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

\begin{aligned} &\cos^2 A + \cos^2 B \\ &\quad {}+ 2 \sin A \sin B \cos C = \frac{15}{8} \end{aligned}

以及

\begin{aligned} &\cos^2 B + \cos^2 C \\ &\quad {}+ 2 \sin B \sin C \cos A = \frac{14}{9} \end{aligned}\text{。}

存在正整数

p

、

q

、

r

、

s

，使得

\begin{aligned} &\cos^2 C + \cos^2 A \\ &\quad {}+ 2 \sin C \sin A \cos B \\ &= \frac{p - q\sqrt{r}}{s} \end{aligned}\text{，}

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

。

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

小提示：

把第一个方程改写为

\sin^2 A + \sin^2 B

- 2\sin A \sin B \cos C = \frac{1}{8}

；正弦定理和余弦定理会把左边变成

\sin^2 C

。

Rewrite the first equation as

\sin^2 A + \sin^2 B

- 2\sin A \sin B \cos C = \frac{1}{8};

the laws of sines and cosines turn the left side into

\sin^2 C

大提示：

因此

\sin^2 C = \frac{1}{8}

，且

\sin^2 A = \frac{4}{9}

，而所求量为

2 - \sin^2 B

，其中

\sin B = \sin(A + C)

。

So

\sin^2 C = \frac{1}{8}

and

\sin^2 A = \frac{4}{9},

and the requested quantity is

2 - \sin^2 B

with

\sin B = \sin(A + C)

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

\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}\text{，}

其中最后一步用到了余弦定理。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

。

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

\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}\text{，}

所以

\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}

。

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

小提示：

大提示：

解答：

其他年份的第 15 题