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

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http://img443.imageshack.us/img443/8034/circlenc1.png
 
http://img443.imageshack.us/img443/8034/circlenc1.png
  
<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> of <math>\triangle ADB</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2</math> = <math>\frac{\sqrt{2}}{3}</math>. Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. Hence the height of <math>\triangle DCE</math> is <math>\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}</math> = <math>\frac{\sqrt{2}}{9}</math>.
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<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> of <math>\triangle ADB</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2</math> = <math>\frac{\sqrt{2}}{3}</math>. Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. Hence the height of <math>\triangle DCE</math> is <math>\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}</math> = <math>\frac{\sqrt{2}}{9}</math>. The diameter is the base of both the triangles <math>\triangle DCE and \triangle ABD</math>. Hence, the ratio of area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is
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<math>\frac{\frac{\sqrt{2}}{36}}{\frac{\sqrt{2}}{9}}</math> is $\frac{1}{3}
  
 
==See also==
 
==See also==

Revision as of 22:35, 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 $\triangle ABD$?

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

Solution

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

$AC$ 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 $CD$ of $\triangle ADB$ is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. Unknown error_msg) = $\frac{\sqrt{2}}{3}$. Area of the $\triangle DCO$ is $\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}$ = $\frac{\sqrt{2}}{36}$. The height of $\triangle DCE$ can be found using the area of $\triangle DCO$ and $DO$ as base. Hence the height of $\triangle DCE$ is $\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}$ = $\frac{\sqrt{2}}{9}$. The diameter is the base of both the triangles $\triangle DCE and \triangle ABD$. Hence, the ratio of area of $\triangle DCE$ to the area of $\triangle ABD$ is $\frac{\frac{\sqrt{2}}{36}}{\frac{\sqrt{2}}{9}}$ is $\frac{1}{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