Problem B. 3832. (September 2005)
B. 3832. P is an arbitrary point of the hypotenuse AB of a rightangled triangle ABC. The foot of the altitude drawn from vertex C is C_{1}. The projection of P onto the leg AC is A_{1}, and its projection onto the leg BC is B_{1}.
a) Prove that the points P, A_{1}, C, B_{1}, C_{1} lie on a circle.
b) Prove that the triangles A_{1}B_{1}C_{1} and ABC are similar.
(3 pont)
Deadline expired on 17 October 2005.
Solution. (a) The angles PA_{1}C, PB_{1}C, PC_{1}C are all right angles, so points A_{1}, B_{1}, C_{1} lie on the circle of diameter PC. Due to the right angle at C, another diameter of this circle is A_{1}B_{1}.
(b) From the triangles ABC and CBC_{1}, BAC=90^{o}angleABC=BCC_{1}. Since quadrilateral CA_{1}C_{1}B_{1} is cyclic, B_{1}CC_{1}=B_{1}A_{1}C_{1}. Therefore, the red angles in the Figure are equal.
Similarly, also the blue angles are equal.
Triangles ABC and A_{1}B_{1}C_{1} are similar because they have equal angles, respectively.
Statistics:
384 students sent a solution.  
3 points:  205 students. 
2 points:  64 students. 
1 point:  98 students. 
0 point:  14 students. 
Unfair, not evaluated:  3 solutions. 
