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

m (See also)
Line 1: Line 1:
 +
==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 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==
 +
 +
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>.
 +
 +
Using law of sines on <math>\triangle ABC</math>, we can create the following equation:
 +
 +
<math>\frac{24}{\sin(\angle ABC)}</math> <math>=</math> <math>\frac{BC}{\sin(\angle BAC)}</math>
 +
 +
<math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqrt{6}</math>.
 +
 +
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.
 +
 +
<math>\sin(75)</math> can be found through the sin addition formula.
 +
 +
<math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math>
 +
 +
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>
 +
 +
<math>72\sqrt{3} + 216</math>
 +
 +
<math>72 + 3 + 216 =</math> <math>\boxed{291}</math>
 +
 +
==See also==
 +
{{AIME box|year=2001|n=I|num-b=3|num-a=5}}
 +
 +
{{MAA Notice}}

Revision as of 15:14, 23 May 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 $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$

$72\sqrt{3} + 216$

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS