Question

在一个等角五边形中，边长平方和等于 $308$ ，对角线长度平方和等于 $800$ 。该五边形周长的平方可表示为 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 为正整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

In an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be expressed as $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

答案

答案：681

知识点：等角多边形向量三角学代数变形

难度评级：3270

解答：

在等角五边形中，每条边的方向转过外角 $72^\circ$ ，所以边向量为 $s_k u_k$ ，其中 $k = 1, \ldots, 5$ ， $u_k = (\cos 72k^\circ, \sin 72k^\circ)$ ，且 $\sum_k s_k u_k = 0$ 。记 $Q = \sum s_k^2 = 308$ 、 $P_1 = \sum_{k} s_k s_{k+1}$ 、 $P_2 = \sum_k s_k s_{k+2}$ （下标循环）。每条对角线都是两个相邻边向量之和，所以其平方为 $s_{k+1}^2 + s_{k+2}^2 + 2 s_{k+1} s_{k+2} \cos 72^\circ$ ，把五条相加得因而 $800 = 2Q + 2\cos 72^\circ \, P_1,$ $\begin{aligned} &2\cos 72^\circ \, P_1 = 800 - 616 \\ &= 184. \end{aligned}$

展开 $\left|\sum_k s_k u_k\right|^2 = 0$ ， $u_k$ 与 $u_{k+1}$ 的夹角为 $72^\circ$ ， $u_k$ 与 $u_{k+2}$ 的夹角为 $144^\circ$ ：所以 $2\cos 36^\circ \, P_2 = 492$ 。利用 $\cos 72^\circ = \frac{\sqrt{5} - 1}{4}$ 和 $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$ ，得到 $2P_1 = \frac{184}{\cos 72^\circ} = 184\left(\sqrt{5} + 1\right)$ ，且 $2P_2 = \frac{492}{\cos 36^\circ} = 492\left(\sqrt{5} - 1\right)$ 。 $\begin{aligned} &0 = Q + 2\cos 72^\circ \, P_1 \\ &\quad {}+ 2\cos 144^\circ \, P_2 \\ &= 308 + 184 - 2\cos 36^\circ \, P_2, \end{aligned}$

周长的平方为因此 $m + n = 676 + 5 = 681$ 。 $\begin{aligned} &\left(\sum s_k\right)^2 = Q + 2P_1 + 2P_2 \\ &= 308 + 184\sqrt{5} + 184 \\ &\quad {}+ 492\sqrt{5} - 492 \\ &= 676\sqrt{5}. \end{aligned}$

In an equiangular pentagon each side direction turns by the exterior angle $72^\circ,$ so the sides are the vectors $s_k u_k$ for $k = 1, \ldots, 5,$ where $u_k = (\cos 72k^\circ, \sin 72k^\circ)$ and $\sum_k s_k u_k = 0.$ Write $Q = \sum s_k^2 = 308,$ $P_1 = \sum_{k} s_k s_{k+1},$ and $P_2 = \sum_k s_k s_{k+2}$ (indices cyclic). Each diagonal is a sum of two consecutive side vectors, so its square is $s_{k+1}^2 + s_{k+2}^2 + 2 s_{k+1} s_{k+2} \cos 72^\circ,$ and summing all five gives $800 = 2Q + 2\cos 72^\circ \, P_1,$ so $\begin{aligned} &2\cos 72^\circ \, P_1 = 800 - 616 \\ &= 184. \end{aligned}$

Expanding $\left|\sum_k s_k u_k\right|^2 = 0,$ the angle between $u_k$ and $u_{k+1}$ is $72^\circ$ and between $u_k$ and $u_{k+2}$ is $144^\circ:$ $\begin{aligned} &0 = Q + 2\cos 72^\circ \, P_1 \\ &\quad {}+ 2\cos 144^\circ \, P_2 \\ &= 308 + 184 - 2\cos 36^\circ \, P_2, \end{aligned}$ so $2\cos 36^\circ \, P_2 = 492.$ Using $\cos 72^\circ = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4},$ we get $2P_1 = \frac{184}{\cos 72^\circ} = 184\left(\sqrt{5} + 1\right)$ and $2P_2 = \frac{492}{\cos 36^\circ} = 492\left(\sqrt{5} - 1\right).$

The square of the perimeter is $\begin{aligned} &\left(\sum s_k\right)^2 = Q + 2P_1 + 2P_2 \\ &= 308 + 184\sqrt{5} + 184 \\ &\quad {}+ 492\sqrt{5} - 492 \\ &= 676\sqrt{5}. \end{aligned}$ Therefore $m + n = 676 + 5 = 681.$

2026 AIME I 第 14 题

答案：681

难度评级：3270

解答：

其他年份的第 14 题