Question

Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8,$ $PB=6,$ and $PC=10.$ To the nearest integer the area of triangle $ABC$ is:

$159$

$131$

$95$

$79$

$50$

Answer: D

Concepts:equilateral triangle transformation Pythagorean Triple law of cosines area

Difficulty rating: 2370

Small Hint:

Rotate $P$ by $60^\circ$ about $A$ so that $C$ maps to $B$

Big Hint:

The resulting $6$ - $8$ - $10$ triangle determines $\angle APB$

Solution:

Rotate $P$ by $60^\circ$ about $A$ to $P'.$ Since this rotation sends $C$ to $B,$ we have $P'B=PC=10,$ $PP'=PA=8,$ and $PB=6.$ Thus triangle $PP'B$ is right. Also triangle $APP'$ is equilateral, so $\angle APB=60^\circ+90^\circ=150^\circ.$

If the equilateral triangle has side $s,$ the law of cosines in triangle $APB$ gives $\begin{aligned} s^2 &=6^2+8^2-2(6)(8)\cos150^\circ\\ &=100+48\sqrt3. \end{aligned}$ Its area is $\frac{\sqrt3}{4}s^2=36+25\sqrt3,$ whose nearest integer is $79.$

Therefore, the correct answer is D.

1967 AMC 12 Problem 40

Answer: D

Difficulty rating: 2370

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years