Question

一个棱锥的底面是边长为 $20$ 、 $20$ 和 $24$ 的三角形。从底面三个顶点到棱锥第四个顶点的三条棱长都为 $25$ 。该棱锥的体积为 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 是正整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

A pyramid has a triangular base with side lengths $20,$ $20,$ and $24.$ The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25.$ The volume of the pyramid is $m\sqrt{n},$ 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：803

知识点：棱锥外接圆、外心与外接圆半径勾股定理体积

难度评级：2390

解答：

由于棱锥顶点到底面三个顶点等距，它在底面上的垂足是底面三角形的外心。底面是边长为 $20, 20, 24$ 的等腰三角形，到底边 $24$ 的高为 $\sqrt{20^2 - 12^2} = 16$ ，所以面积为 $K = \frac{1}{2} \cdot 24 \cdot 16 = 192$ ，外接圆半径为 $R = \frac{abc}{4K} = \frac{20 \cdot 20 \cdot 24}{4 \cdot 192} = \frac{25}{2}.$

棱锥的高为 $\sqrt{25^2 - \left(\frac{25}{2}\right)^2} = \frac{25\sqrt{3}}{2}$ ，所以体积为 $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$ 。因此 $m + n = 800 + 3 = 803$ 。

Since the apex is equidistant from all three base vertices, its foot is the circumcenter of the base. The base is isosceles with sides $20, 20, 24:$ its altitude to the side of length $24$ is $\sqrt{20^2 - 12^2} = 16,$ so its area is $K = \frac{1}{2} \cdot 24 \cdot 16 = 192,$ and its circumradius is $R = \frac{abc}{4K} = \frac{20 \cdot 20 \cdot 24}{4 \cdot 192} = \frac{25}{2}.$

The height of the pyramid is $\sqrt{25^2 - \left(\frac{25}{2}\right)^2} = \frac{25\sqrt{3}}{2},$ so the volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}.$ Then $m + n = 800 + 3 = 803.$

2017 AIME I 第 4 题

答案：803

难度评级：2390

解答：

其他年份的第 4 题