AMC12 2017 A

AMC12 2017 A · Q19

AMC12 2017 A · Q19. 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 $\dfrac{x}{3-x}=\dfrac{4-x}{x}$ and $x=\dfrac{12}{7}$. 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 $\dfrac{3}{5}y+\dfrac{5}{4}y=3$ yields $y=\dfrac{60}{37}$. Thus $\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}$.

答案（D）：在第一个图中，$\triangle FEB \sim \triangle DCE$，所以 $\dfrac{x}{3-x}=\dfrac{4-x}{x}$，并且 $x=\dfrac{12}{7}$。在第二个图中，小三角形与大三角形相似，因此长度为 $3$ 的边被分成的各段长度如图所示。解方程 $\dfrac{3}{5}y+\dfrac{5}{4}y=3$ 得 $y=\dfrac{60}{37}$。因此 $\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}$。

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