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

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== Problem ==
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==Problem==
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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>.
 
  
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== Solution 1==
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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 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>.
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<center><asy> size(180);
 
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pointpen = black; pathpen = black+linewidth(0.7);
 
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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);
 
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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));
 
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MP("24",(A+3*T)/4,SE);
 
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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));
 
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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>AD = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}</math>. The area of
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==Solution==
  
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<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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After chasing angles, <math>\angle ATC=75^{\circ}</math> and <math>\angle TCA=75^{\circ}</math>, meaning <math>\triangle TAC</math> is an isosceles triangle and <math>AC=24</math>.
  
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and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>.
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Using law of sines on <math>\triangle ABC</math>, we can create the following equation:
  
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==Solution 2==
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<math>\frac{24}{\sin(\angle ABC)}</math> <math>=</math> <math>\frac{BC}{\sin(\angle BAC)}</math>
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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
 
  
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<math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
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<math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqrt{6}</math>.
  
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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>.
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We can then use the Law of Sines area formula <math>\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)</math> to find the area of the triangle.
  
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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>.
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<math>\sin(75)</math> can be found through the sin addition formula.
  
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Using the [[Law of Sines]] on <math>ATB</math>:
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<math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math>
  
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<math>\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}</math>
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Therefore, the area of the triangle is <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> <math>\cdot</math> <math>24</math> <math>\cdot</math> <math>12\sqrt{6}</math> <math>\cdot</math> <math>\frac{1}{2}</math>
  
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<math>24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB</math>
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<math>72\sqrt{3} + 216</math>
  
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<math>24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB</math>
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<math>72 + 3 + 216 =</math> <math>\boxed{291}</math>
  
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<math>24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB</math>
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==Solution 2 (no trig)==
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First, draw a good diagram.
  
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<math>24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB</math>
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We realize that the measure of angle C is 75 degrees.  The measure of angle CAT is 30 degrees.  Therefore, the measure of angle CTA must also be 75 degrees, making CAT an isosceles triangle.  AT and AC are congruent, so AC must equal 24.
  
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<math>AB = 12 \cdot (\sqrt{3} + 1)</math>
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We now drop an altitude from C, and call the foot this altitude point D.  By 30-60-90 triangles, AD equals twelve and CD equals <math>12\sqrt{3}</math>. 
  
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We now plug in <math>AC</math>, <math>AB</math> and <math>\sin \angle A</math> into the formula for the area:
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We also notice that CDB is an isosceles right triangle.  CD is congruent to BD, which makes BD also equal to <math>12\sqrt{3}</math>.  The base AB is 12+<math>12\sqrt{3}</math>, and the altitude CD is <math>12\sqrt{3}</math>.  We can easily find that the area triangle ABC is 216+<math>72\sqrt{3}</math>, so a+b+c=<math>\boxed{291}</math>.
  
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<math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
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-youyanli
  
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<math>[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}</math>
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==See also==
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<math>[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)</math>
 
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<math>[ABC] = 216 + 72\sqrt{3}</math>
 
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Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math>
 
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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.
 
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== See also ==
 
 
{{AIME box|year=2001|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2001|n=I|num-b=3|num-a=5}}
  
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[[Category:Intermediate Geometry Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:18, 5 December 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

After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$, meaning $\triangle TAC$ is an isosceles triangle and $AC=24$.

Using law of sines on $\triangle ABC$, we can create the following equation:

$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$

$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$, so $BC = 12\sqrt{6}$.

We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.

$\sin(75)$ can be found through the sin addition formula.

$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$

Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$

$72\sqrt{3} + 216$

$72 + 3 + 216 =$ $\boxed{291}$

Solution 2 (no trig)

First, draw a good diagram.

We realize that the measure of angle C is 75 degrees. The measure of angle CAT is 30 degrees. Therefore, the measure of angle CTA must also be 75 degrees, making CAT an isosceles triangle. AT and AC are congruent, so AC must equal 24.

We now drop an altitude from C, and call the foot this altitude point D. By 30-60-90 triangles, AD equals twelve and CD equals $12\sqrt{3}$.

We also notice that CDB is an isosceles right triangle. CD is congruent to BD, which makes BD also equal to $12\sqrt{3}$. The base AB is 12+$12\sqrt{3}$, and the altitude CD is $12\sqrt{3}$. We can easily find that the area triangle ABC is 216+$72\sqrt{3}$, so a+b+c=$\boxed{291}$.

-youyanli

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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