AMC10 2017 A

AMC10 2017 A · Q21

AMC10 2017 A · Q21. It mainly tests Triangles (properties), Similarity.

A square with side length x is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length y is inscribed in another right triangle with sides of length 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle. What is $\frac{x}{y}$?

一个边长为 $x$ 的正方形内接于一个边长为 3、4、5 的直角三角形中，使得正方形的一个顶点与三角形的直角顶点重合。另一个边长为 $y$ 的正方形内接于另一个边长为 3、4、5 的直角三角形中，使得正方形的一条边位于三角形的斜边上。求 $\frac{x}{y}$？

(A) $\frac{12}{13}$ $\frac{12}{13}$

(B) $\frac{35}{37}$ $\frac{35}{37}$

(D) $\frac{37}{35}$ $\frac{37}{35}$

(E) $\frac{13}{12}$ $\frac{13}{12}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): In the first figure $\triangle FEB \sim \triangle DCE$, so $\frac{x}{3-x}=\frac{4-x}{x}$ and $x=\frac{12}{7}$. In the second figure, the small triangles are similar to the large one, so the lengths of the portions of the side of length $3$ are as shown. Solving $\frac{3}{5}y+\frac{5}{4}y=3$ yields $y=\frac{60}{37}$. Thus $\frac{x}{y}=\frac{12}{7}\cdot\frac{37}{60}=\frac{37}{35}$.

答案（D）：在第一幅图中，$\triangle FEB \sim \triangle DCE$，因此 $\frac{x}{3-x}=\frac{4-x}{x}$，且 $x=\frac{12}{7}$。在第二幅图中，小三角形与大三角形相似，因此边长为 $3$ 的那条边被分成的各段长度如图所示。解方程 $\frac{3}{5}y+\frac{5}{4}y=3$ 得 $y=\frac{60}{37}$。因此 $\frac{x}{y}=\frac{12}{7}\cdot\frac{37}{60}=\frac{37}{35}$。

Topics