Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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<cmath>9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.</cmath> | <cmath>9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.</cmath> | ||
− | Thus, since <math>r</math> and <math>s</math> are rational, <math>s^2 = 9</math> and <math>sr = 5</math>. So <math>s = 3</math> | + | Thus, since <math>r</math> and <math>s</math> are rational, <math>s^2 = 9</math> and <math>sr = 5</math>. So <math>s = 3</math>, <math>r = \frac{5}{3}</math>, and <math>\frac{r}{s} = \frac{5}{9}</math>. |
== See also == | == See also == |
Revision as of 19:59, 8 December 2013
Contents
[hide]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 draw the line . Since is tangent to the circle, is a right angle. Using the Pythagorean Theorem on , then, we see
But it can also be seen that since lies on . Therefore, . Using the Law of Cosines on , we see
\[AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)\] (Error compiling LaTeX. Unknown error_msg)
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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