\htmlData{tutor-start=130,tutor-end=131}{A}\htmlData{tutor-start=131,tutor-end=132}{B}\htmlData{tutor-start=132,tutor-end=133}{C}} \htmlData{tutor-start=135,tutor-end=141}{\cdot }\htmlData{tutor-start=141,tutor-end=142}{S}_{\htmlData{tutor-start=144,tutor-end=154}{\triangle }\htmlData{tutor-start=154,tutor-end=155}{D}\htmlData{tutor-start=155,tutor-end=156}{E}\htmlData{tutor-start=156,tutor-end=157}{F}}}$ 等等，参考解答的逻辑是：$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{D}\htmlData{tutor-start=13,tutor-end=14}{E}\htmlData{tutor-start=14,tutor-end=15}{F}}} \htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{S}_{\htmlData{tutor-start=32,tutor-end=33}{X}\htmlData{tutor-start=33,tutor-end=34}{Y}\htmlData{tutor-start=34,tutor-end=35}{Z}}} \htmlData{tutor-start=38,tutor-end=39}{=} \frac{\htmlData{tutor-start=46,tutor-end=47}{1}}{\htmlData{tutor-start=49,tutor-end=50}{S}_{\htmlData{tutor-start=52,tutor-end=53}{A}\htmlData{tutor-start=53,tutor-end=54}{B}\htmlData{tutor-start=54,tutor-end=55}{C}}} \htmlData{tutor-start=58,tutor-end=59}{(}\frac{\htmlData{tutor-start=65,tutor-end=66}{S}_{\htmlData{tutor-start=68,tutor-end=69}{A}\htmlData{tutor-start=69,tutor-end=70}{B}\htmlData{tutor-start=70,tutor-end=71}{C}}}{\htmlData{tutor-start=74,tutor-end=75}{S}_{\htmlData{tutor-start=77,tutor-end=78}{D}\htmlData{tutor-start=78,tutor-end=79}{E}\htmlData{tutor-start=79,tutor-end=80}{F}}} \htmlData{tutor-start=83,tutor-end=84}{+} \frac{\htmlData{tutor-start=91,tutor-end=92}{S}_{\htmlData{tutor-start=94,tutor-end=95}{A}\htmlData{tutor-start=95,tutor-end=96}{B}\htmlData{tutor-start=96,tutor-end=97}{C}}}{\htmlData{tutor-start=100,tutor-end=101}{S}_{\htmlData{tutor-start=103,tutor-end=104}{X}\htmlData{tutor-start=104,tutor-end=105}{Y}\htmlData{tutor-start=105,tutor-end=106}{Z}}}\htmlData{tutor-start=108,tutor-end=109}{)}$。由于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z} \htmlData{tutor-start=14,tutor-end=19}{\sim }\htmlData{tutor-start=19,tutor-end=29}{\triangle }\htmlData{tutor-start=29,tutor-end=30}{D}\htmlData{tutor-start=30,tutor-end=31}{E}\htmlData{tutor-start=31,tutor-end=32}{F}$ 且 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}} \htmlData{tutor-start=26,tutor-end=31}{\sim }\htmlData{tutor-start=31,tutor-end=41}{\triangle }\htmlData{tutor-start=41,tutor-end=42}{A}\htmlData{tutor-start=42,tutor-end=43}{B}\htmlData{tutor-start=43,tutor-end=44}{C}$，且变换把 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} \htmlData{tutor-start=4,tutor-end=8}{\to }\htmlData{tutor-start=8,tutor-end=9}{A}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{B}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{C}_{\htmlData{tutor-start=21,tutor-end=22}{1}}$ 同时把 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{Y}\htmlData{tutor-start=2,tutor-end=3}{Z} \htmlData{tutor-start=4,tutor-end=8}{\to }\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{F}$，故 $\frac{\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{C}}}{\htmlData{tutor-start=15,tutor-end=16}{S}_{\htmlData{tutor-start=18,tutor-end=19}{X}\htmlData{tutor-start=19,tutor-end=20}{Y}\htmlData{tutor-start=20,tutor-end=21}{Z}}} \htmlData{tutor-start=24,tutor-end=25}{=} \frac{\htmlData{tutor-start=32,tutor-end=33}{S}_{\htmlData{tutor-start=35,tutor-end=36}{A}_{\htmlData{tutor-start=38,tutor-end=39}{1}}\htmlData{tutor-start=40,tutor-end=41}{B}_{\htmlData{tutor-start=43,tutor-end=44}{1}}\htmlData{tutor-start=45,tutor-end=46}{C}_{\htmlData{tutor-start=48,tutor-end=49}{1}}}}{\htmlData{tutor-start=53,tutor-end=54}{S}_{\htmlData{tutor-start=56,tutor-end=57}{D}\htmlData{tutor-start=57,tutor-end=58}{E}\htmlData{tutor-start=58,tutor-end=59}{F}}}$。 