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Difference between revisions of "2023 AMC 12B Problems/Problem 17"

(Created page with "==Solution== The length of the side opposite to the angle with <math>120^\circ</math> is longest. We denote its value as <math>x</math>. Because three side lengths form an...")
 
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Therefore, the area of the triangle is
 
Therefore, the area of the triangle is
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<math></math>
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<cmath>
 
\begin{align*}
 
\begin{align*}
 
\frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ
 
\frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ
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= \boxed{\textbf{(E) <math>15 \sqrt{3}</math>}} .
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= \boxed{\textbf{(E) } 15 \sqrt{3}} .
 
\end{align*}
 
\end{align*}
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<math></math>
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</cmath>
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 18:33, 15 November 2023

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 \begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) } 15 \sqrt{3}} . \end{align*}

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

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