2004 AMC 10A Problem 22

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}

Tangents from a point have equal length:

CF = CB

and

EF = EA

With

AE = x,

apply the Pythagorean theorem to

\triangle CDE:

(2 - x)^2 + 2^2 = (2 + x)^2

Let

F

be the point where

CE

touches the semicircle and let

x = AE.

Since tangents from a point are equal,

CF = CB = 2

and

EF = EA = x,

so

CE = 2 + x.

In right triangle

CDE,

we have

DE = 2 - x

and

DC = 2,

so

(2 - x)^2 + 2^2 = (2 + x)^2.

This gives

x = \dfrac{1}{2},

hence

CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}.

Thus, the correct answer is D.

Problem 22 in other years