Problem:
Square ABCD has sides of length 1. Points E and F are on BC and CD, respectively, so that △AEF is equilateral. A square with vertex B has sides that are parallel to those of ABCD and a vertex on AE. The length of a side of this smaller square is ca−b, where a,b, and c are positive integers and b is not divisible by the square of any prime. Find a+b+c.
Solution:
Let CF=x. Then, because △ADF≅△ABE, it follows that DF=BE=1−x, and CE=x. Hence 2x2=EF2=AE2=(1−x)2+1, and so x=3−1. Let P and Q be the vertices of the smaller square that are on AE and AB, respectively. Then
PQAB−PQPQABPQ1=PQAB−BQ=PQAQ=BEAB, so =1+BEAB, and =AB1+BE1.
Thus PQ1=1+1−(3−1)1=1+2−31=1+2+3=3+3. Consequently PQ=3+31=63−3, and a+b+c=12
OR
Let BOPQ be the smaller square, where Q is between A and B, and let BQ=y. Then QP=y, and AQ=ytan75∘. Thus 1=AB=AQ+QB=ytan75∘+y, so y=1+tan75∘1. But tan75∘=tan(45∘+30∘)=1−(1/3)1+(1/3)=2+3. Therefore y=3+31=63−3.
OR
Place a coordinate system so that A is the origin, and the coordinates of B,C, and D are (1,0),(1,1), and (0,1), respectively. Let BE=p. Then, as in the first solution, p=1−(3−1)=2−3. Hence line AE has slope 2−3 and contains the origin. Thus line AE has equation y=(2−3)x. Let q be the length of a side of the smaller square. Then one vertex of that square has coordinates (1−q,q) and is on line AE. Therefore q=(2−3)(1−q), which yields q=(3−3)/6.
