Question

一个 $30$ - $60$ - $90^\circ$ 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 $30$ - $60$ - $90^\circ$ 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 $\dfrac{x}{x + y}?$

$\dfrac{3}{7}$

$\dfrac{\sqrt3}{4}$

$\dfrac{4}{9}$

$\dfrac{5}{11}$

$\dfrac{4\sqrt3}{15}$

答案：A

知识点：特殊直角三角形高线中线（几何）

难度评级：1660

解答：

令直角顶点 $C = (0,0)$ ，短边 $CB = 1$ ，其中 $B = (1, 0)$ ，长边 $CA = \sqrt3$ ，其中 $A = (0, \sqrt3)$ 。从 $C$ 到斜边 $AB$ 的高的垂足为 $H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$ ，且这条高在直线 $x = \sqrt3\,y$ 上。从 $A$ 到 $CB$ 中点 $\left(\tfrac12, 0\right)$ 的中线与这条高相交于 $\left(\tfrac37, \tfrac{\sqrt3}{7}\right)$ 。这把 $CH$ （长度 $\tfrac{\sqrt3}{2}$ ）分成 $\tfrac{4\sqrt3}{14}$ 和 $\tfrac{3\sqrt3}{14}$ ，所以 $x = \tfrac{3\sqrt3}{14}$ ，于是 $\dfrac{x}{x + y} = \dfrac{3\sqrt3/14}{\sqrt3/2} = \dfrac{3}{7}$ 。因此正确答案是 A。

Place the right angle at $C = (0,0),$ the short leg $CB = 1$ with $B = (1, 0),$ and the long leg $CA = \sqrt3$ with $A = (0, \sqrt3).$ The altitude from $C$ to hypotenuse $AB$ has foot $H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$ and runs along $x = \sqrt3\,y.$ The median from $A$ to the midpoint $\left(\tfrac12, 0\right)$ of $CB$ meets that altitude at $\left(\tfrac37, \tfrac{\sqrt3}{7}\right).$ This cuts $CH$ (length $\tfrac{\sqrt3}{2}$ ) into $\tfrac{4\sqrt3}{14}$ and $\tfrac{3\sqrt3}{14},$ so $x = \tfrac{3\sqrt3}{14}$ and $\dfrac{x}{x + y} = \dfrac{3\sqrt3/14}{\sqrt3/2} = \dfrac{3}{7}.$ Thus, A is the correct answer.

2025 AMC 10B 第 13 题

答案：A

难度评级：1660

解答：

其他年份的第 13 题