Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus | We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus | ||
− | <cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4 | + | <cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}</cmath> |
This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>. | This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>. | ||
Revision as of 22:07, 1 November 2015
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, 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
.
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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