Question

设 $ABCD$ 是等腰梯形，满足 $AD = BC$ 且 $AB \lt CD$ 。已知从 $A$ 到直线 $BC$ 、 $CD$ ，和 $BD$ 的距离分别为 $15$ 、 $18$ ，和 $10$ ，设 $K$ 为 $ABCD$ 的面积。求 $\sqrt{2} \cdot K$ 。

Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB \lt CD.$ Suppose that the distances from $A$ to the lines $BC,$ $CD,$ and $BD$ are $15,$ $18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt{2} \cdot K.$

答案

答案：567

知识点：梯形坐标几何距离公式

难度评级：2990

解答：

因为 $AB \parallel CD$ ，从 $A$ 到 $CD$ 的距离 $18$ 就是高。令 $D = (0, 0)$ 、 $C = (c, 0)$ 、 $A = (a, 18)$ 、 $B = (c - a, 18)$ ，并设 $u = AB = c - 2a \gt 0$ 点到直线距离公式给出所以 $324 + a^2 = \left(\frac{6u}{5}\right)^2$ ，且 $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2$ 。 $\begin{aligned} \frac{18u}{\sqrt{324 + a^2}} &= 15 \quad (\text{to } BC), \\ \small \frac{18u}{\sqrt{324 + (u + a)^2}} \\ &= 10 \quad (\text{to } BD), \end{aligned}$

两式相减，得 $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2$ ，因此 $u + 2a = \frac{9}{5}u$ ，即 $a = \frac{2}{5}u$ 。代回可得 $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25}$ ，所以 $u^2 = \frac{2025}{8}$ ， $u = \frac{45\sqrt{2}}{4}$ 。

因此 $CD = c = u + 2a = \frac{9}{5}u$ $= \frac{81\sqrt{2}}{4}$ ，并且所以 $\sqrt{2} \cdot K = 567$ 。 $\begin{aligned} K &= \frac{AB + CD}{2} \cdot 18 \\ &= 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) \\ &= \frac{567\sqrt{2}}{2}, \end{aligned}$

Since $AB \parallel CD,$ the distance $18$ from $A$ to $CD$ is the height. Put $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18),$ and let $u = AB = c - 2a \gt 0.$ The point-to-line distance formulas give $\begin{aligned} \frac{18u}{\sqrt{324 + a^2}} &= 15 \quad (\text{to } BC), \\ \small \frac{18u}{\sqrt{324 + (u + a)^2}} \\ &= 10 \quad (\text{to } BD), \end{aligned}$ so $324 + a^2 = \left(\frac{6u}{5}\right)^2$ and $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2.$

Subtracting, $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2,$ hence $u + 2a = \frac{9}{5}u$ and $a = \frac{2}{5}u.$ Substituting back, $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25},$ so $u^2 = \frac{2025}{8}$ and $u = \frac{45\sqrt{2}}{4}.$

Then $CD = c = u + 2a = \frac{9}{5}u$ $= \frac{81\sqrt{2}}{4},$ and $\begin{aligned} K &= \frac{AB + CD}{2} \cdot 18 \\ &= 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) \\ &= \frac{567\sqrt{2}}{2}, \end{aligned}$ so $\sqrt{2} \cdot K = 567.$

2021 AIME I 第 9 题

答案：567

难度评级：2990

解答：

其他年份的第 9 题