Difference between revisions of "2013 AIME II Problems/Problem 10"
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Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math> | Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math> | ||
− | So the | + | So the area <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math> |
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math> | Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math> |
Revision as of 13:52, 2 March 2014
Contents
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
Now we put the figure in the Cartesian plane, let the center of the circle , then , and
The equation for Circle O is , and let the slope of the line be , then the equation for line is
Then we get , according to Vieta's Formulas, we get
, and
So,
Also, the distance between and is
So the area
Then the maximum value of is
So the answer is .
Solution 2
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) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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