AMC10 2025 B

AMC10 2025 B · Q13

AMC10 2025 B · Q13. It mainly tests Triangles (properties), Pythagorean theorem.

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

一个$30^\circ{-}60^\circ{-}90^\circ$三角形的斜边的高被到最短边中点的中线分成两段$x<y$。求$\tfrac{x}{x+y}$的比率。

(A) \dfrac{3}{7} \dfrac{3}{7}

(B) \dfrac{\sqrt3}{4} \dfrac{\sqrt3}{4}

(D) \dfrac{5}{11} \dfrac{5}{11}

(E) \dfrac{4\sqrt3}{15} \dfrac{4\sqrt3}{15}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle BDC$ are both $30^\circ{-}60^\circ{-}90^\circ$ triangles. From the side-length ratio, we get $AD=1$ and $DC=3.$ We obtain the following diagram: From here, we will proceed with mass points. Throughout this solution, we will use $W_P$ to denote the weight of point $P.$ Let $W_C=1.$ Since $3AD=DC,$ it follows that $W_A=3$ and $W_D=W_C+W_A=4.$ Since $AE=EB$ and $W_A=3,$ it follows that $W_B=3.$ Now we focus on $\overline{BD}:$ Since $W_B=3$ and $W_D=4,$ we have $\frac{DF}{FB}=\frac xy=\frac34.$ Therefore, the answer is \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]

不失一般性，令$\triangle ABC$的边长$AB=2, BC=2\sqrt{3},$ 和$AC=4.$ 令$D$为从$B$到$\overline{AC}$的垂足，$E$为$\overline{AB}$的中点，$F$为$\overline{CE}$与$\overline{BD}$的交点。注意$\triangle ADB$和$\triangle BDC$都是$30^\circ{-}60^\circ{-}90^\circ$三角形。从边长比率，得$AD=1$和$DC=3$。我们得到如下图：从这里，我们使用质点法。在整个解法中，用$W_P$表示点$P$的重量。令$W_C=1.$ 因为$3AD=DC$，故$W_A=3$和$W_D=W_C+W_A=4.$ 因为$AE=EB$且$W_A=3$，故$W_B=3$。现在关注$\overline{BD}$：因为$W_B=3$和$W_D=4$，我们有$\frac{DF}{FB}=\frac xy=\frac34.$ 因此，答案是\[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]

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