2001 AIME I Problems/Problem 4

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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)

Firstly, 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 of it a point D on side AB. 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

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