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

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== Problem ==
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==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>.
 
  
== See also ==
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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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==Solution==
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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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Using law of sines on <math>\triangle ABC</math>, we can create the following equation:
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<math>\frac{24}{\sin(\angle ABC)}</math> <math>=</math> <math>\frac{BC}{\sin(\angle BAC)}</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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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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<math>\sin(75)</math> can be found through the sin addition formula.
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<math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</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>72\sqrt{3} + 216</math>
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<math>72 + 3 + 216 =</math> <math>\boxed{291}</math>
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==Solution 2 (no trig)==
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First, draw a good diagram.
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We realize that <math>\angle C = 75^\circ</math>, and <math>\angle CAT = 30^\circ</math>.  Therefore, <math>\angle CTA = 75^\circ</math> as well, making <math>\triangle CAT</math> an isosceles triangle.  <math>AT</math> and <math>AC</math> are congruent, so <math>AC=24</math>. We now drop an altitude from <math>C</math>, and call the foot this altitude point <math>D</math>. 
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<center><asy>
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size(200);
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defaultpen(linewidth(0.4)+fontsize(8));
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pair A,B,C,D,T,F;
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A = origin;
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T = scale(24)*dir(30);
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C = scale(24)*dir(60);
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B = extension(C,T,A,(1,0));
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F = foot(T,A,B);
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D = foot(C,A,B);
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draw(A--B--C--A--T, black+0.8);
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draw(C--D, dashed);
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label(rotate(degrees(T-A))*"$24$", A--T, N);
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label(rotate(degrees(C-A))*"$24$", A--C, 2*NW);
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label("$12\sqrt 3$", C--D, E);
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label("$12\sqrt 3$", D--B, S);
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label("$12$", A--D, S);
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pen p = fontsize(8)+red;
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MA("45^\circ", C,B,A,2);
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MA("30^\circ", B,A,T,2.5);
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MA("30^\circ", T,A,C,3.5);
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dot("$A$", A, SW);
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dot("$B$", B, SE);
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dot("$C$", C, N);
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dot("$T$", T, NE);
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dot("$D$", D, S);
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</asy></center>
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By 30-60-90 triangles, <math>AD=12</math> and <math>CD=12\sqrt{3}</math>. 
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We also notice that <math>\triangle CDB</math> is an isosceles right triangle.  <math>CD</math> is congruent to <math>BD</math>, which makes <math>BD=12\sqrt{3}</math>.  The base <math>AB</math> is <math>12+12\sqrt{3}</math>, and the altitude <math>CD=12\sqrt{3}</math>.  We can easily find that the area of triangle <math>ABC</math> is <math>216+72\sqrt{3}</math>, so <math>a+b+c=\boxed{291}</math>.
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-youyanli
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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}}
  
[[Category:Intermediate Geometry Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:56, 24 June 2021

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 $\angle C = 75^\circ$, and $\angle CAT = 30^\circ$. Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$.

[asy] size(200); defaultpen(linewidth(0.4)+fontsize(8));  pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*"$24$", A--T, N); label(rotate(degrees(C-A))*"$24$", A--C, 2*NW);  label("$12\sqrt 3$", C--D, E); label("$12\sqrt 3$", D--B, S); label("$12$", A--D, S); pen p = fontsize(8)+red; MA("45^\circ", C,B,A,2); MA("30^\circ", B,A,T,2.5); MA("30^\circ", T,A,C,3.5);  dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$T$", T, NE); dot("$D$", D, S); [/asy]

By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$.

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

-youyanli

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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