2021 Fall AMC 12A Problems/Problem 14
Contents
Problem
In the figure, equilateral hexagon has three nonadjacent acute interior angles that each measure
. The enclosed area of the hexagon is
. What is the perimeter of the hexagon?
Solution 1 (Law of Cosines and Equilateral Triangle Area)
Isosceles triangles ,
, and
are congruent by SAS congruence. By CPCTC,
, so triangle
is equilateral.
Let the side length of the hexagon be .
The area of each isosceles triangle is by the fourth formula here.
By the Law of Cosines on triangle ,
. Hence, the area of the equilateral triangle
is
.
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or . Hence,
and the perimeter is
.
Solution 2
We will be referring to the following diagram.
Observe that
Letting
the perimeter will be
We know that and using such, we have
and
Thus, we have
\[\begin{align*}[ACE]&=\frac{\sqrt3}{4}\left(2\cdotCG\right)^2\\
&=\sqrt3(2+sqrt3)\end{align*}\]
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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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