Question

假设 $\triangle ABC$ 的角满足 $\cos(3A) + \cos(3B)$ $+ \cos(3C) = 1$ 。该三角形的两条边长为 $10$ 和 $13$ 。存在正整数 $m$ ，使得 $\triangle ABC$ 剩余一边的最大可能长度为 $\sqrt{m}$ 。求 $m$ 。

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A) + \cos(3B)$ $+ \cos(3C) = 1.$ Two sides of the triangle have lengths $10$ and $13.$ There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

答案

答案：399

知识点：三角恒等式因式分解余弦定理

难度评级：2990

小提示：

写成 $1 - \cos 3A = 2\sin^2 \frac{3A}{2}$ ，再用和差化积处理 $\cos 3B + \cos 3C$ ；整个条件会分解

Write $1 - \cos 3A = 2\sin^2 \frac{3A}{2}$ and combine $\cos 3B + \cos 3C$ by sum-to-product; the whole condition factors

大提示：

必有一个角等于 $120^\circ$ 。当这个角夹在长为 $10$ 和 $13$ 的两边之间时，剩余边最长。

One angle must equal $120^\circ.$ The remaining side is longest when that angle lies between the sides of lengths $10$ and $13.$

解答：

使用 $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ 和 $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2}$ ，又因为 $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ ，所以 $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2}$ ，于是条件变为 $\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2} \end{aligned}\text{。}$

对三角形的任意角 $X$ ， $\frac{3X}{2}$ 严格位于 $0^\circ$ 和 $270^\circ$ 之间，所以 $\sin\frac{3X}{2} = 0$ 当且仅当 $X = 120^\circ$ 。因此该三角形有一个角为 $120^\circ$ 。

当 $120^\circ$ 角夹在长为 $10$ 和 $13$ 的两边之间时，剩余边最长（如果 $120^\circ$ 角对着其中一条已知边，则剩余边会比那条边短）。由余弦定理，该边长为 $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399}$ ，所以 $m = 399$ 。

Using $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ and $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2},$ together with $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ so that $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2},$ the condition becomes $\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2}. \end{aligned}$

For an angle $X$ of a triangle, $\frac{3X}{2}$ lies strictly between $0^\circ$ and $270^\circ,$ so $\sin\frac{3X}{2} = 0$ exactly when $X = 120^\circ.$ Hence one angle of the triangle is $120^\circ.$

The remaining side is longest when the $120^\circ$ angle sits between the sides of lengths $10$ and $13$ (if $120^\circ$ were opposite one of them, the remaining side would be shorter than that side). By the law of cosines its length is $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399},$ so $m = 399.$

2014 AIME II 第 12 题

答案：399

难度评级：2990

小提示：

大提示：

解答：

其他年份的第 12 题