Difference between revisions of "2009 AIME I Problems/Problem 15"
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− | First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So <math>\angle{BAC} = 60^\circ{}</math>. Therefore, let <math>\angle{I_B AB} = \angle{I_B AD} = \alpha | + | First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So <math>\angle{BAC} = 60^\circ{}</math>. Therefore, let <math>\angle{I_B AB} = \angle{I_B AD} = \alpha</math> and <math>\angle{I_C AD} = \angle{I_C AC} = 30 - \alpha.</math> Therefore, in triangle <math>ABD</math>, we know that <math>\angle{BI_BD} = 90^\circ{} + \frac{\angle{BAD}}{2} = 90 + \alpha</math> and <math>\angle{CI_CD} = 90^\circ{}+ \frac{\angle{CAD}}{2} = 90^\circ{} + (30^\circ{} - \alpha) = 120^\circ{} - \alpha</math>. Now note that quadrilaterals <math>BI_BDP</math> and <math>CI_CDP</math> are both cyclic. This means that <math>\angle{BPD} = 180^\circ{} - \angle{BI_BD} = 90^\circ{} + \alpha</math> and <math>\angle{DPC} = 180^\circ{} - \angle{DI_CC} = 60^\circ + \alpha</math>. Therefore, <math>\angle{BPC} = 150^\circ{}</math>. |
+ | |||
+ | Now note that in order to maximize the area of <math>BPC</math>, we have to maximize the distance from <math>P</math> to line <math>BC</math>. | ||
(I will work on finishing solution maybe sometime later today...) | (I will work on finishing solution maybe sometime later today...) | ||
Revision as of 10:11, 9 August 2019
Contents
[hide]Problem
In triangle , , , and . Let be a point in the interior of . Let 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 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
First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So . Therefore, let and Therefore, in triangle , we know that and . Now note that quadrilaterals and are both cyclic. This means that and . Therefore, .
Now note that in order to maximize the area of , we have to maximize the distance from to line . (I will work on finishing solution maybe sometime later today...)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.