The following interesting geometry problem was recently (2020) brought to a class of a colleague and friend of mine, Hans Humenberger, from the University of Vienna, Austria by a student with the surname Lux:
Let c1 and c2 be two circles intersecting in A and B and a straight line through A is drawn, intersecting the two circles in M and N. Further let K be the midpoint of MN, P the intersection point of the angle bisector of ∠MAB with c1, R the intersection point of the angle bisector of ∠ BAN with c2. Prove that ∠ PKR = 90o.
The student got the problem from his grandfather, a retired mathematics teacher from France, but without a solution.
The Lux Problem
1) Can you explain why (prove that) this result is true?
2) If you get stuck, press the Hint button in the dynamic sketch at the top. Then view Pompe's Hexagon Theorem.
3) Alternatively, have a look at this sketch for a hint at a dynamic proof of the result.
4) If still stuck, view two other equivalent versions of the same problem, namely, Klingens' Theorem.
If still stuck for a solution, or to compare your own solution(s), read the joint paper of mine with colleague Hans Humenberger, University of Vienna, in the free online journal At Right Angles in the March 2021 issue at Ghosts of a Problem Past. This paper gives 4 different proofs as well as links to several other proofs, including one showing the result as a special case of a similar rectangle generalization of Van Aubel's quadrilateral theorem.
Created 30 March 2020 by Michael de Villiers; updated 12 April 2021.