2010 AMC 10A Problems/Problem 19
Problem
Equiangular hexagon has side lengths
and
. The area of
is
of the area of the hexagon. What is the sum of all possible values of
?
Solution
Solution 1
It is clear that is an equilateral triangle. From the Law of Cosines on triangle ABC, we get that
. Therefore, the area of
is
.
If we extend ,
and
so that
and
meet at
,
and
meet at
, and
and
meet at
, we find that hexagon
is formed by taking equilateral triangle
of side length
and removing three equilateral triangles,
,
and
, of side length
. The area of
is therefore
.
Based on the initial conditions,
Simplifying this gives us . By Vieta's Formulas (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of
is
.
Solution 2
As above, we find that the area of is
.
We also find by the sine triangle area formula that , and thus
This simplifies to
.
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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