AMC10 2023 A

AMC10 2023 A · Q11

AMC10 2023 A · Q11. It mainly tests Quadratic equations, Pythagorean theorem.

A square of area $2$ is inscribed in a square of area $3$, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?

一个面积为$2$的正方形内接在一个面积为$3$的正方形中，形成了四个全等的三角形，如下图所示。阴影直角三角形的短腿与长腿的比率为多少？

(A) \frac15 \frac15

(B) \frac14 \frac14

(D) \sqrt3-\sqrt2 \sqrt3-\sqrt2

(E) \sqrt2-1 \sqrt2-1

Answer

Correct choice: (C)

正确答案：（C）

Solution

The side lengths of the inner square and outer square are $\sqrt{2}$ and $\sqrt{3}$ respectively. Let the shorter side of our triangle be $x$, thus the longer leg is $\sqrt{3}-x$. Hence, by the Pythagorean Theorem, we have \begin{align*} (\sqrt{3}-x)^2+x^2&=(\sqrt{2})^2 \\ 3-2\sqrt{3}x+x^2+x^2&=2 \\ 2x^2-2\sqrt{3}x+1&=0. \end{align*} By the quadratic formula, we find that $x=\frac{\sqrt{3}\pm1}{2}$, so the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

内正方形的边长为$\sqrt{2}$，外正方形的边长为$\sqrt{3}$。设三角形的短边为$x$，则长腿为$\sqrt{3}-x$。因此，根据勾股定理，我们有 \begin{align*} (\sqrt{3}-x)^2+x^2&=(\sqrt{2})^2 \\ 3-2\sqrt{3}x+x^2+x^2&=2 \\ 2x^2-2\sqrt{3}x+1&=0. \end{align*} 使用二次公式，得到$x=\frac{\sqrt{3}\pm1}{2}$，因此答案为$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}$。

Topics