Difference between revisions of "2005 AMC 10A Problems/Problem 23"

(Solution)
Line 8: Line 8:
  
 
{{solution}}
 
{{solution}}
<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}\-\frac{1}{3}\=\frac{1}{6}</math>.
+
<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}</math> - <math>\frac{1}{3}</math> = <math>\frac{1}{6}</math>.
 
<math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> is \sqrt{(\frac{1}{2}\)^2-(\frac{1}{6}\)^2)=\frac(\sqrt{2}\){3}\
 
<math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> is \sqrt{(\frac{1}{2}\)^2-(\frac{1}{6}\)^2)=\frac(\sqrt{2}\){3}\
  

Revision as of 22:01, 24 December 2008

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}\

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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
All AMC 10 Problems and Solutions
Invalid username
Login to AoPS