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

Revision as of 21:35, 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}

@@ Line 7: / Line 7: @@
 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>.
+<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
+<math>\frac{\frac{\sqrt{2}}{36}}{\frac{\sqrt{2}}{9}}</math> is $\frac{1}{3}
 ==See also==

2005 AMC 10A (Problems • Answer Key • Resources)
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
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Difference between revisions of "2005 AMC 10A Problems/Problem 23"

Revision as of 21:35, 24 December 2008

Problem

Solution

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