Difference between revisions of "2001 AIME I Problems/Problem 4"
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In [[triangle]] <math>ABC</math>, angles <math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The [[angle bisector|bisector]] of angle <math>A</math> intersects <math>\overline{BC}</math> at <math>T</math>, and <math>AT=24</math>. The area of triangle <math>ABC</math> can be written in the form <math>a+b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. | In [[triangle]] <math>ABC</math>, angles <math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The [[angle bisector|bisector]] of angle <math>A</math> intersects <math>\overline{BC}</math> at <math>T</math>, and <math>AT=24</math>. The area of triangle <math>ABC</math> can be written in the form <math>a+b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. | ||
− | == Solution == | + | == Solution 1== |
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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>. | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Using the [[Law of Sines]] on <math>ATB</math>: | ||
+ | |||
+ | <math>\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}</math> | ||
+ | |||
+ | <math>24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB</math> | ||
+ | |||
+ | <math>24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB</math> | ||
+ | |||
+ | <math>24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB</math> | ||
+ | |||
+ | <math>24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB</math> | ||
+ | |||
+ | <math>AB = 12 \cdot (\sqrt{3} + 1)</math> | ||
+ | |||
+ | We now plug in <math>AC</math>, <math>AB</math> and <math>\sin \angle A</math> into the formula for the area: | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math> | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}</math> | ||
+ | |||
+ | <math>[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)</math> | ||
+ | |||
+ | <math>[ABC] = 216 + 72\sqrt{3}</math> | ||
+ | |||
+ | Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math> | ||
== See also == | == See also == |
Revision as of 22:02, 27 June 2011
Contents
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 angle-chasing, we find that , and thus . Now is a right triangle and is a right triangle, so . The area of
and the answer is .
Solution 2
Since has a measure of , and thus has sines and cosines that are easy to compute, we attempt to find and , and use the formula that
By angle chasing, we find that is a triangle with and . Thus .
Switching to the lower triangle , , and , with .
Using the Law of Sines on :
We now plug in , and into the formula for the area:
Thus the answer is
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |