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2001 AIME I Problems/Problem 4

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Problem

In triangle ABC, angles A and B measure 60 degrees and 45 degrees, respectively. The bisector of angle A intersects \overline{BC} at T, and AT=24. The area of triangle ABC can be written in the form a+b\sqrt{c}, where a, b, and c are positive integers, and c is not divisible by the square of any prime. Find a+b+c.

Solution

size(180);pointpen = black; pathpen = black+linewidth(0.7);pair A=(0,0),B=(12+12*3^.5,0),C=(12,12*3^.5),D=foot(C,A,B),T=IP(CR...

Let D be the foot of the altitude from C to \overline{AB}. By simple angle-chasing, we find that \angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT, and thus AC = AT = 24. Now \triangle ADC is a 30-60-90 right triangle and BDC is a 45-45-90 right triangle, so AC = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}. The area of

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{...

and the answer is a+b+c = 216 + 72 + 3 = \boxed{291}.

See also

2001 AIME I (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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