2006 AIME II Problems/Problem 6
Solution 1
![[asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); [/asy]](http://latex.artofproblemsolving.com/f/f/d/ffd7cfe7abb8965aaa343e619177c5c073a82b6c.png)
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.
