Difference between revisions of "2010 AMC 12A Problems/Problem 17"
(→Solution 2) |
(→Solution 2) |
||
Line 23: | Line 23: | ||
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}}{4}</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}}{4}</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=0</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>. |
+ | |||
+ | Note: Since <math>r</math> has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is. | ||
== See also == | == See also == |
Revision as of 09:03, 5 February 2018
Contents
[hide]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 previous 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 .
Note: Since has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.
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.