AoPSWiki
Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!
Personal tools

2004 AMC 12A Problems/Problem 18

From AoPSWiki

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}?

Image:AMC10_2004A_22.png

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

Contents

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{eqna...

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

Solution 2

Image:2004_AMC12A-18.png

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.

See also

2004 AMC 12A (ProblemsResources)
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 (ProblemsResources)
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
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us