2012 AMC 12B Problems/Problem 21
Problem 21
Square is inscribed in equiangular hexagon
with
on
,
on
, and
on
. Suppose that
, and
. What is the side-length of the square?
Solution (Long)
Extend and
so that they meet at
. Then
, so
and therefore
is parallel to
. Also, since
is parallel and equal to
, we get
, hence
is congruent to
. We now get
.
Let ,
, and
.
Drop a perpendicular line from to the line of
that meets line
at
, and a perpendicular line from
to the line of
that meets
at
, then
is congruent to
since
is complementary to
. Then we have the following equations:
The sum of these two yields that
So, we can now use the law of cosines in :
Therefore
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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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