2023 AMC 12B Problem 17

Attempt Problem 17 of the 2023 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

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17.

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?

12\sqrt{3}

8\sqrt{6}

14\sqrt{2}

20\sqrt{2}

15\sqrt{3}

Let the sides be

6,\ 6+d,\ 6+2d;

the

120^\circ

angle is opposite the longest side

Apply the law of cosines with

\cos 120^\circ=-\tfrac12

to find

d,

then area

=\tfrac12ab\sin 120^\circ

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.

Problem 17 in other years