AMC10 2004 A

AMC10 2004 A · Q22

AMC10 2004 A · Q22. It mainly tests Linear equations, Pythagorean theorem.

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

正方形ABCD边长为2。以AB为直径在正方形内作半圆，从C引半圆的切线与边AD交于点E。求CE的长度。

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

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

(D) 5/2 $\frac{5}{2}$

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $F$ be the point at which $\overline{CE}$ is tangent to the semicircle, and let $G$ be the midpoint of $\overline{AB}$. Because $\overline{CF}$ and $\overline{CB}$ are both tangents to the semicircle, $CF=CB=2$. Similarly, $EA=EF$. Let $x=AE$. The Pythagorean Theorem applied to $\triangle CDE$ gives $$(2-x)^2+2^2=(2+x)^2.$$ It follows that $x=\frac12$ and $CE=2+x=\frac52.$

（D）设点 $F$ 为 $\overline{CE}$ 与半圆相切处的切点，设 $G$ 为 $\overline{AB}$ 的中点。因为 $\overline{CF}$ 和 $\overline{CB}$ 都是该半圆的切线，所以 $CF=CB=2$。同理，$EA=EF$。令 $x=AE$。对 $\triangle CDE$ 应用勾股定理得 $$(2-x)^2+2^2=(2+x)^2.$$ 因此 $x=\frac12$，且 $CE=2+x=\frac52$。

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