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2025 AIME II 第 14 题

先试着解答 2025 AIME II 第 14 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2025 AIME II 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

14.

令 △ABC\triangle ABC 为直角三角形，∠A=90∘\angle A = 90^\circ，且 BC=38BC = 38。三角形内部存在点 KK 和 LL，满足 AK=AL=BK=CL=KL=14。 \begin{gathered} AK = AL = BK \\ = CL = KL = 14 \end{gathered}\text{。} 四边形 BKLCBKLC 的面积可表示为 n3n\sqrt{3}，其中 nn 是正整数。求 nn。
Let △ABC\triangle ABC be a right triangle with ∠A=90∘\angle A = 90^\circ and BC=38.BC = 38. There exist points KK and LL inside the triangle such that AK=AL=BK=CL=KL=14. \begin{gathered} AK = AL = BK \\ = CL = KL = 14. \end{gathered} The area of the quadrilateral BKLCBKLC can be expressed as n3n\sqrt{3} for some positive integer n.n. Find n.n.
答案：104
知识点：等边三角形三角恒等式面积分割
难度评级：3270
小提示：
三角形 AKLAKL 是等边三角形，所以 ∠KAL=60∘\angle KAL = 60^\circ，且 ∠BAK+∠LAC=30∘\angle BAK + \angle LAC = 30^\circ；同时 AB=28cos⁡∠BAKAB = 28\cos\angle BAK
Triangle AKLAKL is equilateral, so ∠KAL=60∘\angle KAL = 60^\circ and ∠BAK+∠LAC=30∘;\angle BAK + \angle LAC = 30^\circ; also AB=28cos⁡∠BAKAB = 28\cos\angle BAK
大提示：
在 α+β=30∘\alpha + \beta = 30^\circ 下，关系 AB2+AC2=382AB^2 + AC^2 = 38^2 会确定 cos⁡(α−β)\cos(\alpha - \beta)；从 ABCABC 中减去三角形 ABKABK、ACLACL、AKLAKL
With α+β=30∘,\alpha + \beta = 30^\circ, the relation AB2+AC2=382AB^2 + AC^2 = 38^2 determines cos⁡(α−β);\cos(\alpha - \beta); subtract triangles ABK,ABK, ACL,ACL, AKLAKL from ABCABC
解答：
因为 AK=AL=KL=14AK = AL = KL = 14，三角形 AKLAKL 是等边三角形，且 ∠KAL=60∘\angle KAL = 60^\circ。令 α=∠BAK\alpha = \angle BAK，β=∠LAC\beta = \angle LAC，所以 α+β=30∘\alpha + \beta = 30^\circ。因为 AK=KBAK = KB，点 KK 在 AB‾\overline{AB} 的垂直平分线上，所以 AB=2⋅14cos⁡α=28cos⁡αAB = 2 \cdot 14\cos\alpha = 28\cos\alpha；同理 AC=28cos⁡βAC = 28\cos\beta。于是 AB2+AC2=382AB^2 + AC^2 = 38^2 给出 cos⁡2α+cos⁡2β=361196\cos^2\alpha + \cos^2\beta = \frac{361}{196}，即 cos⁡2α+cos⁡2β=16598\cos 2\alpha + \cos 2\beta = \frac{165}{98}。由和化积，2cos⁡(α+β)cos⁡(α−β)2\cos(\alpha+\beta)\cos(\alpha-\beta) =3cos⁡(α−β)= \sqrt{3}\cos(\alpha - \beta) =16598= \frac{165}{98}，所以 cos⁡(α−β)=55398\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}。
