2017 AMC 12B 第 8 题

Question

某个长方形的短边与长边之比，等于长边与对角线之比。这个长方形的短边与长边之比的平方是多少？

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?

\dfrac{\sqrt{3} - 1}{2}

\dfrac{1}{2}

\dfrac{\sqrt{5} - 1}{2}

\dfrac{\sqrt{2}}{2}

\dfrac{\sqrt{6} - 1}{2}

答案：C

知识点：二次方程勾股定理比与比例

难度评级：1440

小提示：

设短边为

x

，长边为

y

，则对角线为

\sqrt{x^2 + y^2}

；写出

\dfrac{x}{y} = \dfrac{y}{\sqrt{x^2+y^2}}

。

With short side

x

and long side

y,

the diagonal is

\sqrt{x^2 + y^2};

write

\dfrac{x}{y} = \dfrac{y}{\sqrt{x^2+y^2}}

大提示：

令

r = \dfrac{x^2}{y^2}

。将比例方程平方，得到

\dfrac{1}{r} = r + 1

。

Let

r = \dfrac{x^2}{y^2}.

Squaring the ratio equation gives

\dfrac{1}{r} = r + 1

解答：

设

x

和

y

分别为短边和长边，则对角线为

\sqrt{x^2 + y^2}

，并且

\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}

。令

r = \dfrac{x^2}{y^2}

，右边为

\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1}

，因此

r = \dfrac{1}{r+1}

，得

r^2 + r - 1 = 0

。正根为

r = \dfrac{\sqrt{5} - 1}{2}

。

所以正确答案是 C。

Let

x

and

y

be the short and long sides, so the diagonal is

\sqrt{x^2 + y^2}

and

\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}.

Writing

r = \dfrac{x^2}{y^2},

the right side is

\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1},

so

r = \dfrac{1}{r+1},

giving

r^2 + r - 1 = 0.

The positive root is

r = \dfrac{\sqrt{5} - 1}{2}.

Thus, the correct answer is C.

2017 AMC 12B 第 8 题

答案：C

难度评级：1440

小提示：

大提示：

解答：

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