Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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− | One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(s, 0)</math> and <math>E</math> will be <math>\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)</math> since <math>B, D</math>, and <math>E</math> are [[collinear]] and | + | One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(s, 0)</math> and <math>E</math> will be <math>\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)</math> since <math>B, D</math>, and <math>E</math> are [[collinear]] and contain the diagonal of <math>ABCD</math>. The [[Pythagorean theorem]] results in |
<cmath>AF^2 + EF^2 = AE^2</cmath> | <cmath>AF^2 + EF^2 = AE^2</cmath> |
Revision as of 12:42, 30 October 2007
Problem
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Square has side length , a circle centered at has radius , and and are both rational. The circle passes through , and lies on . Point lies on the circle, on the same side of as . Segment is tangent to the circle, and . What is ?
Solution
One possibility is to use the coordinate plane, setting at the origin. Point will be and will be since , and are collinear and contain the diagonal of . The Pythagorean theorem results in
This implies that and ; dividing gives us .
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |