Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath> | <cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath> | ||
− | 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>. | + | Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 13:31, 13 August 2017
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 | |
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 |
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