2006 AIME II Problems/Problem 6
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
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.
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 .
Suppose Note that since the triangle is equilateral, and by symmetry, Note that if and , then Also note that Using the fact , this yields
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: .
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