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> | ||
− | ==Solution== | + | ==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) | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | [[File:PJ_2023_12B_Q17.png]] | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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 20:41, 15 November 2023
Contents
[hide]Problem
Triangle ABC has side lengths in arithmetic progression, and the smallest side has length If the triangle has an angle of what is the area of ?
Solution 1
The length of the side opposite to the angle with is longest. We denote its value as .
Because three side lengths form an arithmetic sequence, the middle-valued side length is .
Following from the law of cosines, we have
By solving this equation, we get . Thus, .
Therefore, the area of the triangle is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Since the triangle's longest side must correspond to the angle, the triangle is unique. By analytic geometry, we construct the following plot.
We know the coordinates of point being the origin and being . Constructing the line which point can lay on, here since , is on the line
I denote as the perpendicular line from to , and assume . Here we know is a triangle. Hence and .
Furthermore, due to the arithmetic progression, we know . Hence, in ,
Thus, the area is equal to .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.