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>. | ||
− | ==Solution== | + | ==Solution 1== |
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math> | Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math> | ||
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So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math> | So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math> | ||
− | Also, the distance between <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 ares <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math> | So the ares <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>. | ||
+ | |||
+ | <cmath>\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}</cmath> | ||
+ | |||
+ | Therefore, to maximize area of <math>\triangle BKL</math>, we need to maximize area of <math>\triangle OKL</math>. | ||
+ | |||
+ | <cmath>\triangle OKL = \frac12 r^2 \sin{\angle KOL}</cmath> | ||
+ | |||
+ | So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>. | ||
+ | |||
+ | 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 | ||
==See Also== | ==See Also== | ||
{{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 08:56, 31 January 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 ares
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 . So the answer is
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 |
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