AMC12 2012 A

AMC12 2012 A · Q12

AMC12 2012 A · Q12. It mainly tests Quadratic equations, Circle theorems.

A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square?

正方形区域 $ABCD$ 在边 $CD$ 上的点 $(0,1)$ 处与圆 $x^2+y^2=1$ 外切。顶点 $A$ 和 $B$ 在圆 $x^2+y^2=4$ 上。求该正方形的边长。

(A) $\dfrac{\sqrt{10} + \sqrt{5}}{10}$ $\dfrac{\sqrt{10} + \sqrt{5}}{10}$

(B) $\dfrac{2\sqrt{5}}{5}$ $\dfrac{2\sqrt{5}}{5}$

(D) $\dfrac{2\sqrt{19} - 4}{5}$ $\dfrac{2\sqrt{19} - 4}{5}$

(E) $\dfrac{9 - \sqrt{17}}{5}$ $\dfrac{9 - \sqrt{17}}{5}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem. \begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*} Use the quadratic formula. \[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]

两个圆的半径分别为 $1$ 和 $2$。画出上图所示的三角形，并用 $s$（正方形边长）表示该三角形的边。因为 $AO$ 是大圆的半径，等于 $2$，可用勾股定理： \begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*} 用二次公式： \[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]

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