Difference between revisions of "2013 AIME II Problems/Problem 10"
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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> | ||
− | The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math> | + | The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>. |
− | Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math> | + | Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get |
<math>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math> | <math>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math> |
Revision as of 00:07, 5 November 2016
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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