2013 AIME II Problems/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 .
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
Also, the distance between and is
So the area
Then the maximum value of is
So the answer is .
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:
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
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 .
in the case It is possible – if we rotate such triangle, we can find position when lies on email@example.com, vvsss
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