2013 AIME II Problems/Problem 10
Contents
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 (Coordbash)
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/
Solution 4
Let les on such that , call We call By similar triangle, we have . Then, we realize the area is just As . Now, we have to maximize , which is obviously reached when , the answer is leads to
~bluesoul
Solution 5
Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. is the base of triangles and const The maximum possible area for and are at the same position of point .
has sides
in the case It is possible – if we rotate such triangle, we can find position when lies on vladimir.shelomovskii@gmail.com, vvsss
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 | ||
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