2010 AMC 12A Problems/Problem 17
- The following problem is from both the 2010 AMC 12A #17 and 2010 AMC 10A #19, so both problems redirect to this page.
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 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 (Edited by CoolMath8)
Proof Triangle ACE is Equilateral.
We know triangles ABC, CDE, and EFA are the same by SAS congruence, so the side opposite the 120 degree is also the same (since the triangles are congruent). Thus ACE is congruent. Q.E.D. ~mathboy282
Video Solution by the Beauty of Math
https://youtu.be/rsURe5Xh-j0?t=961
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 |
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.