Problem:
In the figure, ABCD is a square of side length 1 . The rectangles JKHG and EBCF are congruent. What is BE ?
Answer Choices:
A. 21β(6ββ2)
B. 41β
C. 2β3β
D. 63ββ
E. 1β22ββ Solution:
Let x=BE=GH=CF, and let ΞΈ=β DHG=β AGJ=β FKH. Note that AD=GJ=HK=1. In right triangle GDH,xsinΞΈ=DG=1βAG=1βcosΞΈ, so x=sinΞΈ1βcosΞΈβ. Then 1=CD=CF+FH+HD=x+sinΞΈ+xcosΞΈ. Substituting for x gives
Let a=EK,b=EJ, and c=JK=BE. Then triangles KEJ,GDH, and JAG are similar right triangles and it follows that a2+b2=c2,caβ=1βbβc, and cbβ=1βa. The first equation is equivalent to a2=(c+b)(cβb), and the last equation is equivalent to ac=cβb. Multiplying by c+b and equating to the first equation gives ac(c+b)=(c+b)(cβb)=a2. Because a>0, it follows that a=c(c+b). Plugging into the second equation gives c(1βbβc)=c(c+b). Because c>0, it follows that c+b=21β. Thus a=2cβ and
c2=a2+b2=4c2β+(21ββc)2
Solving for c gives c=2Β±3β. If c=2+3β, then b=21ββc=β23ββ3β<0. Thus BE=c=2β3ββ.
