2001 AIME I Problems/Problem 4

Revision as of 23:00, 15 February 2021 by Xiaoxiong12345 (talk | contribs) (Solution 2 (no trig))


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 $a+b+c$.


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 $AC=24$.

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 $a+b+c=\boxed{291}$.


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