Lesson Nice geometry problem of a Math Olympiad level

Nice geometry problem of a Math Olympiad level

Problem 1

Before to move forward, make a sketch of the problem. 
I will assume that the sketch is in front of your eyes.


       Let O be the intersection point of PR and AQ.

       Let   be the angle ∠AOP = ∠QOR :   = ∠AOP = ∠QOR
       (these angles are congruent since they are vertical angles).


Since the areas of triangle ABQ and quadrilateral PBQR are equal, we conclude from this fact 
that the areas of triangles AOP and ROQ are equal. The equation for these equal areas is

     = .    (1)


After reducing, from (1) we have 

    OA*OP = OQ*OR.    (2)


It leads to proportion

     = .    (3)


Thus we see that triangles  AOR  and  QOP  have congruent angles  AOR  and  QOP  and proportional sides
that conclude these angles.  So, these triangles  AOR  and  QOP  are  SIMILAR.


    It is the key idea of the solution.
    What follows, is the direct consequence of this idea.



So, the triangles  AOR  and  QOP  are similar.
From it, we conclude that these triangles have congruent corresponding angles.

In particular, angles  OAR  and  OQP  are congruent.

It implies that line  AR  is parallel to   PQ.


It is the same as to say that  PQ  is parallel to  AC.


Hence, triangle  PBQ  is similar to triangle  ABC.



From this similarity, we have   =  =  = .


Thus we proved that under given condition, point Q divides the side BC in the same proportion 
as the point P divides side AB.


Hence, the area of triangle  AQC  is    of the area of triangle  ABC,  i.e.   = 3*5 = 15 square units.


At this point, the problem is solved completely.