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 and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>.   and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>. 
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−  ==Solution 2==
 
−  Since <math>\angle CAB</math> has a measure of <math>60^{\circ}</math>, and thus has sines and cosines that are easy to compute, we attempt to find <math>AC</math> and <math>AB</math>, and use the formula that
 
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−  <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
 
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−  By angle chasing, we find that <math>ACT</math> is a triangle with <math>\angle A = 30^{\circ}, \angle C = 75^{\circ}</math> and <math>\angle T = 75^{\circ}</math>. Thus <math>AC = AT = 24</math>.
 
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−  Switching to the lower triangle <math>ATB</math>, <math>\angle A = 30^{\circ}, \angle T = 105^{\circ}</math>, and <math>\angle B = 45^{\circ}</math>, with <math>AT = 24</math>.
 
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−  Using the [[Law of Sines]] on <math>ATB</math>:
 
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−  <math>\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}</math>
 
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−  <math>24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB</math>
 
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−  <math>24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB</math>
 
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−  <math>24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB</math>
 
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−  <math>24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB</math>
 
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−  <math>AB = 12 \cdot (\sqrt{3} + 1)</math>
 
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−  We now plug in <math>AC</math>, <math>AB</math> and <math>\sin \angle A</math> into the formula for the area:
 
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−  <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
 
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−  <math>[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}</math>
 
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−  <math>[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)</math>
 
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−  <math>[ABC] = 216 + 72\sqrt{3}</math>
 
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−  Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math>
 
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−  Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from <math>T</math> to <math>AB</math>, forming a <math>306090</math> and a <math>454590</math> triangle.
 
   
 == See also ==   == See also == 
Revision as of 16:27, 29 April 2018
Problem
In triangle , angles and measure degrees and degrees, respectively. The bisector of angle intersects at , and . The area of triangle can be written in the form , where , , and are positive integers, and is not divisible by the square of any prime. Find .
Solution 1
Let be the foot of the altitude from to . By simple anglechasing, we find that , and thus . Now is a right triangle and is a right triangle, so . The area of
and the answer is .
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.