# Problem A. 544. (October 2011)

**A. 544.** A circle *k* with center *O* and four distinct fixed points *A*, *B*, *C*, *D* lying on it are given. The circle *k*' intersects *k* perpendicularly at *A* and *B*. Let *X* be a variable point on the line *OA*. Let *U*, other than *A*, be the second intersection of the circles *ACX* and *k*'. Let *V*, other than *A*, be the second intersection of the circles *ADX *and *k*'. Let *W*, other than *B*, be the second intersection of the circle *BDU* and the line *OB*. Finally, let *E*, other than *B*, be the second intersection of the circles *BVW *and *k*. Prove that the location of the point *E* is independent from the choice of the point *X*.

(5 pont)

**Deadline expired on November 10, 2011.**

**Solution. **We will use the basic properties of cross ratios of quadruplets of points on lines and circles.

Let *A*_{1} and *B*_{1} be the points on the circle *k*, diametrically opposite to *A* and *B*, respectively. We show that (*A*_{1},*B*,*C*,*D*)=(*A*,*B*,*U*,*V*)=(*A*,*B*_{1},*D*,*E*). Then the statement follows since the points *A*,*B*_{1},*D* and the cross ratio (*A*,*B*_{1},*D*,*E*) uniquely determine the point *E* on the circle *k*.

By symmetry it is sufficient to show (*A*_{1},*B*,*C*,*D*)=(*A*,*B*,*U*,*V*). Interchanging *A*,*A*_{1},*C*,*X* and *U* by *B*,*B*_{1},*E*,*W* and *V*, respectively, the same proof works for (*A*,*B*,*U*,*V*)=(*A*,*B*,*D*,*E*).

Place the figure in the inversive plane with the ideal point *I*, and apply an inversion with pole *A*. Denote the images of the points by *A*'=*I*,*I*'=*A*,*A*_{1}',*B*',*C*',*D*',*X*',*U*',*V*'. Since the images of circles through *A* are lines, the triplets of points (*B*',*C*',*D*'), (*U*',*V*',*D*'), (*X*',*U*',*C*') and (*X*',*V*',*D*') are collinear. The line *AA*_{1}*X* is tangent to *k*', therefore their images, the lines *AA*_{1}'*X*' and *B*'*U*'*V*' are parallel.

Since the inversion preserves the cross ratios and the quadruplet *A*_{1}',*B*',*C*',*X*' can be projected to *A*',*B*',*U*',*V*' from *X*', we have

(*A*_{1},*B*,*C*,*D*)=(*A*_{1}',*B*',*C*',*D*')=(*A*',*B*',*U*',*V*')=(*A*,*B*,*U*,*V*).

### Statistics:

6 students sent a solution. 5 points: Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter.

Problems in Mathematics of KöMaL, October 2011