Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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− | <math>AC</math> is \frac{1}{3} of diameter and <math>CO</math> is \frac{1}{2}\-\frac{1}{3}\=\frac{1}{6}. | + | <math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is \frac{1}{2}\-\frac{1}{3}\=\frac{1}{6}. |
<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 21:57, 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 ?
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. is of diameter and is \frac{1}{2}\-\frac{1}{3}\=\frac{1}{6}. is the radius of the circle, so using the Pythagorean theorem height is \sqrt{(\frac{1}{2}\)^2-(\frac{1}{6}\)^2)=\frac(\sqrt{2}\){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 | ||
All AMC 10 Problems and Solutions |