2025 AMC 12B 第 10 题

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。

\dfrac{x}{x+y}

是多少？

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

小提示：

把直角放在原点，两条直角边沿坐标轴，然后给每个关键点设坐标。

Place the right angle at the origin with the legs along the axes and coordinatize every point

大提示：

求到最短边的中线与斜边上的高的交点，再看这个交点如何分割高。

Find where the median to the shortest side crosses the altitude to the hypotenuse, then split the altitude at that point

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

小提示：

大提示：

解答：

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