2006 AIME II Problems/Problem 6
Problem
Square has sides of length 1. Points and are on and respectively, so that is equilateral. A square with vertex has sides that are parallel to those of and a vertex on The length of a side of this smaller square is where and are positive integers and is not divisible by the square of any prime. Find
Solution 1
Call the vertices of the new square A', B', C', and D', in relation to the vertices of , and define to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles and are similar. Thus, the sides are proportional: . Simplifying, we get that .
is degrees, so . Thus, , so . Since is equilateral, . is a , so . Substituting back into the equation from the beginning, we get , so . Therefore, , and .
Here's an alternative geometric way to calculate (as opposed to trigonometric): The diagonal is made of the altitude of the equilateral triangle and the altitude of the . The former is , and the latter is ; thus . The solution continues as above.
Solution 2
Since is equilateral, . It follows that . Let . Then, and .
.
Square both sides and combine/move terms to get . Therefore and . The second solution is obviously extraneous, so .
Now, consider the square ABCD to be on the Cartesian Coordinate Plane with . Then, the line containing has slope and equation .
The distance from to is the distance from to .
Similarly, the distance from to is the distance from to .
For some value , these two distances are equal.
Solving for s, , and .
Elegant Solution
Why not solve in terms of the side only (single-variable beauty)? By similar triangles we obtain that , therefore . Then . Using Pythagorean Theorem on yields . This means , and it's clear we take the smaller root: . Answer: .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.