Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | ||
− | Drop a perpendicular line from <math>A</math> to the line of <math>EF</math> that meets line <math>EF</math> at <math>K</math>, and a perpendicular line from <math>Y</math> to the line of <math>EF</math> that meets <math>EF</math> at <math>L</math>, then <math>\triangle AKZ</math> is congruent to <math>\triangle ZLY</math> since <math>\angle | + | Drop a perpendicular line from <math>A</math> to the line of <math>EF</math> that meets line <math>EF</math> at <math>K</math>, and a perpendicular line from <math>Y</math> to the line of <math>EF</math> that meets <math>EF</math> at <math>L</math>, then <math>\triangle AKZ</math> is congruent to <math>\triangle ZLY</math> since <math>\angle YZL</math> is complementary to <math>\angle KZA</math>. Then we have the following equations: |
<cmath>\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1</cmath> | <cmath>\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1</cmath> |
Revision as of 12:56, 28 December 2012
Problem
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