Difference between revisions of "2009 AIME I Problems/Problem 15"
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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>. | 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 |
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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>. | 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>. |
Revision as of 18:27, 2 January 2023
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 .
Diagram
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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