Solution to problem 28 submitted by Michael Shtilman

Problem 28

Quadrilateral ABCD is inscribed into a circle with center O. Point E belongs to AB and point F to CD. Point K lies on EF. It is known that EF is perpendicular to OK. Prove that |EK| = |KF|.


It is easy to construct a counterexample to this problem. Namely, suppose ABCD is a square, and choose a point E on AB very close to A, and a point F on CD such that |CF|=|FD|. It is easy to see that the line through O which is parallel to AB and CD devides EF in the middle. At the same time it can't be perpendicular to EF by the choice of points E and F.

Comment: Many of these problems are deliberately confusing. This one asks one to prove a false statement.

Last revised August 2003