Difference between revisions of "2023 AMC 12B Problems/Problem 17"
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By the law of Cosine | 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>(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 + d^2 | + | <cmath>4d^2 + 24d + 36 = 36 + 36 + 12 d + d^2 + 36 + 6d</cmath> |
<cmath>3d^2 + 6d - 72 = 0</cmath> | <cmath>3d^2 + 6d - 72 = 0</cmath> | ||
<cmath>d^2 + 2d - 24 = 0</cmath> | <cmath>d^2 + 2d - 24 = 0</cmath> |
Revision as of 23:09, 15 November 2023
Contents
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,
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,
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.