2004 AMC 12A Problems/Problem 18

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(Redirected from 2004 AMC 10A Problems/Problem 22)

The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.

Problem

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 $\overline{CE}$ ?

$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$

Solution

Solution 1

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

$\begin{eqnarray*}(2-x)^2 + 2^2 &=& (2+x)^2\\x^2 - 4x + 8 &=& x^2 + 4x + 4\\x &=& \frac{1}{2}\end{eqnarray*}$

Hence $CE = FC + x = \frac{5}{2} \Rightarrow \mathrm{(D)}$ .

Solution 2

Clearly, $EA = EF = BG$ . Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$ , $CE = 5/2$ .

2004 AMC 12A (Problems)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

2004 AMC 10A (Problems)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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