Problem:
In equiangular octagon CAROLINE, CA=RO=LI=NE=2​ and AR=OL=IN=EC=1. The self-intersecting octagon CORNELIA encloses six non-overlapping triangular regions. Let K be the area enclosed by CORNELIA, that is, the total area of the six triangular regions. Then K=ba​, where a and b are relatively prime positive integers. Find a+b.
Solution:
The interior angles of CAROLINE are equal to 135∘. Lines AR,OL,IN, and EC enclose a square, and CAROLINE is inscribed in the square, with CO∥AR∥NI∥EL and AN∥CE∥OL∥RI, as shown.
In particular, CARO is a trapezoid with bases 1 and 3 and height 1 , and ARIN is a 1×3 rectangle. Segment CO intersects segments AN,AI, and RN at X,Y, and Z, respectively, and segments AI and RN intersect at W. Then CX=AX=1. Because △AYX∼△AIN, it follows that AXXY​=ANNI​=31​, so XY=31​ and CY=34​. For any region R, let [R] denote the area of R. Then [CAY]=32​. By symmetry, OZ=CY=34​ and YZ=CO−CY−OZ=3−38​=31​. The distance from W to YZ is 21​, so [YZW]=21​⋅31​⋅21​=121​. By symmetry, [ CORNELIA ]=4[CAY]+2[YZW]=38​+61​=617​. The requested sum is 17+6=23​.
