Difference between revisions of "2013 AIME II Problems/Problem 10"
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Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | ||
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So the answer is <math>104+26+13+3=\boxed{146}</math>. | So the answer is <math>104+26+13+3=\boxed{146}</math>. | ||
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+ | ==Solution 3 (simpler solution)== | ||
+ | A rather easier solution is presented in the Girls' Angle WordPress: | ||
+ | |||
+ | http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/ | ||
==See Also== | ==See Also== | ||
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{{AIME box|year=2013|n=II|num-b=9|num-a=11}} | {{AIME box|year=2013|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:42, 4 September 2021
Problem
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 .
Solution 3 (simpler solution)
A rather easier solution is presented in the Girls' Angle WordPress:
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.