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

Revision as of 19:32, 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)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 27: / Line 27: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==See Also==
+{{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
+{{MAA Notice}}

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

Revision as of 19:32, 15 November 2023

Solution

See Also