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2001 AIME I Problems/Problem 4

Revision as of 11:46, 18 August 2017 by Xiej (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 $a+b+c = 216 + 72 + 3 = \boxed{291}$.

Solution 2

Since $\angle CAB$ has a measure of $60^{\circ}$, and thus has sines and cosines that are easy to compute, we attempt to find $AC$ and $AB$, and use the formula that

$[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A$

By angle chasing, we find that $ACT$ is a triangle with $\angle A = 30^{\circ}, \angle C = 75^{\circ}$ and $\angle T = 75^{\circ}$. Thus $AC = AT = 24$.

Switching to the lower triangle $ATB$, $\angle A = 30^{\circ}, \angle T = 105^{\circ}$, and $\angle B = 45^{\circ}$, with $AT = 24$.

Using the Law of Sines on $ATB$:

$\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}$

$24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB$

$24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB$

$24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB$

$24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB$

$AB = 12 \cdot (\sqrt{3} + 1)$

We now plug in $AC$, $AB$ and $\sin \angle A$ into the formula for the area:

$[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A$

$[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}$

$[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)$

$[ABC] = 216 + 72\sqrt{3}$

Thus the answer is $216 + 72 + 3 = \boxed{291}$

Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from $T$ to $AB$, forming a $30-60-90$ and a $45-45-90$ triangle.

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

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