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2004 AMC 10A Problem 22

Below is the professionally curated solution for Problem 22 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:tangent linePythagorean Theorem

Difficulty rating: 1790

22.

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}?

2+52\dfrac{2 + \sqrt{5}}{2}

5\sqrt{5}

6\sqrt{6}

52\dfrac{5}{2}

5−55 - \sqrt{5}

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=2−xDE = 2 - x and DC=2,DC = 2, so (2−x)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.

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