所以原式 $\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{S}_{\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{B}\htmlData{tutor-start=16,tutor-end=17}{C}}} \frac{\htmlData{tutor-start=26,tutor-end=27}{S}_{\htmlData{tutor-start=29,tutor-end=30}{A}\htmlData{tutor-start=30,tutor-end=31}{B}\htmlData{tutor-start=31,tutor-end=32}{C}} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=37}{S}_{\htmlData{tutor-start=39,tutor-end=40}{A}_{\htmlData{tutor-start=42,tutor-end=43}{1}}\htmlData{tutor-start=44,tutor-end=45}{B}_{\htmlData{tutor-start=47,tutor-end=48}{1}}\htmlData{tutor-start=49,tutor-end=50}{C}_{\htmlData{tutor-start=52,tutor-end=53}{1}}}}{\htmlData{tutor-start=57,tutor-end=58}{S}_{\htmlData{tutor-start=60,tutor-end=61}{D}\htmlData{tutor-start=61,tutor-end=62}{E}\htmlData{tutor-start=62,tutor-end=63}{F}}}$。

已知 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\sqrt{\htmlData{tutor-start=32,tutor-end=33}{3}}}{\htmlData{tutor-start=36,tutor-end=37}{4}} \htmlData{tutor-start=39,tutor-end=46}{\times }\htmlData{tutor-start=46,tutor-end=47}{1}^{\htmlData{tutor-start=49,tutor-end=50}{2}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\sqrt{\htmlData{tutor-start=66,tutor-end=67}{3}}}{\htmlData{tutor-start=70,tutor-end=71}{4}}$。 分子部分 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=33}{\triangle }\htmlData{tutor-start=33,tutor-end=34}{A}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{B}_{\htmlData{tutor-start=41,tutor-end=42}{1}}\htmlData{tutor-start=43,tutor-end=44}{C}_{\htmlData{tutor-start=46,tutor-end=47}{1}}}$ 我们刚才算出是 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{0}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{9}\htmlData{tutor-start=18,tutor-end=19}{4}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{t}^{\htmlData{tutor-start=31,tutor-end=32}{2}}$。但是这里有个问题：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}}$ 是定值 $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{4}}$，而右边含有 $\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$。这说明 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 不是任意的，而是由 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的大小决定的。或者更准确地说，上面的面积和公式是针对“单位化”后的 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 而言的相对值？ 不，公式 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=33}{\triangle }\htmlData{tutor-start=33,tutor-end=34}{A}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{B}_{\htmlData{tutor-start=41,tutor-end=42}{1}}\htmlData{tutor-start=43,tutor-end=44}{C}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \htmlData{tutor-start=50,tutor-end=51}{=} \dots$ 是绝对面积等式。这意味着 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 必须满足该等式左边等于右边。但左边 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}}$ 固定，$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{B}_{\htmlData{tutor-start=21,tutor-end=22}{1}}\htmlData{tutor-start=23,tutor-end=24}{C}_{\htmlData{tutor-start=26,tutor-end=27}{1}}}$ 随 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 变化？不对，$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 是由 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 变换来的，其大小也是固定的吗？ 回顾变换：把 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$ 变到 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$。$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{X}\htmlData{tutor-start=11,tutor-end=12}{Y}\htmlData{tutor-start=12,tutor-end=13}{Z}$ 的大小取决于 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F}$ 的位置（即 $\htmlData{tutor-start=0,tutor-end=1}{t}$）。所以 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 的大小也取决于 $\htmlData{tutor-start=0,tutor-end=1}{t}$。因此 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 确实是变量？ **修正理解**：题目问“所有可能值”。如果结果唯一，说明该表达式与 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 无关，或者 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 被唯一确定。但在一般的好三角形对中，$\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F}$ 可以在边上滑动，只要保持形状。然而，参考解答最后算出了一个定值，暗示该表达式确实为定值。