2009 AIME I Problems/Problem 15
Problem
In triangle , , , and . Let be a point in the interior of . Let points and denote the incenters of triangles and , respectively. The circumcircles of triangles and meet at distinct points and . The maximum possible area of can be expressed in the form , where , , and are positive integers and is not divisible by the square of any prime. Find .
Solution 1
First, by the Law of Cosines, we have so .
Let and be the circumcenters of triangles and , respectively. We first compute Because and are half of and , respectively, the above expression can be simplified to Similarly, . As a result
Therefore is constant (). Also, is or when is or . Let point be on the same side of as with ; is on the circle with as the center and as the radius, which is . The shortest distance from to is .
When the area of is the maximum, the distance from to has to be the greatest. In this case, it's . The maximum area of is and the requested answer is .
Solution 2
From Law of Cosines on , Now, Since and are cyclic quadrilaterals, it follows that Next, applying Law of Cosines on , By AM-GM, , so Finally, and the maximum area would be so the answer is .
Solution 3
Proceed as in Solution 2 until you find . The locus of points that give is a fixed arc from to ( will move along this arc as moves along ) and we want to maximise the area of []. This means we want to be farthest distance away from as possible, so we put in the middle of the arc (making isosceles). We know that and , so . Let be the foot of the perpendicular from to line . Then the area of [] is the same as because base has length . We can split into two triangles and , with and . Then, the area of [] is equal to , and so the answer is .
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
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