Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</math>, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see | But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</math>, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see | ||
− | <cmath>AE^2 | + | <cmath>AE^2 = s^2 + r^2 - 2sr\cos(135^\circ)</cmath> |
<cmath>AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)</cmath> | <cmath>AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)</cmath> | ||
<cmath>AE^2 = s^2 + r^2 + \sqrt{2}sr</cmath> | <cmath>AE^2 = s^2 + r^2 + \sqrt{2}sr</cmath> |
Revision as of 22:24, 17 March 2015
Problem
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
Solution 1
One possibility is to use the coordinate plane, setting at the origin. Point will be and will be since , and are collinear and contain a diagonal of . The Pythagorean theorem results in
This implies that and ; dividing gives us .
Solution 2
First note that angle is right since is tangent to the circle. Using the Pythagorean Theorem on , then, we see
But it can also be seen that . Therefore, since lies on , . Using the Law of Cosines on , we see
Thus, since and are rational, and . So , , and .
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 |
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