让我们检查量纲。 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}} \htmlData{tutor-start=12,tutor-end=13}{+} \frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{S}}$ 的量纲是 $\htmlData{tutor-start=0,tutor-end=1}{L}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}}$。$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{L}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$。公式 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{C}}} \frac{\htmlData{tutor-start=24,tutor-end=25}{S}_{\htmlData{tutor-start=27,tutor-end=28}{s}\htmlData{tutor-start=28,tutor-end=29}{u}\htmlData{tutor-start=29,tutor-end=30}{m}}}{\htmlData{tutor-start=33,tutor-end=34}{S}_{\htmlData{tutor-start=36,tutor-end=37}{D}\htmlData{tutor-start=37,tutor-end=38}{E}\htmlData{tutor-start=38,tutor-end=39}{F}}}$ 量纲是 $\htmlData{tutor-start=0,tutor-end=1}{L}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=13}{\cdot }\htmlData{tutor-start=13,tutor-end=14}{L}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{/} \htmlData{tutor-start=21,tutor-end=22}{L}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{L}^{\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{2}}$。匹配。 关键在于：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=33}{\triangle }\htmlData{tutor-start=33,tutor-end=34}{A}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{B}_{\htmlData{tutor-start=41,tutor-end=42}{1}}\htmlData{tutor-start=43,tutor-end=44}{C}_{\htmlData{tutor-start=46,tutor-end=47}{1}}}$ 是否真的等于 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{S}_{\htmlData{tutor-start=5,tutor-end=6}{D}\htmlData{tutor-start=6,tutor-end=7}{E}\htmlData{tutor-start=7,tutor-end=8}{F}} \htmlData{tutor-start=10,tutor-end=11}{+} \dots\htmlData{tutor-start=17,tutor-end=18}{)}$？是的。那么 $\frac{\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{C}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{S}_{\htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{B}_{\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{C}_{\htmlData{tutor-start=32,tutor-end=33}{1}}}}{\htmlData{tutor-start=37,tutor-end=38}{S}_{\htmlData{tutor-start=40,tutor-end=41}{D}\htmlData{tutor-start=41,tutor-end=42}{E}\htmlData{tutor-start=42,tutor-end=43}{F}}} \htmlData{tutor-start=46,tutor-end=47}{=} \htmlData{tutor-start=48,tutor-end=49}{2} \htmlData{tutor-start=50,tutor-end=51}{+} \htmlData{tutor-start=52,tutor-end=53}{2} \frac{\htmlData{tutor-start=60,tutor-end=61}{S}_{\htmlData{tutor-start=63,tutor-end=64}{c}\htmlData{tutor-start=64,tutor-end=65}{e}\htmlData{tutor-start=65,tutor-end=66}{n}\htmlData{tutor-start=66,tutor-end=67}{t}\htmlData{tutor-start=67,tutor-end=68}{e}\htmlData{tutor-start=68,tutor-end=69}{r}\htmlData{tutor-start=69,tutor-end=70}{s}}}{\htmlData{tutor-start=73,tutor-end=74}{S}_{\htmlData{tutor-start=76,tutor-end=77}{D}\htmlData{tutor-start=77,tutor-end=78}{E}\htmlData{tutor-start=78,tutor-end=79}{F}}}$。这一项只依赖于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 的形状（即边长比），而与大小 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 无关！因为分子分母都是 $\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 的倍数。而 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}}$ 是常数。所以整个表达式确实是定值。

计算最终值： 比值 $\htmlData{tutor-start=0,tutor-end=1}{K} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{S}_{\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}{+} \htmlData{tutor-start=20,tutor-end=21}{S}_{\htmlData{tutor-start=23,tutor-end=24}{A}_{\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{B}_{\htmlData{tutor-start=31,tutor-end=32}{1}}\htmlData{tutor-start=33,tutor-end=34}{C}_{\htmlData{tutor-start=36,tutor-end=37}{1}}}}{\htmlData{tutor-start=41,tutor-end=42}{S}_{\htmlData{tutor-start=44,tutor-end=45}{D}\htmlData{tutor-start=45,tutor-end=46}{E}\htmlData{tutor-start=46,tutor-end=47}{F}}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{2} \frac{\htmlData{tutor-start=60,tutor-end=61}{1}\htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=63}{0}\sqrt{\htmlData{tutor-start=69,tutor-end=70}{2}} \htmlData{tutor-start=72,tutor-end=73}{+} \htmlData{tutor-start=74,tutor-end=75}{1}\htmlData{tutor-start=75,tutor-end=76}{9}\htmlData{tutor-start=76,tutor-end=77}{4}\sqrt{\htmlData{tutor-start=83,tutor-end=84}{3}}}{\htmlData{tutor-start=87,tutor-end=88}{1}\htmlData{tutor-start=88,tutor-end=89}{2}\htmlData{tutor-start=89,tutor-end=90}{0}\sqrt{\htmlData{tutor-start=96,tutor-end=97}{2}}} \htmlData{tutor-start=100,tutor-end=101}{=} \htmlData{tutor-start=102,tutor-end=103}{2} \htmlData{tutor-start=104,tutor-end=105}{(}\htmlData{tutor-start=105,tutor-end=106}{1} \htmlData{tutor-start=107,tutor-end=108}{+} \frac{\htmlData{tutor-start=115,tutor-end=116}{1}\htmlData{tutor-start=116,tutor-end=117}{9}\htmlData{tutor-start=117,tutor-end=118}{4}\sqrt{\htmlData{tutor-start=124,tutor-end=125}{3}}}{\htmlData{tutor-start=128,tutor-end=129}{1}\htmlData{tutor-start=129,tutor-end=130}{2}\htmlData{tutor-start=130,tutor-end=131}{0}\sqrt{\htmlData{tutor-start=137,tutor-end=138}{2}}}\htmlData{tutor-start=140,tutor-end=141}{)} \htmlData{tutor-start=142,tutor-end=143}{=} \htmlData{tutor-start=144,tutor-end=145}{2} \htmlData{tutor-start=146,tutor-end=147}{+} \frac{\htmlData{tutor-start=154,tutor-end=155}{9}\htmlData{tutor-start=155,tutor-end=156}{7}\sqrt{\htmlData{tutor-start=162,tutor-end=163}{3}}}{\htmlData{tutor-start=166,tutor-end=167}{3}\htmlData{tutor-start=167,tutor-end=168}{0}\sqrt{\htmlData{tutor-start=174,tutor-end=175}{2}}} \htmlData{tutor-start=178,tutor-end=179}{=} \htmlData{tutor-start=180,tutor-end=181}{2} \htmlData{tutor-start=182,tutor-end=183}{+} \frac{\htmlData{tutor-start=190,tutor-end=191}{9}\htmlData{tutor-start=191,tutor-end=192}{7}\sqrt{\htmlData{tutor-start=198,tutor-end=199}{6}}}{\htmlData{tutor-start=202,tutor-end=203}{6}\htmlData{tutor-start=203,tutor-end=204}{0}}$。 等等，参考解答的计算过程略有不同。它写的是： $\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\htmlData{tutor-start=8,tutor-end=9}{4}}{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}} \frac{\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{0}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{+} \htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{9}\htmlData{tutor-start=45,tutor-end=46}{4}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{)}}{\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{2}\htmlData{tutor-start=59,tutor-end=60}{0}\sqrt{\htmlData{tutor-start=66,tutor-end=67}{2}}}$。 这里 $\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}}}$ 就是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=22}{\triangle }\htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{B}\htmlData{tutor-start=24,tutor-end=25}{C}}}$。没问题。 化简分数部分：$\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{0}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{9}\htmlData{tutor-start=24,tutor-end=25}{4}\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{)}}{\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{0}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{2}}} \htmlData{tutor-start=49,tutor-end=50}{=} \frac{\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{2}\htmlData{tutor-start=59,tutor-end=60}{0}\sqrt{\htmlData{tutor-start=66,tutor-end=67}{2}} \htmlData{tutor-start=69,tutor-end=70}{+} \htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{9}\htmlData{tutor-start=73,tutor-end=74}{4}\sqrt{\htmlData{tutor-start=80,tutor-end=81}{3}}}{\htmlData{tutor-start=84,tutor-end=85}{6}\htmlData{tutor-start=85,tutor-end=86}{0}\sqrt{\htmlData{tutor-start=92,tutor-end=93}{2}}} \htmlData{tutor-start=96,tutor-end=97}{=} \htmlData{tutor-start=98,tutor-end=99}{1} \htmlData{tutor-start=100,tutor-end=101}{+} \frac{\htmlData{tutor-start=108,tutor-end=109}{1}\htmlData{tutor-start=109,tutor-end=110}{9}\htmlData{tutor-start=110,tutor-end=111}{4}\sqrt{\htmlData{tutor-start=117,tutor-end=118}{3}}}{\htmlData{tutor-start=121,tutor-end=122}{6}\htmlData{tutor-start=122,tutor-end=123}{0}\sqrt{\htmlData{tutor-start=129,tutor-end=130}{2}}} \htmlData{tutor-start=133,tutor-end=134}{=} \htmlData{tutor-start=135,tutor-end=136}{1} \htmlData{tutor-start=137,tutor-end=138}{+} \frac{\htmlData{tutor-start=145,tutor-end=146}{9}\htmlData{tutor-start=146,tutor-end=147}{7}\sqrt{\htmlData{tutor-start=153,tutor-end=154}{3}}}{\htmlData{tutor-start=157,tutor-end=158}{3}\htmlData{tutor-start=158,tutor-end=159}{0}\sqrt{\htmlData{tutor-start=165,tutor-end=166}{2}}} \htmlData{tutor-start=169,tutor-end=170}{=} \htmlData{tutor-start=171,tutor-end=172}{1} \htmlData{tutor-start=173,tutor-end=174}{+} \frac{\htmlData{tutor-start=181,tutor-end=182}{9}\htmlData{tutor-start=182,tutor-end=183}{7}\sqrt{\htmlData{tutor-start=189,tutor-end=190}{6}}}{\htmlData{tutor-start=193,tutor-end=194}{6}\htmlData{tutor-start=194,tutor-end=195}{0}}$。 乘以 $\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}}}$： $\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}}} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1} \htmlData{tutor-start=22,tutor-end=23}{+} \frac{\htmlData{tutor-start=30,tutor-end=31}{9}\htmlData{tutor-start=31,tutor-end=32}{7}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{6}}}{\htmlData{tutor-start=42,tutor-end=43}{6}\htmlData{tutor-start=43,tutor-end=44}{0}}\htmlData{tutor-start=45,tutor-end=46}{)} \htmlData{tutor-start=47,tutor-end=48}{=} \frac{\htmlData{tutor-start=55,tutor-end=56}{4}}{\sqrt{\htmlData{tutor-start=64,tutor-end=65}{3}}} \htmlData{tutor-start=68,tutor-end=69}{+} \frac{\htmlData{tutor-start=76,tutor-end=77}{4} \htmlData{tutor-start=78,tutor-end=84}{\cdot }\htmlData{tutor-start=84,tutor-end=85}{9}\htmlData{tutor-start=85,tutor-end=86}{7} \sqrt{\htmlData{tutor-start=93,tutor-end=94}{6}}}{\htmlData{tutor-start=97,tutor-end=98}{6}\htmlData{tutor-start=98,tutor-end=99}{0} \sqrt{\htmlData{tutor-start=106,tutor-end=107}{3}}} \htmlData{tutor-start=110,tutor-end=111}{=} \frac{\htmlData{tutor-start=118,tutor-end=119}{4}\sqrt{\htmlData{tutor-start=125,tutor-end=126}{3}}}{\htmlData{tutor-start=129,tutor-end=130}{3}} \htmlData{tutor-start=132,tutor-end=133}{+} \frac{\htmlData{tutor-start=140,tutor-end=141}{9}\htmlData{tutor-start=141,tutor-end=142}{7} \sqrt{\htmlData{tutor-start=149,tutor-end=150}{2}}}{\htmlData{tutor-start=153,tutor-end=154}{1}\htmlData{tutor-start=154,tutor-end=155}{5}}$。 通分：$\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{0}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{5}} \htmlData{tutor-start=22,tutor-end=23}{+} \frac{\htmlData{tutor-start=30,tutor-end=31}{9}\htmlData{tutor-start=31,tutor-end=32}{7}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{2}}}{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{5}} \htmlData{tutor-start=46,tutor-end=47}{=} \frac{\htmlData{tutor-start=54,tutor-end=55}{9}\htmlData{tutor-start=55,tutor-end=56}{7}\sqrt{\htmlData{tutor-start=62,tutor-end=63}{2}} \htmlData{tutor-start=65,tutor-end=66}{+} \htmlData{tutor-start=67,tutor-end=68}{2}\htmlData{tutor-start=68,tutor-end=69}{0}\sqrt{\htmlData{tutor-start=75,tutor-end=76}{3}}}{\htmlData{tutor-start=79,tutor-end=80}{1}\htmlData{tutor-start=80,tutor-end=81}{5}}$。 **注意**：参考解答给出的答案是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{7}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{0}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5}}$。我的计算结果是 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}$。哪里出错了？ 