分解面积：[BKLC]=[ABC]−[ABK][BKLC] = [ABC] - [ABK] −[ACL]−[AKL]- [ACL] - [AKL]。首先， [ABC]=12AB⋅AC=392cos⁡αcos⁡β=196(cos⁡(α−β)+cos⁡(α+β))=1103+983=2083。 \begin{gathered} [ABC] = \tfrac{1}{2} AB \cdot AC \\ = 392\cos\alpha\cos\beta \\ = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) \\ = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3} \end{gathered}\text{。} 接着，KK 到 AB‾\overline{AB} 的高为 14sin⁡α14\sin\alpha，所以 [ABK]=12⋅28cos⁡α[ABK] = \frac{1}{2} \cdot 28\cos\alpha ⋅14sin⁡α\cdot 14\sin\alpha =98sin⁡2α= 98\sin 2\alpha，同理 [ACL]=98sin⁡2β[ACL] = 98\sin 2\beta；二者之和为 196sin⁡(α+β)cos⁡(α−β)196\sin(\alpha + \beta)\cos(\alpha - \beta) =98⋅55398= 98 \cdot \frac{55\sqrt{3}}{98} =553= 55\sqrt{3}。最后 [AKL]=34⋅142=493[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}。
因此 [BKLC]=2083−553[BKLC] = 208\sqrt{3} - 55\sqrt{3} −493=1043- 49\sqrt{3} = 104\sqrt{3}，所以 n=104n = 104。
Since AK=AL=KL=14,AK = AL = KL = 14, triangle AKLAKL is equilateral and ∠KAL=60∘.\angle KAL = 60^\circ. Let α=∠BAK\alpha = \angle BAK and β=∠LAC,\beta = \angle LAC, so α+β=30∘.\alpha + \beta = 30^\circ. Because AK=KB,AK = KB, point KK lies on the perpendicular bisector of AB‾,\overline{AB}, so AB=2⋅14cos⁡α=28cos⁡α;AB = 2 \cdot 14\cos\alpha = 28\cos\alpha; similarly AC=28cos⁡β.AC = 28\cos\beta. Then AB2+AC2=382AB^2 + AC^2 = 38^2 gives cos⁡2α+cos⁡2β=361196,\cos^2\alpha + \cos^2\beta = \frac{361}{196}, i.e. cos⁡2α+cos⁡2β=16598.\cos 2\alpha + \cos 2\beta = \frac{165}{98}. By sum-to-product, 2cos⁡(α+β)cos⁡(α−β)2\cos(\alpha+\beta)\cos(\alpha-\beta) =3cos⁡(α−β)= \sqrt{3}\cos(\alpha - \beta) =16598,= \frac{165}{98}, so cos⁡(α−β)=55398.\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}.
Decompose [BKLC]=[ABC]−[ABK][BKLC] = [ABC] - [ABK] −[ACL]−[AKL].- [ACL] - [AKL]. First, [ABC]=12AB⋅AC=392cos⁡αcos⁡β=196(cos⁡(α−β)+cos⁡(α+β))=1103+983=2083. \begin{gathered} [ABC] = \tfrac{1}{2} AB \cdot AC \\ = 392\cos\alpha\cos\beta \\ = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) \\ = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3}. \end{gathered} Next, KK has height 14sin⁡α14\sin\alpha over AB‾,\overline{AB}, so [ABK]=12⋅28cos⁡α[ABK] = \frac{1}{2} \cdot 28\cos\alpha ⋅14sin⁡α\cdot 14\sin\alpha =98sin⁡2α,= 98\sin 2\alpha, and likewise [ACL]=98sin⁡2β;[ACL] = 98\sin 2\beta; their sum is 196sin⁡(α+β)cos⁡(α−β)196\sin(\alpha + \beta)\cos(\alpha - \beta) =98⋅55398= 98 \cdot \frac{55\sqrt{3}}{98} =553.= 55\sqrt{3}. Finally [AKL]=34⋅142=493.[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}.
Therefore [BKLC]=2083−553[BKLC] = 208\sqrt{3} - 55\sqrt{3} −493=1043,- 49\sqrt{3} = 104\sqrt{3}, so n=104.n = 104.

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