2023 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebra Geometry

Triangle

ABC

has side lengths in arithmetic progression, and the smallest side has length

6.

If the triangle has an angle of

120^\circ,

what is the area of

ABC?

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Solution

Let the sides be

6,\ 6+d,\ 6+2d.

The

120^\circ

angle faces the longest side, so

(6+2d)^2=6^2+(6+d)^2

-2\cdot 6\cdot(6+d)\cos 120^\circ.

Using

\cos 120^\circ=-\tfrac12

gives

3d^2+6d-72=0,

d=4

and the sides are

6,10,14.

The area is

\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ

=30\cdot\tfrac{\sqrt3}{2}

=15\sqrt3.

Thus, the correct answer is E.