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

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</asy></center>
 
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Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math>AC = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}</math>. The area of  
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Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math>AD = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}</math>. The area of  
  
 
<cmath>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},</cmath>
 
<cmath>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},</cmath>
  
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and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>.
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and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 07:51, 15 March 2010

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

[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 (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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