Question

在等腰三角形 $ABC$ 中， $A$ 位于原点， $B$ 位于 $(20, 0)$ 点 $C$ 在第一象限，且 $AC = BC$ ， $\angle BAC = 75^\circ$ 。将 $\triangle ABC$ 绕点 $A$ 逆时针旋转，直到 $C$ 的像落在正 $y$ 轴上。原三角形与旋转后三角形的公共区域面积可写成 $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ ，其中 $p$ ， $q$ ， $r$ ， $s$ 为整数。求 $\frac{p - q + r - s}{2}$ 。

In isosceles triangle $ABC,$ $A$ is located at the origin and $B$ is located at $(20, 0).$ Point $C$ is in the first quadrant with $AC = BC$ and $\angle BAC = 75^\circ.$ If $\triangle ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original triangle and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s,$ where $p,$ $q,$ $r,$ $s$ are integers. Find $\frac{p - q + r - s}{2}.$

答案

答案：875

知识点：变换正弦定理相似面积分割

难度评级：3270

解答：

因为 $AC$ 与正 $x$ 轴成 $75^\circ$ 角，所以旋转角为 $15^\circ$ 。设 $B'$ 和 $C'$ 分别为 $B$ 和 $C$ 的像。由于 $\angle B'AB = 15^\circ$ 且 $\angle ABC = 75^\circ$ ，线段 $AB'$ 垂直于 $BC$ ；令 $D$ 为它们的交点，并令 $E = BC \cap B'C'$ ， $F = AC \cap B'C'$ 。公共区域为四边形 $ADEF$ ，其面积为 $[AB'F] - [EB'D]$ 。

在三角形 $AB'F$ 中， $\angle FAB' = 75^\circ - 15^\circ = 60^\circ$ ，且 $\angle AB'F = 75^\circ$ ，所以 $\angle AFB' = 45^\circ$ ，正弦定理给出 $B'F = 20\sin 60^\circ/\sin 45^\circ$ $= 10\sqrt{6}$ 。又 $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ ， $\begin{aligned} [AB'F] &= \tfrac{1}{2} \cdot 20 \cdot 10\sqrt{6}\,\sin 75^\circ \\ &= 50(3 + \sqrt{3}). \end{aligned}$

在直角三角形 $ABD$ 中， $AD = 20\cos 15^\circ$ ，且 $BD = 20\sin 15^\circ$ ，所以 $[ABD] = 200\sin 15^\circ\cos 15^\circ$ $= 100\sin 30^\circ = 50$ ，并且 $B'D = 20(1 - \cos 15^\circ)$ 。三角形 $EB'D$ 与 $ABD$ 相似（都在 $D$ 处为直角，且 $\angle EB'D = \angle ABD = 75^\circ$ ），所以利用 $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ ， $\begin{aligned} [EB'D] &= 50\left(\frac{1 - \cos 15^\circ}{\sin 15^\circ}\right)^2 \\ &= 50 \\ &\quad {}\cdot \left(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}\right). \end{aligned}$ $\begin{aligned} [ADEF] &= 50(3 + \sqrt{3}) \\ &\quad {}- 50 \\ &{}\cdot (15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}) \\ &= 500\sqrt{2} - 350\sqrt{3} \\ &\quad {}+ 300\sqrt{6} - 600, \end{aligned}$ 因此所以 $(p, q, r, s)$ $= (500, -350, 300, -600)$ ，且 $\frac{p - q + r - s}{2} = \frac{1750}{2} = 875$ 。

Since $AC$ makes a $75^\circ$ angle with the positive $x$ -axis, the rotation is by $15^\circ.$ Let $B'$ and $C'$ be the images of $B$ and $C.$ Because $\angle B'AB = 15^\circ$ and $\angle ABC = 75^\circ,$ segment $AB'$ is perpendicular to $BC;$ let $D$ be their intersection, and let $E = BC \cap B'C'$ and $F = AC \cap B'C'.$ The common region is the quadrilateral $ADEF,$ whose area is $[AB'F] - [EB'D].$

In triangle $AB'F,$ $\angle FAB' = 75^\circ - 15^\circ = 60^\circ$ and $\angle AB'F = 75^\circ,$ so $\angle AFB' = 45^\circ,$ and the law of sines gives $B'F = 20\sin 60^\circ/\sin 45^\circ$ $= 10\sqrt{6}.$ With $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},$ $\begin{aligned} [AB'F] &= \tfrac{1}{2} \cdot 20 \cdot 10\sqrt{6}\,\sin 75^\circ \\ &= 50(3 + \sqrt{3}). \end{aligned}$

In right triangle $ABD,$ $AD = 20\cos 15^\circ$ and $BD = 20\sin 15^\circ,$ so $[ABD] = 200\sin 15^\circ\cos 15^\circ$ $= 100\sin 30^\circ = 50,$ and $B'D = 20(1 - \cos 15^\circ).$ Triangles $EB'D$ and $ABD$ are similar (right angles at $D,$ and $\angle EB'D = \angle ABD = 75^\circ$ ), so, using $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},$ $\begin{aligned} [EB'D] &= 50\left(\frac{1 - \cos 15^\circ}{\sin 15^\circ}\right)^2 \\ &= 50 \\ &\quad {}\cdot \left(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}\right). \end{aligned}$ Therefore $\begin{aligned} [ADEF] &= 50(3 + \sqrt{3}) \\ &\quad {}- 50 \\ &{}\cdot (15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}) \\ &= 500\sqrt{2} - 350\sqrt{3} \\ &\quad {}+ 300\sqrt{6} - 600, \end{aligned}$ so $(p, q, r, s)$ $= (500, -350, 300, -600)$ and $\frac{p - q + r - s}{2} = \frac{1750}{2} = 875.$

2007 AIME I 第 12 题

答案：875

难度评级：3270

解答：

其他年份的第 12 题