Difference between revisions of "2009 AIME I Problems/Problem 15"
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== Solution 1 == | == Solution 1 == | ||
− | First, by [[Law of Cosines]], we have <cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},</cmath> so <math>\angle BAC = 60^\circ</math>. | + | First, by the [[Law of Cosines]], we have <cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},</cmath> so <math>\angle BAC = 60^\circ</math>. |
Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>BI_BD</math> and <math>CI_CD</math>, respectively. We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> Because <math>\angle BDI_B</math> and <math>\angle I_BBD</math> are half of <math>\angle BDA</math> and <math>\angle ABD</math>, respectively, the above expression can be simplified to <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.</cmath> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>. As a result <cmath> | Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>BI_BD</math> and <math>CI_CD</math>, respectively. We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> Because <math>\angle BDI_B</math> and <math>\angle I_BBD</math> are half of <math>\angle BDA</math> and <math>\angle ABD</math>, respectively, the above expression can be simplified to <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.</cmath> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>. As a result <cmath> | ||
− | Therefore <math>\angle CPB</math> is constant (<math>150^\circ</math>). Also, <math>P</math> is <math>B</math> or <math>C</math> when <math>D</math> is <math>B</math> or <math>C</math>. Let point <math>L</math> be on the same side of <math>\overline{BC}</math> as <math>A</math> with <math> | + | Therefore <math>\angle CPB</math> is constant (<math>150^\circ</math>). Also, <math>P</math> is <math>B</math> or <math>C</math> when <math>D</math> is <math>B</math> or <math>C</math>. Let point <math>L</math> be on the same side of <math>\overline{BC}</math> as <math>A</math> with <math>LC = LB = BC = 14</math>; <math>P</math> is on the circle with <math>L</math> as the center and <math>\overline{LC}</math> as the radius, which is <math>14</math>. The shortest distance from <math>L</math> to <math>\overline{BC}</math> is <math>7\sqrt {3}</math>. |
When the area of <math>\triangle BPC</math> is the maximum, the distance from <math>P</math> to <math>\overline{BC}</math> has to be the greatest. In this case, it's <math>14 - 7\sqrt {3}</math>. The maximum area of <math>\triangle BPC</math> is | When the area of <math>\triangle BPC</math> is the maximum, the distance from <math>P</math> to <math>\overline{BC}</math> has to be the greatest. In this case, it's <math>14 - 7\sqrt {3}</math>. The maximum area of <math>\triangle BPC</math> is |
Revision as of 18:42, 16 July 2020
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 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 .
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.