## The Lux Problem

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 *c*_{1} and *c*_{2} 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 *c*_{1}, *R* the intersection point of the angle bisector of ∠ *BAN* with *c*_{2}. Prove that ∠ *PKR* = 90^{o}.

The student got the problem from his grandfather, a retired mathematics teacher from France, but without a solution.

####

The Lux Problem

**Challenge**:

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.

Copyright © 2019 KCP Technologies, a McGraw-Hill Education Company. All rights reserved.

Release: 2019Q4, Semantic Version: 4.6.2, Build Number: 1039, Build Stamp: dn.kcptech.com/20191209222013

*Back
to "Dynamic Geometry Sketches"*

*Back
to "Student Explorations"*

Created 30 March 2020 by Michael de Villiers; updated 12 April 2021.