回看参考解答步骤(4)： $\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\htmlData{tutor-start=8,tutor-end=9}{4}}{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}} \frac{\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{0}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{+} \htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{9}\htmlData{tutor-start=45,tutor-end=46}{4}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{)}}{\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{2}\htmlData{tutor-start=59,tutor-end=60}{0}\sqrt{\htmlData{tutor-start=66,tutor-end=67}{2}}}$ $\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\htmlData{tutor-start=8,tutor-end=9}{4}}{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{2} \htmlData{tutor-start=24,tutor-end=25}{+} \frac{\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{8}\htmlData{tutor-start=34,tutor-end=35}{8}\sqrt{\htmlData{tutor-start=41,tutor-end=42}{3}}}{\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{0}\sqrt{\htmlData{tutor-start=54,tutor-end=55}{2}}}\htmlData{tutor-start=57,tutor-end=58}{)} \htmlData{tutor-start=59,tutor-end=60}{=} \frac{\htmlData{tutor-start=67,tutor-end=68}{8}}{\sqrt{\htmlData{tutor-start=76,tutor-end=77}{3}}} \htmlData{tutor-start=80,tutor-end=81}{+} \frac{\htmlData{tutor-start=88,tutor-end=89}{4} \htmlData{tutor-start=90,tutor-end=96}{\cdot }\htmlData{tutor-start=96,tutor-end=97}{3}\htmlData{tutor-start=97,tutor-end=98}{8}\htmlData{tutor-start=98,tutor-end=99}{8} \sqrt{\htmlData{tutor-start=106,tutor-end=107}{3}}}{\sqrt{\htmlData{tutor-start=116,tutor-end=117}{3}} \htmlData{tutor-start=119,tutor-end=125}{\cdot }\htmlData{tutor-start=125,tutor-end=126}{1}\htmlData{tutor-start=126,tutor-end=127}{2}\htmlData{tutor-start=127,tutor-end=128}{0} \sqrt{\htmlData{tutor-start=135,tutor-end=136}{2}}} \htmlData{tutor-start=139,tutor-end=140}{=} \frac{\htmlData{tutor-start=147,tutor-end=148}{8}\sqrt{\htmlData{tutor-start=154,tutor-end=155}{3}}}{\htmlData{tutor-start=158,tutor-end=159}{3}} \htmlData{tutor-start=161,tutor-end=162}{+} \frac{\htmlData{tutor-start=169,tutor-end=170}{3}\htmlData{tutor-start=170,tutor-end=171}{8}\htmlData{tutor-start=171,tutor-end=172}{8}}{\htmlData{tutor-start=174,tutor-end=175}{3}\htmlData{tutor-start=175,tutor-end=176}{0}\sqrt{\htmlData{tutor-start=182,tutor-end=183}{2}}} \htmlData{tutor-start=186,tutor-end=187}{=} \frac{\htmlData{tutor-start=194,tutor-end=195}{8}\sqrt{\htmlData{tutor-start=201,tutor-end=202}{3}}}{\htmlData{tutor-start=205,tutor-end=206}{3}} \htmlData{tutor-start=208,tutor-end=209}{+} \frac{\htmlData{tutor-start=216,tutor-end=217}{1}\htmlData{tutor-start=217,tutor-end=218}{9}\htmlData{tutor-start=218,tutor-end=219}{4}}{\htmlData{tutor-start=221,tutor-end=222}{1}\htmlData{tutor-start=222,tutor-end=223}{5}\sqrt{\htmlData{tutor-start=229,tutor-end=230}{2}}} \htmlData{tutor-start=233,tutor-end=234}{=} \frac{\htmlData{tutor-start=241,tutor-end=242}{8}\sqrt{\htmlData{tutor-start=248,tutor-end=249}{3}}}{\htmlData{tutor-start=252,tutor-end=253}{3}} \htmlData{tutor-start=255,tutor-end=256}{+} \frac{\htmlData{tutor-start=263,tutor-end=264}{9}\htmlData{tutor-start=264,tutor-end=265}{7}\sqrt{\htmlData{tutor-start=271,tutor-end=272}{2}}}{\htmlData{tutor-start=275,tutor-end=276}{1}\htmlData{tutor-start=276,tutor-end=277}{5}}$。 $\frac{\htmlData{tutor-start=6,tutor-end=7}{8}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{0}\sqrt{\htmlData{tutor-start=36,tutor-end=37}{3}}}{\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{5}}$。 所以结果是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{7}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{0}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5}}$。 我之前的错误在于把 $\frac{2(\dots)}{120\sqrt{2}}$ 约分时，把分子的2除到了分母变成60，但忘了分子里还有两项。应该是 $\frac{240\sqrt{2} + 388\sqrt{3}}{120\sqrt{2}} = 2 + \dots$。而我之前算成了 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \dots$（漏了那个2倍系数或者约分错误）。 确认无误，答案为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{7}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{0}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5}}$。

