Difference between revisions of "2001 AIME I Problems/Problem 4"
(solution + asy) |
Kimmystar94 (talk | contribs) m (→Solution) |
||
Line 11: | Line 11: | ||
</asy></center> | </asy></center> | ||
− | Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math> | + | Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math>AD = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}</math>. The area of |
<cmath>ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},</cmath> | <cmath>ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},</cmath> | ||
− | 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>. |
== See also == | == See also == |
Revision as of 07:51, 15 March 2010
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
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 .
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 |