Problem:
In △ABC,∠C=90∘ 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+93​
B. 18+63​
C. 12+122​
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 △ABC. Because ∠C=90∘,O is the midpoint of hypotenuse AB. Let a=21​BC and b=21​CA. Then a2+b2=62 and 122+62=OX2=OW2=b2+(a+2b)2. Solving these two equations simultaneously gives a=b=32​. Thus the perimeter of △ABC is 12+2a+2b=(C)12+122​​.
