2006 AIME I 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
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.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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