2004 AMC 12A Problem 18

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}

Two tangent segments from a common external point to a circle are equal

With

x = AE,

the tangent lengths give

CE = 2 + x,

then apply the Pythagorean theorem to

\triangle CDE

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.

Problem 18 in other years