Difference between revisions of "2001 AIME I Problems/Problem 4"

m (Solution 2)
m (Solution 2)
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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>.
 
==Solution 2==
 
Since <math>\angle CAB</math> has a measure of <math>60^{\circ}</math>, and thus has sines and cosines that are easy to compute, we attempt to find <math>AC</math> and <math>AB</math>, and use the formula that
 
 
<math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
 
 
By angle chasing, we find that <math>ACT</math> is a triangle with <math>\angle A = 30^{\circ}, \angle C = 75^{\circ}</math> and <math>\angle T = 75^{\circ}</math>.  Thus <math>AC = AT = 24</math>.
 
 
Switching to the lower triangle <math>ATB</math>, <math>\angle A = 30^{\circ}, \angle T =  105^{\circ}</math>, and <math>\angle B = 45^{\circ}</math>, with <math>AT = 24</math>.
 
 
Using the [[Law of Sines]] on <math>ATB</math>:
 
 
<math>\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}</math>
 
 
<math>24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB</math>
 
 
<math>24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB</math>
 
 
<math>24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB</math>
 
 
<math>24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB</math>
 
 
<math>AB = 12 \cdot (\sqrt{3} + 1)</math>
 
 
We now plug in <math>AC</math>, <math>AB</math> and <math>\sin \angle A</math> into the formula for the area:
 
 
<math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
 
 
<math>[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}</math>
 
 
<math>[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)</math>
 
 
<math>[ABC] = 216 + 72\sqrt{3}</math>
 
 
Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math>
 
 
Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from <math>T</math> to <math>AB</math>, forming a <math>30-60-90</math> and a <math>45-45-90</math> triangle.
 
  
 
== See also ==
 
== See also ==

Revision as of 16:27, 29 April 2018

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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