Question

一个 $30\text{-}60\text{-}90$ 直角三角形斜边上的高，被到最短边的中线分成长度为 $x \lt y$ 的两段。求 $\dfrac{x}{x+y}$ ？

The altitude to the hypotenuse of a $30\text{-}60\text{-}90$ right triangle is divided into two segments of lengths $x \lt y$ by the median to the shortest side of the triangle. What is the ratio $\dfrac{x}{x+y}?$

$\dfrac{3}{7}$

$\dfrac{\sqrt{3}}{4}$

$\dfrac{4}{9}$

$\dfrac{5}{11}$

$\dfrac{4\sqrt{3}}{15}$

答案：A

知识点：特殊直角三角形坐标几何中线（几何）

难度评级：1580

解答：

取 $C = (0,0)$ 、 $A = (\sqrt{3}, 0)$ 、 $B = (0, 1)$ ，则 $AB$ 是斜边， $BC$ 是最短边。从 $C$ 到 $AB$ 的高交点为 $H = \left(\tfrac{\sqrt{3}}{4}, \tfrac{3}{4}\right)$ 。从 $A$ 到最短边中点 $M = \left(0, \tfrac{1}{2}\right)$ 的中线与高交于 $P = \left(\tfrac{\sqrt{3}}{7}, \tfrac{3}{7}\right)$ 。于是 $CP = \tfrac{2\sqrt{3}}{7} = \tfrac{4\sqrt{3}}{14}$ ， $PH = \tfrac{3\sqrt{3}}{14}$ ，所以 $x = \tfrac{3\sqrt{3}}{14}$ ， $\dfrac{x}{x+y} = \dfrac{3}{7}$ 。

所以正确答案是 A。

Take $C = (0,0),$ $A = (\sqrt{3}, 0),$ $B = (0, 1),$ so $AB$ is the hypotenuse and $BC$ is the shortest side. The altitude from $C$ meets $AB$ at $H = \left(\tfrac{\sqrt{3}}{4}, \tfrac{3}{4}\right).$ The median from $A$ to $M = \left(0, \tfrac{1}{2}\right)$ crosses the altitude at $P = \left(\tfrac{\sqrt{3}}{7}, \tfrac{3}{7}\right).$ This splits the altitude into $CP = \tfrac{2\sqrt{3}}{7} = \tfrac{4\sqrt{3}}{14}$ and $PH = \tfrac{3\sqrt{3}}{14},$ so $x = \tfrac{3\sqrt{3}}{14}$ and $\dfrac{x}{x+y} = \dfrac{3}{7}.$

Thus, the correct answer is A.

2025 AMC 12B 第 10 题

答案：A

难度评级：1580

解答：

其他年份的第 10 题