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

~isabelchen

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

~isabelchen

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 1 by OmegaLearn

https://youtu.be/uVHCLHBWWJM

Video Solution 2

https://youtu.be/3w59Hikefqs

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

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