2010 AIME II Problems/Problem 9
Problem
Let be a regular hexagon. Let , , , , , and be the midpoints of sides , , , , , and , respectively. The segments , , , , , and bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of be expressed as a fraction where and are relatively prime positive integers. Find .
Contents
Solution
Let be the intersection of and
and be the intersection of and .
Let be the center.
Solution 1
Let (without loss of generality).
Note that is the vertical angle to an angle of regular hexagon, and so has degree .
Because and are rotational images of one another, we get that and hence .
Using a similar argument, , and
Applying the Law of cosines on ,
Thus, the answer is 4 + 7 = .
Solution 2 (Coordinate Bash)
We can use coordinates. Let be at with at ,
then is at ,
is at ,
is at ,
Line has the slope of and the equation of
Line has the slope of and the equation
Let's solve the system of equation to find
Finally,
Thus, the answer is .
Solution 3
Use the diagram. Now notice that all of the "overlapping triangles" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.