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

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== Problem ==
 
== Problem ==
 
In [[triangle]] <math>ABC</math>, angles <math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The [[angle bisector|bisector]] of angle <math>A</math> intersects <math>\overline{BC}</math> at <math>T</math>, and <math>AT=24</math>. The area of triangle <math>ABC</math> can be written in the form <math>a+b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>.
 
In [[triangle]] <math>ABC</math>, angles <math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The [[angle bisector|bisector]] of angle <math>A</math> intersects <math>\overline{BC}</math> at <math>T</math>, and <math>AT=24</math>. The area of triangle <math>ABC</math> can be written in the form <math>a+b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>.
 
== Solution 1==
 
<center><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></center>
 
 
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>
 
 
and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>.
 
  
 
== 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 $a+b+c$.

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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