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Brazilian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

ABCDABCD is any convex quadrilateral. Squares center E,F,G,HE, F, G, H are constructed on the outside of the edges AB,BC,CDAB, BC, CD and DADA respectively. Show that EGEG and FHFH are equal and perpendicular.

Solution

Consider A,B,C,DA, B, C, D to be the points a,b,c,da, b, c, d in the complex plane. Then E,F,G,HE, F, G, H are the points a+b2+ib−a2\frac{a+b}{2} + i\frac{b-a}{2}, b+c2+ic−b2\frac{b+c}{2} + i\frac{c-b}{2}, c+d2+id−c2\frac{c+d}{2} + i\frac{d-c}{2}, d+a2+ia−d2\frac{d+a}{2} + i\frac{a-d}{2}. Hence vector GE⃗\vec{GE} is represented by the complex number h=a+b−c−d2+i−a+b+c−d2h = \frac{a+b-c-d}{2} + i\frac{-a+b+c-d}{2} and the vector FH⃗\vec{FH} by the k=a−b−c+d2+ia+b−c−d2k = \frac{a-b-c+d}{2} + i\frac{a+b-c-d}{2}. But k=ihk = ih, so EGEG and FHFH are equal and perpendicular.

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