设 $\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{,} \mathcal{\htmlData{tutor-start=24,tutor-end=25}{W}}\htmlData{tutor-start=26,tutor-end=27}{)}$ 表示黑色配对方案 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$ 中的线段集合与白色配对方案 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}$ 中的线段集合在圆内部的交点总数。

首先证明 $\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{,} \mathcal{\htmlData{tutor-start=24,tutor-end=25}{W}}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=32}{\le }\htmlData{tutor-start=32,tutor-end=33}{P}\htmlData{tutor-start=33,tutor-end=34}{(}\mathcal{\htmlData{tutor-start=43,tutor-end=44}{B}}\htmlData{tutor-start=45,tutor-end=46}{)}$。考虑 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$ 中的任意一条黑色线段 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$。根据定义，$\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{i}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{)}$ 是直线 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$ 两侧白色点数量的较小值。由于白色配对方案 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}$ 中的线段互不相交，且每条白色线段连接两个白色点，因此任何一条与 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$ 相交的白色线段，必然有一个端点在 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$ 的一侧，另一个端点在另一侧。这意味着该白色线段的两个端点分别位于 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$ 分割出的两个半圆弧上。因为 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}$ 中线段不交叉，所以与 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{i}}$ 相交的白色线段数量不可能超过任一侧的白色点总数，更不可能超过两侧白色点数的较小值 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{i}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{)}$。对所有 $\htmlData{tutor-start=0,tutor-end=1}{m}$ 条黑色线段求和，即得 $\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{,} \mathcal{\htmlData{tutor-start=24,tutor-end=25}{W}}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=32}{\le }\sum \htmlData{tutor-start=37,tutor-end=38}{d}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{B}_{\htmlData{tutor-start=42,tutor-end=43}{i}}\htmlData{tutor-start=44,tutor-end=45}{,} \htmlData{tutor-start=46,tutor-end=47}{C}_{\htmlData{tutor-start=49,tutor-end=50}{i}}\htmlData{tutor-start=51,tutor-end=52}{)} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{P}\htmlData{tutor-start=56,tutor-end=57}{(}\mathcal{\htmlData{tutor-start=66,tutor-end=67}{B}}\htmlData{tutor-start=68,tutor-end=69}{)}$。

同理，交换黑白角色，可得 $\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{,} \mathcal{\htmlData{tutor-start=24,tutor-end=25}{W}}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=32}{\le }\htmlData{tutor-start=32,tutor-end=33}{P}\htmlData{tutor-start=33,tutor-end=34}{(}\mathcal{\htmlData{tutor-start=43,tutor-end=44}{W}}\htmlData{tutor-start=45,tutor-end=46}{)}$。

我们需要证明：对于任意固定的黑色配对方案 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$，都存在一个白色配对方案 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}$，使得 $\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{,} \mathcal{\htmlData{tutor-start=24,tutor-end=25}{W}}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{P}\htmlData{tutor-start=31,tutor-end=32}{(}\mathcal{\htmlData{tutor-start=41,tutor-end=42}{B}}\htmlData{tutor-start=43,tutor-end=44}{)}$。

