1967 AMC 12 Problem 40

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

Rotate

P

by

60^\circ

about

A

so that

C

maps to

B

The resulting

6

-

8

-

10

triangle determines

\angle APB

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.

Problem 40 in other years