# 2023 AMC 12B Problems/Problem 17

## 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

Let the side lengths of $\triangle ABC$ be $6$, $x$, and $2x-6$, where $6 \le x \le 2x-6$. As $2x-6$ is the longest side, the angle opposite to it will be $120^{\circ}$.

By the law of Cosine $$(2x-6)^2 = 6^2 + x^2 - 2 \cdot 6 \cdot x \cdot \cos 120^{\circ}$$ $$4x^2 - 24x + 36 = 36 + x^2 + 6x$$$$3x^2 - 30x = 0$$ $$x^2 - 10x = 0$$ As $x \neq 0$, $x = 10$.

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

## Solution 3

Let the side lengths of $\triangle ABC$ be $6$, $6+d$, and $6+2d$, where $6 \le 6+d \le 6+2d$. 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 $d \ge 0$, $d = 4$, $6+d = 10$

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

## Solution 4 (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.

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

## Video Solution 2

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