Question

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 $\overline{CE}?$

$\dfrac{2 + \sqrt{5}}{2}$

$\sqrt{5}$

$\sqrt{6}$

$\dfrac{5}{2}$

$5 - \sqrt{5}$

Solution:

Let $F$ be the point where $CE$ touches the semicircle. Since $CB$ and $CF$ are both tangents from $C,$ we have $CF = CB = 2.$ Similarly, with $x = AE,$ the tangents from $E$ give $EF = EA = x.$

Thus $CE = CF + FE = 2 + x.$ In right triangle $CDE,$ where $CD = 2$ and $DE = 2 - x,$ $(2 - x)^2 + 2^2 = (2 + x)^2.$

Expanding gives $8x = 4,$ so $x = \tfrac12$ and $CE = 2 + \tfrac12 = \tfrac52.$

Thus, the correct answer is D.

2004 AMC 12A Problem 18

Solution:

Problem 18 in Other Years