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

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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution==
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==Solution 1==
  
 
The length of the side opposite to the angle with <math>120^\circ</math> is longest.  
 
The length of the side opposite to the angle with <math>120^\circ</math> is longest.  
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution 2==
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Since the triangle's longest side must correspond to the <math>120^\circ</math> angle, the triangle is unique. By analytic geometry, we construct the following plot.
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[[File:PJ_2023_12B_Q17.png]]
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We know the coordinates of point <math>A</math> being the origin and <math>B</math> being <math>(6,0)</math>. Constructing the line which point <math>C</math> can lay on, here since <math>\angle B=120^\circ</math>, <math>C</math> is on the line <cmath>y=\sqrt{3}\left(x-6\right).</cmath>
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I denote <math>D</math> as the perpendicular line from <math>C</math> to <math>AB</math>, and assume <math>CD=k</math>. Here we know <math>\triangle BCD</math> is a <math>30^\circ-60^\circ-90-^\circ</math> triangle. Hence <math>DC=\sqrt{3}k</math> and <math>BC=2k</math>.
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Furthermore, due to the arithmetic progression, we know <math>AC=4k-6</math>. Hence, in <math>\triangle ACD</math>, <cmath>\left(4k-6\right)^{2}=\left(6+k\right)^{2}+3k^{2},</cmath> <cmath>k=5.</cmath>
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Thus, the area is equal to <math>\frac{1}{2}\cdot 6\cdot \sqrt{3} x=15\sqrt{3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2023|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:41, 15 November 2023

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

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 $\frac{1}{2}\cdot 6\cdot \sqrt{3} x=15\sqrt{3}$.

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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