Difference between revisions of "2010 AMC 12A Problems/Problem 17"
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==Solution 3== | ==Solution 3== | ||
− | Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent triangles <math>ABC, CDE, EFA</math> as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-7}{1}=7\text{or} \textbf{E}</math>. | + | Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent triangles <math>ABC, CDE, EFA</math> as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-7}{1}=7\text{or} \textbf{(E)}</math>. |
-vsamc | -vsamc | ||
Revision as of 18:38, 10 March 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 1
It is clear that is an equilateral triangle. From the Law of Cosines on
, 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
Step 1: Use Law of Cosines in the same manner as the previous solution to get .
Step 2: ~
~
via SAS congruency. Using the formula
. The area of the hexagon is equal to
. We are given that
of this area is equal to
; solving for
in terms of
gives
.
Step 3: and by Vieta's Formulas , we get
.
Note: Since has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.
Solution 3
Find the area of the triangle as how it was done in solution 1. Find the sum of the areas of the congruent triangles
as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles
is
of the area of the hexagon. Hence
times the latter is equal to the triangle
. Hence
. We can simplify this to
. By Vieta's, we get the sum of all possible values of
is
.
-vsamc
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.