Difference between revisions of "2013 AIME II Problems/Problem 10"
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So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>. | So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>. | ||
− | Eventually, we get <cmath>\triangle BKL= | + | Eventually, we get <cmath>\triangle BKL= \frac12 \cdot (\sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath> |
So the answer is <math>104+26+13+3=\boxed{146}</math>. | So the answer is <math>104+26+13+3=\boxed{146}</math>. |
Revision as of 20:35, 4 June 2020
Problem 10
Given a circle of radius , let be a point at a distance from the center of the circle. Let be the point on the circle nearest to point . A line passing through the point intersects the circle at points and . The maximum possible area for can be written in the form , where , , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Draw perpendicular to at . Draw perpendicular to at .
Therefore, to maximize area of , we need to maximize area of .
So when area of is maximized, .
Eventually, we get
So the answer is .
See Also
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
2013 AIME II (Problems • Answer Key • Resources) | ||
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