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2023 AMC 12B Problems/Problem 17

Revision as of 22:35, 15 November 2023 by Prof joker (talk | contribs)

Problem

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 $ABC$?

$\textbf{(A) }12\sqrt{3}\qquad\textbf{(B) }8\sqrt{6}\qquad\textbf{(C) }14\sqrt{2}\qquad\textbf{(D) }20\sqrt{2}\qquad\textbf{(E) }15\sqrt{3}$

Solution 1

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)

Solution 2 (Analytic Geometry)

Since the triangle's longest side must correspond to the $120^\circ$ angle, the triangle is unique. By analytic geometry, we construct the following plot.

PJ 2023 12B Q17.png

We know the coordinates of point $A$ being the origin and $B$ being $(6,0)$. Constructing the line which point $C$ can lay on, here since $\angle B=120^\circ$, $C$ is on the line \[y=\sqrt{3}\left(x-6\right).\]

I denote $D$ as the perpendicular line from $C$ to $AB$, and assume $CD=k$. Here we know $\triangle BCD$ is a $30^\circ-60^\circ-90-^\circ$ triangle. Hence $DC=\sqrt{3}k$ and $BC=2k$.

Furthermore, due to the arithmetic progression, we know $AC=4k-6$. Hence, in $\triangle ACD$, \[\left(4k-6\right)^{2}=\left(6+k\right)^{2}+3k^{2},\] \[k=5.\]

Thus, the area is equal to $\frac{1}{2}\cdot 6\cdot \sqrt{3} k=\boxed{\textbf{(E) } 15 \sqrt{3}}$.

~Prof. Joker

Solution 3

Let the side lengths be $6$, $6+d$, and $6+2d$. As $6+2d$ is the longest side, the angle opposite to it will be $120^{\circ}$.

By the law of Cosine \[(6+2d)^2 = 6^2 + (6+d)^2 - 2 \cdot 6 \cdot (6+d) \cdot \cos 120^{\circ}\] \[4d^2 + 24d + 36 = 36 + 36 + 12 d + d^2 + 36 + 6d\] \[3d^2 + 6d - 72 = 0\] \[d^2 + 2d - 24 = 0\] \[(d+6)(d-4)=0\]

As $d>0$, $d = 4$, $6+d = 10$

Therefore, $[ABC] = \frac{ 6 \cdot 10 \cdot \sin 120^{\circ} }{2} = \boxed{\textbf{(E) } 15 \sqrt{3}}$

~isabelchen

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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