Problem A. 572. (November 2012)
A. 572. Two circles k1 and k2 with centres O1 and O2, respectively, intersect perpendicularly at P and Q. Their external homothety center is H. The line t is tangent to k1 at T1 and tangent to k2 at T2. Let X be a point in the interior of the two circles such that HX=HP=HQ, and let X' be the reflection of X about t. Let the circle XX'T2 and the shorter arc PQ of k1 meet at U1, and let the circle XX'T1 and the shorter arc PQ of k2 meet at U2. Finally, let the lines O1U1 and O2U2 meet at V. Show that VU1=VU2.
(5 pont)
Deadline expired on December 10, 2012.
Statistics:
10 students sent a solution. 5 points: Bodnár Levente, Cyril Letrouit, Ioan Laurentiu Ploscaru, Janzer Olivér, Kabos Eszter, Machó Bónis, Maga Balázs, Nagy Róbert, Omer Cerrahoglu, Szabó 789 Barnabás.
Problems in Mathematics of KöMaL, November 2012