Difference between revisions of "2010 AMC 12A Problems/Problem 17"
m (→Solution) |
|||
Line 19: | Line 19: | ||
<math>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</math> | <math>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</math> | ||
− | <math>r^2-6r+1 = 0</math> | + | Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. |
− | |||
− | |||
− | By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. |
Revision as of 17:45, 11 February 2010
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
It is clear that is an equilateral triangle. From the Law of Cosines, 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 we know that the sum of the possible value of
is
.