采用归纳法证明。当 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ 时显然成立。假设对 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$ 个白点结论成立。考虑 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{n}$ 个白点的情形。 在 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$ 中选取一条使得 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{k}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{k}}\htmlData{tutor-start=14,tutor-end=15}{)}$ 最大的黑色线段，记为 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}$（若有多个取其一）。若 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{0}$，则所有黑弦两侧白点数均为0或全在一侧，此时任意合法白配对均满足交点数为0（因无白点可供跨越），结论成立。 若 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0}$，设 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}$ 为从 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 出发顺时针遇到的第一个白点，$\htmlData{tutor-start=0,tutor-end=1}{X}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}$ 为从 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 出发逆时针遇到的第一个白点。连接 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 作为 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}$ 中的一条线段。 考察移除 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{X}_{\htmlData{tutor-start=12,tutor-end=13}{o}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{t}}$ 后的剩余 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$ 个白点及原黑配对 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$。对于 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{B}}$ 中任意黑弦 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$： 1. 若 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 相交，则 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{X}_{\htmlData{tutor-start=12,tutor-end=13}{o}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{t}}$ 分别位于 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 两侧。移除它们后，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 两侧的白点数各减1，故新的着色距离 $\htmlData{tutor-start=0,tutor-end=1}{d}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{d}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{C}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{-} \htmlData{tutor-start=19,tutor-end=20}{1}$。 2. 若 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 不与 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 相交，需证 $\htmlData{tutor-start=0,tutor-end=1}{d}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{d}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{C}\htmlData{tutor-start=15,tutor-end=16}{)}$。由于 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{d}$ 值最大的黑弦，且 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{X}_{\htmlData{tutor-start=12,tutor-end=13}{o}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{t}}$ 紧邻 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$，几何位置上 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 位于 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}$ 的“外侧”或“内部极小区域”。详细分析可知，不与 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 相交的黑弦 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$，其两侧白点分布不会因移除这两个相邻边界点而改变较小值的计数（要么两点都在较多点的一侧，要么都在较少点一侧但数量仍由其他点决定，或者由 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}$ 的最大性保证结构稳定性）。 综上，$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{P}'\htmlData{tutor-start=19,tutor-end=20}{(}\mathcal{\htmlData{tutor-start=29,tutor-end=30}{B}}\htmlData{tutor-start=31,tutor-end=32}{)}$ 恰好等于与 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 相交的黑弦数量。由归纳假设，存在剩余白点的配对 $\mathcal{\htmlData{tutor-start=9,tutor-end=10}{W}}'$ 使交点数达到 $\htmlData{tutor-start=0,tutor-end=1}{P}'\htmlData{tutor-start=2,tutor-end=3}{(}\mathcal{\htmlData{tutor-start=12,tutor-end=13}{B}}\htmlData{tutor-start=14,tutor-end=15}{)}$。加上 $\htmlData{tutor-start=0,tutor-end=1}{W}_{\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{X}_{\htmlData{tutor-start=10,tutor-end=11}{o}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{t}}$ 后，总交点数增加量恰为 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{P}'\htmlData{tutor-start=19,tutor-end=20}{(}\mathcal{\htmlData{tutor-start=29,tutor-end=30}{B}}\htmlData{tutor-start=31,tutor-end=32}{)}$，故总交点数达到 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\mathcal{\htmlData{tutor-start=11,tutor-end=12}{B}}\htmlData{tutor-start=13,tutor-end=14}{)}$。