Difference between revisions of "2023 AMC 12B Problems/Problem 17"
Prof joker (talk | contribs) m (→Solution 2 (Analytic Geometry)) |
Isabelchen (talk | contribs) (Solution 1 and 2 are different. I put Prof. Joker's solution behind mine because I want to group the solutions that uses law of cosine (solution 1, 2, 3) together.) |
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 2 (Analytic Geometry)== | + | ==Solution 2== |
+ | Let the side lengths be <math>6</math>, <math>x</math>, and <math>2x-6</math>. As <math>2x-6</math> is the longest side, the angle opposite to it will be <math>120^{\circ}</math>. | ||
+ | |||
+ | By the law of Cosine<cmath>(2x-6)^2 = 6^2 + x^2 - 2 \cdot 6 \cdot x \cdot \cos 120^{\circ}</cmath><cmath>4x^2 - 24x + 36 = 36 + x^2 + 6x</cmath><cmath>3x^2 - 30x = 0</cmath><cmath>x^2 - 10x = 0</cmath> | ||
+ | As <math>x \neq 0</math>, <math>x = 10</math>. | ||
+ | |||
+ | Therefore, <math>[ABC] = \frac{ 6 \cdot 10 \cdot \sin 120^{\circ} }{2} = \boxed{\textbf{(E) } 15 \sqrt{3}}</math> | ||
+ | |||
+ | ~isabelchen | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let the side lengths be <math>6</math>, <math>6+d</math>, and <math>6+2d</math>. As <math>6+2d</math> is the longest side, the angle opposite to it will be <math>120^{\circ}</math>. | ||
+ | |||
+ | By the law of Cosine<cmath>(6+2d)^2 = 6^2 + (6+d)^2 - 2 \cdot 6 \cdot (6+d) \cdot \cos 120^{\circ}</cmath><cmath>4d^2 + 24d + 36 = 36 + 36 + 12 d + d^2 + 36 + 6d</cmath><cmath>3d^2 + 6d - 72 = 0</cmath><cmath>d^2 + 2d - 24 = 0</cmath><cmath>(d+6)(d-4)=0</cmath> | ||
+ | As <math>d>0</math>, <math>d = 4</math>, <math>6+d = 10</math> | ||
+ | |||
+ | Therefore, <math>[ABC] = \frac{ 6 \cdot 10 \cdot \sin 120^{\circ} }{2} = \boxed{\textbf{(E) } 15 \sqrt{3}}</math> | ||
+ | |||
+ | ~isabelchen | ||
+ | |||
+ | ==Solution 4 (Analytic Geometry)== | ||
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. | 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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~Prof. Joker | ~Prof. Joker | ||
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==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 23:03, 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
Let the side lengths be , , and . As is the longest side, the angle opposite to it will be .
By the law of Cosine As , .
Therefore,
~isabelchen
Solution 3
Let the side lengths be , , and . As is the longest side, the angle opposite to it will be .
By the law of Cosine As , ,
Therefore,
~isabelchen
Solution 4 (Analytic Geometry)
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 .
~Prof. Joker
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.