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

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

$ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},$

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

2001 AIME I (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2001 AIME I Problems/Problem 4

From AoPSWiki

Problem

Solution

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