Problem:
In β³ABC,β C=90β and AB=12. Squares ABXY and ACWZ are constructed outside of the triangle. The points X,Y,Z, and W lie on a circle. What is the perimeter of the triangle?
Answer Choices:
A. 12+93β
B. 18+63β
C. 12+122β
D. 30
E. 32
Solution:
Let O be the center of the circle on which X,Y,Z, and W lie. Then O lies on the perpendicular bisectors of segments XY and ZW, and OX=OW. Note that segments XY and AB have the same perpendicular bisector and segments ZW and AC have the same perpendicular bisector, from which it follows that O lies on the perpendicular bisectors of segments AB and AC; that is, O is the circumcenter of β³ABC. Because β C=90β,O is the midpoint of hypotenuse AB. Let a=21βBC and b=21βCA. Then a2+b2=62 and 122+62=OX2=OW2=b2+(a+2b)2. Solving these two equations simultaneously gives a=b=32β. Thus the perimeter of β³ABC is 12+2a+2b=(C)12+122ββ.