AMC12 2004 A

AMC12 2004 A · Q18

AMC12 2004 A · Q18. It mainly tests Pythagorean theorem, Circle theorems.

Square $ABCD$ has side length $2$. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

正方形 $ABCD$ 的边长为 $2$。在正方形内部作以 $\overline{AB}$ 为直径的半圆，从 $C$ 点作该半圆的切线，与边 $\overline{AD}$ 交于 $E$。求 $\overline{CE}$ 的长度。

(A) $2 + \sqrt{5}$ $2 + \sqrt{5}$

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

(D) $\dfrac{5}{2}$ $\dfrac{5}{2}$

(E) $5 - \sqrt{5}$ $5 - \sqrt{5}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the point of tangency be $F$. By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The Pythagorean Theorem on $\triangle CDE$ yields \begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*} Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

设切点为 $F$。由两切线定理 $BC = FC = 2$ 且 $AE = EF = x$。因此 $DE = 2-x$。在 $\triangle CDE$ 上用勾股定理得 \begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*} 因此 $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$。

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