Question

在 $\triangle RED$ 中， $RD = 1$ 、 $\angle DRE = 75^\circ$ 、 $\angle RED = 45^\circ$ 。令 $M$ 为线段 $\overline{RD}$ 的中点。点 $C$ 在边 $\overline{ED}$ 上，且 $\overline{RC} \perp \overline{EM}$ 。将线段 $\overline{DE}$ 经过 $E$ 延长到点 $A$ ，使 $CA = AR$ 。于是 $AE = \frac{a - \sqrt{b}}{c}$ ，其中 $a$ 和 $c$ 是互质正整数， $b$ 是正整数。求 $a + b + c$ 。

In $\triangle RED,$ $RD = 1,$ $\angle DRE = 75^\circ$ and $\angle RED = 45^\circ.$ Let $M$ be the midpoint of segment $\overline{RD}.$ Point $C$ lies on side $\overline{ED}$ such that $\overline{RC} \perp \overline{EM}.$ Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA = AR.$ Then $AE = \frac{a - \sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a + b + c.$

答案

答案：56

知识点：坐标几何正弦定理斜率

难度评级：3060

解答：

因为 $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ$ ，令 $D = (0,0)$ ，且 $E$ 在正 $x$ 轴上，则 $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ 。正弦定理给出 $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2}$ ，并且 $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ 。

直线 $EM$ 的斜率为 $\frac{\sqrt{3}/4}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}}$ ，所以直线 $RC$ 的斜率为 $\frac{1 + 2\sqrt{3}}{\sqrt{3}}$ 。从 $R$ 下降 $\frac{\sqrt{3}}{2}$ 到 $x$ 轴，水平向左移动 $\frac{3/2}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22}$ ，所以 $C = (c, 0)$ ，其中 $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}$ 。

对 $A = (t, 0)$ ，条件 $CA = AR$ 写作 $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}$ ，这是关于 $t$ 的一次方程： $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}$ 。因此 $\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}$ 所以 $a + b + c = 7 + 27 + 22 = 56$ 。

Since $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ,$ place $D = (0,0)$ with $E$ on the positive $x$ -axis, so $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The law of sines gives $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2},$ and $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).$

The slope of $EM$ is $\frac{\sqrt{3}/4}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}},$ so line $RC$ has slope $\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Descending from $R$ by $\frac{\sqrt{3}}{2}$ to the $x$ -axis moves us left by $\frac{3/2}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22},$ so $C = (c, 0)$ with $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.$

For $A = (t, 0),$ the condition $CA = AR$ reads $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4},$ which is linear in $t:$ $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}.$ Then $\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}$ so $a + b + c = 7 + 27 + 22 = 56.$

2014 AIME II 第 11 题

答案：56

难度评级：3060

解答：

其他年份的第 11 题