2001 AIME I Problems/Problem 4

Revision as of 17:27, 29 April 2018 by Johnting1021 (talk | contribs) (Solution 2)

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 1

[asy] 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(A,24),B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(D(MP("T",T,NE))--A); D(MP("D",D)--C,linetype("6 6") + linewidth(0.7));  MP("24",(A+3*T)/4,SE); D(anglemark(C,B,A,65)); D(anglemark(B,A,C,65)); D(rightanglemark(C,D,B,50)); MP("30^{\circ}",A,(4,1)); MP("45^{\circ}",B,(-3,1)); [/asy]

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 $AD = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}$. 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 $a+b+c = 216 + 72 + 3 = \boxed{291}$.

See also

2001 AIME I (ProblemsAnswer KeyResources)
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

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