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# Difference between revisions of "1967 AHSME Problems/Problem 40"

## Problem

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:

$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

## Solution

$[asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label("A",(0,10),N); label("B",(-9.5,-5.2),N); label("C",(9.5,-5.2),N); dot((-3,0)); label("P",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5)); dot((-9,7.5)); label("P'",(-9.2,7.5),N); draw((-9,7.5)--(0,10)); draw((-9,7.5)--(-8.66,-5)); draw((-9,7.5)--(-3,0)); [/asy]$

Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle.

Rotate $\triangle APC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$, so $$\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.$$

Therefore, $\triangle APP'$ is equilateral, so $P'P=8$, which means $\angle P'PB = 90^{\circ}.$

Let $\angle BP'P = \alpha .$ Notice that $\cos\alpha = \frac{8}{10}=\frac{4}{5},$ and $\sin\alpha = \frac{3}{5}.$

Applying the Law of Cosines to $\triangle APB$, \begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}

We want to find the area of $\triangle ABC$', which is $$AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{(D) 79}.$$

~pfalcon