Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | ||
| − | <math> \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} </math> | + | <math> \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} </math>. The area of |
==Solution== | ==Solution== | ||
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> 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> = $\frac{\sqrt{2}}{36}. | |
| − | <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 <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2</math> = <math>\frac{\sqrt{2}}{3}</math> | ||
==See also== | ==See also== | ||
Revision as of 21:15, 24 December 2008
Problem
Let 
be a diameter of a circle and let 
be a point on with 
. Let 
and 
be points on the circle such that 
and 
is a second diameter. What is the ratio of the area of 
to the area of 
?
. The area of
Solution
http://img443.imageshack.us/img443/8034/circlenc1.png
is 
of diameter and 
is 
- = 
. 
is the radius of the circle, so using the Pythagorean theorem height 
is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. Unknown error_msg) = 
. Area of the 
is 
= $\frac{\sqrt{2}}{36}.
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 | ||
| All AMC 10 Problems and Solutions | ||
