Question

设 $A, B, C$ 是一个三角形的三个角，其中 $A$ 和 $C$ 为锐角， $B$ 为钝角，并满足以及存在正整数 $p$ 、 $q$ 、 $r$ 、 $s$ ，使得其中 $p + q$ 与 $s$ 互质，且 $r$ 不被任何质数平方整除。求 $p + q + r + s$ 。 $\begin{aligned} &\cos^2 A + \cos^2 B \\ &\quad {}+ 2 \sin A \sin B \cos C = \frac{15}{8} \end{aligned}$ $\begin{aligned} &\cos^2 B + \cos^2 C \\ &\quad {}+ 2 \sin B \sin C \cos A = \frac{14}{9}. \end{aligned}$ $\begin{aligned} &\cos^2 C + \cos^2 A \\ &\quad {}+ 2 \sin C \sin A \cos B \\ &= \frac{p - q\sqrt{r}}{s}, \end{aligned}$

Let $A, B, C$ be angles of a triangle with $A$ and $C$ acute and $B$ greater than a right angle satisfying $\begin{aligned} &\cos^2 A + \cos^2 B \\ &\quad {}+ 2 \sin A \sin B \cos C = \frac{15}{8} \end{aligned}$ and $\begin{aligned} &\cos^2 B + \cos^2 C \\ &\quad {}+ 2 \sin B \sin C \cos A = \frac{14}{9}. \end{aligned}$ There are positive integers $p,$ $q,$ $r,$ and $s$ for which $\begin{aligned} &\cos^2 C + \cos^2 A \\ &\quad {}+ 2 \sin C \sin A \cos B \\ &= \frac{p - q\sqrt{r}}{s}, \end{aligned}$ where $p + q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p + q + r + s.$

答案

答案：222

知识点：正弦定理余弦定理三角恒等式

难度评级：3370

解答：

把每个 $\cos^2$ 替换为 $1 - \sin^2$ ，第一个方程变为 $\sin^2 A + \sin^2 B$ $- 2 \sin A \sin B \cos C = \frac{1}{8}$ 。由正弦定理， $\sin A = \frac{a}{2R}$ ，其他角同理，所以左边等于其中最后一步用余弦定理。因此 $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}$ 。同样的论证把第二个方程化为 $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9}$ ，并说明所求表达式等于 $2 - \sin^2 B$ 。 $\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}$

因为 $A$ 和 $C$ 是锐角， $\cos A = \frac{\sqrt{5}}{3}$ ， $\cos C = \frac{\sqrt{14}}{4}$ ，且 $\sin A = \frac{2}{3}$ ， $\sin C = \frac{\sqrt{2}}{4}$ 。于是所以 $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}$ 。 $\begin{aligned} \sin B &= \sin(A + C) \\ &= \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} \\ &= \frac{2\sqrt{14} + \sqrt{10}}{12}, \end{aligned}$

因此 $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72}$ ，且 $p + q + r + s$ $= 111 + 4 + 35 + 72$ $= 222$ 。

Replacing each $\cos^2$ by $1 - \sin^2,$ the first equation becomes $\sin^2 A + \sin^2 B$ $- 2 \sin A \sin B \cos C = \frac{1}{8}.$ By the law of sines, $\sin A = \frac{a}{2R}$ and so on, so the left side equals $\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}$ by the law of cosines. Hence $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}.$ The same argument turns the second equation into $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9},$ and shows the requested expression equals $2 - \sin^2 B.$

Since $A$ and $C$ are acute, $\cos A = \frac{\sqrt{5}}{3}$ and $\cos C = \frac{\sqrt{14}}{4},$ with $\sin A = \frac{2}{3}$ and $\sin C = \frac{\sqrt{2}}{4}.$ Then $\begin{aligned} \sin B &= \sin(A + C) \\ &= \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} \\ &= \frac{2\sqrt{14} + \sqrt{10}}{12}, \end{aligned}$ so $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}.$

Therefore $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72},$ and $p + q + r + s$ $= 111 + 4 + 35 + 72$ $= 222.$

2013 AIME II 第 15 题

答案：222

难度评级：3370

解答：

其他年份的第 15 题