2023 AMC 12B Problems/Problem 17

Solution

The length of the side opposite to the angle with $120^\circ$ is longest. We denote its value as $x$ .

Because three side lengths form an arithmetic sequence, the middle-valued side length is $\frac{x + 6}{2}$ .

Following from the law of cosines, we have $\begin{align*} 6^2 + \left( \frac{x + 6}{2} \right)^2 - 2 \cdot 6 \cdot \frac{x + 6}{2} \cdot \cos 120^\circ = x^2 . \end{align*}$

By solving this equation, we get $x = 14$ . Thus, $\frac{x + 6}{2} = 10$ .

Therefore, the area of the triangle is $$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) $15 \sqrt{3}$ }} . \end{align*} $$ (Error compiling LaTeX. Unknown error_msg)

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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2023 AMC 12B Problems/Problem 17

Solution