2006 AIME II Problems/Problem 6

Problem

Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$ . Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$ .

Here's an alternative geometric way to calculate $AE$ (as opposed to trigonometric): The diagonal $\overline{AC}$ is made of the altitude of the equilateral triangle and the altitude of the $45-45-90 \triangle$ . The former is $\frac{AE\sqrt{3}}{2}$ , and the latter is $\frac{AE}{2}$ ; thus $\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}$ . The solution continues as above.