Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2023 AMC 12B Problems/Problem 17: Difference between revisions

Newer edit →
Professorchenedu (talk | contribs)
editor
175 edits
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..."
 
Professorchenedu (talk | contribs)
editor
175 edits
Line 19: Line 19:


Therefore, the area of the triangle is
Therefore, the area of the triangle is
<math></math>
<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
= \boxed{\textbf{(E) <math>15 \sqrt{3}</math>}} .
= \boxed{\textbf{(E) } 15 \sqrt{3}} .
\end{align*}
\end{align*}
<math></math>
</cmath>


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

Revision as of 17:33, 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)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_17&oldid=203370"