Difference between revisions of "2010 AMC 12A Problems/Problem 17"
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Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{2}</math> and because of the congruency, the area condition, and the fact <math>\triangle{ACE}</math> is equilateral, <math>AC=\sqrt{7r}</math>. | Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{2}</math> and because of the congruency, the area condition, and the fact <math>\triangle{ACE}</math> is equilateral, <math>AC=\sqrt{7r}</math>. | ||
− | Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=</math> and by [[Vieta's Formulas]] we get <math>\boxed{\textbf{E}}</math>. | + | Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] we get <math>\boxed{\textbf{E}}</math>. |
== See also == | == See also == |
Revision as of 15:18, 28 December 2016
Contents
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, 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 pervious solution to get .
Step 2: ~
~
via SAS congruency. Using the formula
and because of the congruency, the area condition, and the fact
is equilateral,
.
Step 3: and by Vieta's Formulas we get
.
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 |
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