Question

一个等腰梯形有内切圆，且该圆与四条边都相切。圆的半径为 $3$ ，梯形面积为 $72$ 。设梯形的两条平行边长为 $r$ 和 $s$ ，且 $r \ne s$ 。求 $r^2 + s^2$ 。

An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3,$ and the area of the trapezoid is $72.$ Let the parallel sides of the trapezoid have lengths $r$ and $s,$ with $r \ne s.$ Find $r^2 + s^2.$

答案

答案：504

知识点：梯形内切圆、内心与内切圆半径勾股定理

难度评级：2230

解答：

圆同时与两条平行边相切，所以梯形高为 $2 \cdot 3 = 6$ 。由面积得 $\frac{r + s}{2} \cdot 6 = 72$ ，所以 $r + s = 24$ 。由 Pitot 定理，两条腰的长度和也为 $24$ ，又因为梯形等腰，所以每条腰为 $12$ 。

从较短底边的一个端点作垂线，腰是一条直角三角形的斜边，两条直角边为 $6$ 和 $\frac{|r - s|}{2}$ ：所以 $(r - s)^2 = 432$ 。因此 $r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2}$ $= \frac{576 + 432}{2} = 504$ 。 $144 = 36 + \left(\frac{r - s}{2}\right)^2,$

The circle is tangent to both parallel sides, so the height of the trapezoid is $2 \cdot 3 = 6.$ From the area, $\frac{r + s}{2} \cdot 6 = 72,$ so $r + s = 24.$ By the Pitot theorem the legs together also sum to $24,$ and since the trapezoid is isosceles each leg is $12.$

Dropping a perpendicular from an endpoint of the shorter base, the leg is the hypotenuse of a right triangle with legs $6$ and $\frac{|r - s|}{2}:$ $144 = 36 + \left(\frac{r - s}{2}\right)^2,$ so $(r - s)^2 = 432.$ Therefore $r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2}$ $= \frac{576 + 432}{2} = 504.$

2025 AIME I 第 6 题

答案：504

难度评级：2230

解答：

其他年份的第 6 题