AMC12 2017 B

AMC12 2017 B · Q8

AMC12 2017 B · Q8. It mainly tests Quadratic equations, Triangles (properties).

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

某个矩形的短边与长边的比等于长边与对角线的比。这个矩形短边与长边的比的平方是多少？

(A) $\frac{\sqrt{3}-1}{2}$ $\frac{\sqrt{3}-1}{2}$

(B) $\frac{1}{2}$ $\frac{1}{2}$

(D) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$

(E) $\frac{\sqrt{6}-1}{2}$ $\frac{\sqrt{6}-1}{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $x$ be the length of the short side of the rectangle, and let $y$ be the length of the long side. Then the length of the diagonal is $\sqrt{x^2+y^2}$, and \[ \frac{x^2}{y^2}=\frac{y^2}{x^2+y^2},\quad \text{so}\quad \frac{y^2}{x^2}=\frac{x^2+y^2}{y^2}=\frac{x^2}{y^2}+1. \] Let $r=\frac{x^2}{y^2}$ be the requested squared ratio. Then $\frac{1}{r}=r+1$, so $r^2+r-1=0$. By the quadratic formula, the positive solution is \[ r=\frac{\sqrt{5}-1}{2}. \]

答案（C）：设 $x$ 为矩形短边的长度，设 $y$ 为矩形长边的长度。那么对角线的长度为 $\sqrt{x^2+y^2}$，并且 \[ \frac{x^2}{y^2}=\frac{y^2}{x^2+y^2},\quad \text{因此}\quad \frac{y^2}{x^2}=\frac{x^2+y^2}{y^2}=\frac{x^2}{y^2}+1。 \] 令 $r=\frac{x^2}{y^2}$ 为所求的平方比。那么 $\frac{1}{r}=r+1$，所以 $r^2+r-1=0$。由求根公式，正根为 \[ r=\frac{\sqrt{5}-1}{2}。 \]

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