Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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===Solution 1=== | ===Solution 1=== |
Revision as of 04:55, 16 January 2020
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
.
Solution 3 (no trig)
Extend to point M so that it creates right triangle
where
. It is given that the hexagon is equiangular, therefore
. (exterior angles of a polygon add up to 360
)
We can use either Pythagorean theorem or the properties of a triangle to find the length of
and
. The legs of
are
and
.
Using Pythagorean theorem, we get . We can then follow
to solve for
.
.
Alternatively, we can find the area of . We know that the three smaller triangles:
,
, and
are congruent because of
. Therefore one of the smaller triangles accounts for
of the total area. The height of the smaller triangle
is just
so the area is
. We can then find the area of the hexagon using
.
We can even find the area of and
and solve for
because the ratio of the areas is
to
.
~Zeric Hang
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.