2005 AMC 10A Problems/Problem 23

Problem

Let $AB$ be a diameter of a circle and let $C$ be a point on $AB$ with $2\cdot AC=BC$ . Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$ ?

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

Solution

http://img443.imageshack.us/img443/8034/circlenc1.png

This problem needs a solution. If you have a solution for it, please help us out by adding it. $AC$ is \frac{1}{3} of diameter and $CO$ is \frac{1}{2}\-\frac{1}{3}\=\frac{1}{6}. $OD$ is the radius of the circle, so using the Pythagorean theorem height $CD$ is \sqrt{(\frac{1}{2}\)^2-(\frac{1}{6}\)^2)=\frac(\sqrt{2}\){3}\

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2005 AMC 10A Problems/Problem 23

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