Question

令 $\triangle ABC$ 为直角三角形， $\angle A = 90^\circ$ ，且 $BC = 38$ 。三角形内部存在点 $K$ 和 $L$ ，满足四边形 $BKLC$ 的面积可表示为 $n\sqrt{3}$ ，其中 $n$ 是正整数。求 $n$ 。 $\begin{gathered} AK = AL = BK \\ = CL = KL = 14. \end{gathered}$

Let $\triangle ABC$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such that $\begin{gathered} AK = AL = BK \\ = CL = KL = 14. \end{gathered}$ The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt{3}$ for some positive integer $n.$ Find $n.$

答案

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

解答：

因为 $AK = AL = KL = 14$ ，三角形 $AKL$ 是等边三角形，且 $\angle KAL = 60^\circ$ 。令 $\alpha = \angle BAK$ ， $\beta = \angle LAC$ ，所以 $\alpha + \beta = 30^\circ$ 。因为 $AK = KB$ ，点 $K$ 在 $\overline{AB}$ 的垂直平分线上，所以 $AB = 2 \cdot 14\cos\alpha = 28\cos\alpha$ ；同理 $AC = 28\cos\beta$ 。于是 $AB^2 + AC^2 = 38^2$ 给出 $\cos^2\alpha + \cos^2\beta = \frac{361}{196}$ ，即 $\cos 2\alpha + \cos 2\beta = \frac{165}{98}$ 。由和化积， $2\cos(\alpha+\beta)\cos(\alpha-\beta)$ $= \sqrt{3}\cos(\alpha - \beta)$ $= \frac{165}{98}$ ，所以 $\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}$ 。

分解面积： $[BKLC] = [ABC] - [ABK]$ $- [ACL] - [AKL]$ 。首先，接着， $K$ 到 $\overline{AB}$ 的高为 $14\sin\alpha$ ，所以 $[ABK] = \frac{1}{2} \cdot 28\cos\alpha$ $\cdot 14\sin\alpha$ $= 98\sin 2\alpha$ ，同理 $[ACL] = 98\sin 2\beta$ ；二者之和为 $196\sin(\alpha + \beta)\cos(\alpha - \beta)$ $= 98 \cdot \frac{55\sqrt{3}}{98}$ $= 55\sqrt{3}$ 。最后 $[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}$ 。 $\begin{gathered} [ABC] = \tfrac{1}{2} AB \cdot AC \\ = 392\cos\alpha\cos\beta \\ = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) \\ = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3}. \end{gathered}$

因此 $[BKLC] = 208\sqrt{3} - 55\sqrt{3}$ $- 49\sqrt{3} = 104\sqrt{3}$ ，所以 $n = 104$ 。

Since $AK = AL = KL = 14,$ triangle $AKL$ is equilateral and $\angle KAL = 60^\circ.$ Let $\alpha = \angle BAK$ and $\beta = \angle LAC,$ so $\alpha + \beta = 30^\circ.$ Because $AK = KB,$ point $K$ lies on the perpendicular bisector of $\overline{AB},$ so $AB = 2 \cdot 14\cos\alpha = 28\cos\alpha;$ similarly $AC = 28\cos\beta.$ Then $AB^2 + AC^2 = 38^2$ gives $\cos^2\alpha + \cos^2\beta = \frac{361}{196},$ i.e. $\cos 2\alpha + \cos 2\beta = \frac{165}{98}.$ By sum-to-product, $2\cos(\alpha+\beta)\cos(\alpha-\beta)$ $= \sqrt{3}\cos(\alpha - \beta)$ $= \frac{165}{98},$ so $\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}.$

Decompose $[BKLC] = [ABC] - [ABK]$ $- [ACL] - [AKL].$ First, $\begin{gathered} [ABC] = \tfrac{1}{2} AB \cdot AC \\ = 392\cos\alpha\cos\beta \\ = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) \\ = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3}. \end{gathered}$ Next, $K$ has height $14\sin\alpha$ over $\overline{AB},$ so $[ABK] = \frac{1}{2} \cdot 28\cos\alpha$ $\cdot 14\sin\alpha$ $= 98\sin 2\alpha,$ and likewise $[ACL] = 98\sin 2\beta;$ their sum is $196\sin(\alpha + \beta)\cos(\alpha - \beta)$ $= 98 \cdot \frac{55\sqrt{3}}{98}$ $= 55\sqrt{3}.$ Finally $[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}.$

Therefore $[BKLC] = 208\sqrt{3} - 55\sqrt{3}$ $- 49\sqrt{3} = 104\sqrt{3},$ so $n = 104.$

2025 AIME II 第 14 题

答案：104

难度评级：3270

解答：

其他年份的第 14 题