Question

在三角形 $ABC$ 中，角 $B$ 与角 $C$ 相等。点 $P$ 和 $Q$ 分别在 $\overline{AC}$ 和 $\overline{AB}$ 上，且 $AP = PQ = QB = BC$ 。角 $ACB$ 的大小是角 $APQ$ 的 $r$ 倍，其中 $r$ 是正实数。求不超过 $1000r$ 的最大整数。

In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r.$

答案

答案：571

知识点：等腰三角形正弦定理三角恒等式

难度评级：2990

小提示：

设 $\angle A = \alpha$ ，并令 $AP = 1$ 。等腰三角形 $APQ$ 给出 $\angle AQP = \alpha$ 和 $AQ = 2\cos\alpha$ 。

Let $\angle A = \alpha$ and $AP = 1.$ Isosceles triangle $APQ$ gives $\angle AQP = \alpha$ and $AQ = 2\cos\alpha$

大提示：

因为 $BC = 1$ ，正弦定理给出 $AB = \frac{1}{2\sin(\frac{\alpha}{2})}$ ，而 $AQ + QB = AB$ 化为 $\sin\frac{3\alpha}{2} = \frac{1}{2}$ 。

With $BC = 1,$ the law of sines gives $AB = \frac{1}{2\sin(\frac{\alpha}{2})},$ and $AQ + QB = AB$ reduces to $\sin\frac{3\alpha}{2} = \frac{1}{2}$

解答：

设 $\angle A = \alpha$ ，并按比例缩放使 $AP = PQ = QB = BC = 1$ 。在三角形 $APQ$ 中， $AP = PQ$ ，所以 $\angle AQP = \angle A = \alpha$ ，从而 $\angle APQ = 180^\circ - 2\alpha$ ，且 $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha$ 。在三角形 $ABC$ 中， $\angle B = \angle C = 90^\circ - \frac{\alpha}{2}$ ，所以 $\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\frac{\alpha}{2})}{\sin\alpha} = \frac{1}{2\sin(\frac{\alpha}{2})} \end{aligned}\text{。}$

由 $AQ + QB = AB$ ， $\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\frac{\alpha}{2})} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1 \end{aligned}\text{。}$ 由积化和差公式， $4\sin\frac{\alpha}{2}\cos\alpha$ $= 2\sin\frac{3\alpha}{2}$ $- 2\sin\frac{\alpha}{2}$ ，所以方程化为 $\sin\frac{3\alpha}{2} = \frac{1}{2}$ 。于是 $\alpha = 20^\circ$ 或 $\alpha = 100^\circ$ ，但后者会使 $AQ = 2\cos\alpha$ 为负，故 $\alpha = 20^\circ$ 。

此时 $\angle ACB = 80^\circ$ ， $\angle APQ = 140^\circ$ ，所以 $r = \frac{80}{140} = \frac{4}{7}$ 。于是 $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571$ 。

Let $\angle A = \alpha,$ and scale so $AP = PQ = QB = BC = 1.$ In triangle $APQ,$ the equal sides $AP = PQ$ give $\angle AQP = \angle A = \alpha,$ so $\angle APQ = 180^\circ - 2\alpha$ and, by the law of sines, $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.$ In triangle $ABC,$ $\angle B = \angle C = 90^\circ - \frac{\alpha}{2},$ so $\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\frac{\alpha}{2})}{\sin\alpha} = \frac{1}{2\sin(\frac{\alpha}{2})}. \end{aligned}$

Since $AQ + QB = AB,$ $\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\frac{\alpha}{2})} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1. \end{aligned}$ By the product-to-sum identity, $4\sin\frac{\alpha}{2}\cos\alpha$ $= 2\sin\frac{3\alpha}{2}$ $- 2\sin\frac{\alpha}{2},$ so the equation collapses to $\sin\frac{3\alpha}{2} = \frac{1}{2}.$ Then $\alpha = 20^\circ$ or $\alpha = 100^\circ,$ but the latter makes $AQ = 2\cos\alpha$ negative, so $\alpha = 20^\circ.$

Now $\angle ACB = 80^\circ$ and $\angle APQ = 140^\circ,$ so $r = \frac{80}{140} = \frac{4}{7},$ and $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.$

2000 AIME I 第 14 题

答案：571

难度评级：2990

小提示：

大提示：

解答：

其他年份的第 14 题