Difference between revisions of "2009 AIME I Problems/Problem 15"
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== Solution 3 == | == Solution 3 == | ||
− | Proceed as in Solution 2 until you find <math>\angle CPB = 150</math>. The locus of points <math>P</math> that give <math>\angle | + | Proceed as in Solution 2 until you find <math>\angle CPB = 150</math>. The locus of points <math>P</math> that give <math>\angle CPB = 150</math> is a fixed arc from <math>B</math> to <math>C</math> (<math>P</math> will move along this arc as <math>D</math> moves along <math>BC</math>) and we want to maximise the area of [<math>\triangle BPC</math>]. This means we want <math>P</math> to be farthest distance away from <math>BC</math> as possible, so we put <math>P</math> in the middle of the arc (making <math>\triangle BPC</math> isosceles). We know that <math>BC=14</math> and <math>\angle CPB = 150</math>, so <math>\angle PBC = \angle PCB = 15</math>. Let <math>O</math> be the foot of the perpendicular from <math>P</math> to line <math>BC</math>. Then the area of [<math>\triangle BPC</math>] is the same as <math>7OP</math> because base <math>BC</math> has length <math>14</math>. We can split <math>\triangle BPC</math> into two <math>15-75-90</math> triangles <math>BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</math>] is equal to <math>7 \cdot OP=98-49\sqrt{3}</math>, and so the answer is <math>98+49+3=\boxed{150}</math>. |
== See also == | == See also == |
Revision as of 17:34, 20 August 2020
Contents
[hide]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) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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