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2004 AMC 10A Problem 22

Problem 22 of 25HarderGeometry

Square ABCDABCD has side length 2.2. A semicircle with diameter AB\overline{AB} is constructed inside the square, and the tangent to the semicircle from CC intersects side AD\overline{AD} at E.E. What is the length of CE?\overline{CE}?

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Solution

Let FF be the point where CECE touches the semicircle and let x=AE.x = AE. Since tangents from a point are equal, CF=CB=2CF = CB = 2 and EF=EA=x,EF = EA = x, so CE=2+x.CE = 2 + x. In right triangle CDE,CDE, we have DE=2xDE = 2 - x and DC=2,DC = 2, so (2x)2+22=(2+x)2. (2 - x)^2 + 2^2 = (2 + x)^2. This gives x=12,x = \dfrac{1}{2}, hence CE=2+12=52.CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}. Thus, the correct answer is D.

More practice

Concepts: tangent line · Pythagorean Theorem

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