Difference between revisions of "2001 AIME I Problems/Problem 4"
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+ | ==Problem== | ||
+ | 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 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== | ||
+ | |||
+ | After chasing angles, <math>\angle ATC=75^{\circ}</math> and <math>\angle TCA=75^{\circ}</math>, meaning <math>\triangle TAC</math> is an isosceles triangle and <math>AC=24</math>. | ||
+ | |||
+ | Using law of sines on <math>\triangle ABC</math>, we can create the following equation: | ||
+ | |||
+ | <math>\frac{24}{\sin(\angle ABC)}</math> <math>=</math> <math>\frac{BC}{\sin(\angle BAC)}</math> | ||
+ | |||
+ | <math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqrt{6}</math>. | ||
+ | |||
+ | We can then use the Law of Sines area formula <math>\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)</math> to find the area of the triangle. | ||
+ | |||
+ | <math>\sin(75)</math> can be found through the sin addition formula. | ||
+ | |||
+ | <math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> | ||
+ | |||
+ | Therefore, the area of the triangle is <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> <math>\cdot</math> <math>24</math> <math>\cdot</math> <math>12\sqrt{6}</math> <math>\cdot</math> <math>\frac{1}{2}</math> | ||
+ | |||
+ | <math>72\sqrt{3} + 216</math> | ||
+ | |||
+ | <math>72 + 3 + 216 =</math> <math>\boxed{291}</math> | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2001|n=I|num-b=3|num-a=5}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 15:14, 23 May 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
After chasing angles, and , meaning is an isosceles triangle and .
Using law of sines on , we can create the following equation:
and , so .
We can then use the Law of Sines area formula to find the area of the triangle.
can be found through the sin addition formula.
Therefore, the area of the triangle 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.