Question

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 $a + b + c.$

Answer

Solution:

Since $\angle A = 60^\circ$ and $\angle B = 45^\circ,$ we have $\angle C = 75^\circ.$ In triangle $ATC,$ angle $TAC = 30^\circ$ (half of angle $A$ ), so $\angle ATC = 180^\circ - 30^\circ - 75^\circ = 75^\circ.$ Thus triangle $ACT$ is isosceles with $AC = AT = 24.$

Drop the altitude $CH$ to $\overline{AB}.$ Triangle $ACH$ is $30$ - $60$ - $90,$ so $AH = 12$ and $CH = 12\sqrt{3}.$ Triangle $BCH$ is $45$ - $45$ - $90,$ so $BH = CH = 12\sqrt{3}.$

The area is $\frac{1}{2} \cdot CH \cdot AB = \frac{1}{2} \cdot 12\sqrt{3}\,(12 + 12\sqrt{3}) = 216 + 72\sqrt{3}.$ Then $a + b + c = 216 + 72 + 3 = 291.$

2001 AIME I Problem 4

Solution:

Problem 4 in Other Years