Question 1208063
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The area of △ABC is 40. Points P, Q and R lie on sides AB, BC and CA respectively. 
If AP = 3 and PB = 5, and the area of △ABQ is equal to the area of PBQR, determine the area of △AQC.
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It looks like I made an error in my reasoning/solution.


Thanks to Edwin for noticing it.


I think that there is no sense to keep that wrong solution here, so I delete it.


If I will have some other ideas, I will place them here. 


Thanks to Edwin again.


Sorry for inconvenience. 



\\\\\\\\\\\\\ August 20, 2024, approximately one week later \\\\\\\\\\\\\\\



            It took some time for me to find another - correct - solution, 

            but finally I found it and place this my updated solution below.



\\\\\\\\\\\\\\\\\\ August 20, 2024 \\\\\\\\\\\\\\\\\\\\



.
The area of △ABC is 40. Points P, Q and R lie on sides AB, BC and CA respectively. 
If AP = 3 and PB = 5, and the area of △ABQ is equal to the area of PBQR, determine the area of △AQC.


<B>Solution</B>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This problem is very nice.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is nice, because the major idea of the solution is hidden - 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;it is not on the surface and should be dug up.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Due to this reason, &nbsp;it is a typical &nbsp;Math &nbsp;Olympiad problem.


<pre>
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  {{{alpha}}} be the angle ∠AOP = ∠QOR :  {{{alpha}}} = ∠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

    {{{(1/2)*OA*OP*sin(alpha)}}} = {{{(1/2)*OQ*OR*sin(alpha)}}}.    (1)


After reducing, from (1) we have 

    OA*OP = OQ*OR.    (2)


It leads to proportion

    {{{abs(OA)/abs(OR)}}} = {{{abs(OQ)/abs(OP)}}}.    (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  {{{abs(BQ)/abs(BC)}}} = {{{abs(BP)/abs(BA)}}} = {{{5/(5+3)}}} = {{{5/8}}}.


Thus we proved that under given condition, point Q divides the side BC in the same proportion  {{{5/8}}}
as the point P divides side AB.


Hence, the area of triangle  AQC  is  {{{3/8}}}  of the area of triangle  ABC,  i.e.  {{{(3/8)*40}}} = 3*5 = 15 square units.


At this point, the problem is solved completely.
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Nice &nbsp;(&nbsp;!&nbsp;)



\\\\\\\\\\\\\\\\\\\\\\ Sept.18,2024 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\



As &nbsp;I said at the beginning, &nbsp;this solution is the fixed version of my previous incorrect solution.


Edwin's notes relate to that previous, &nbsp;now outdated solution, 
which &nbsp;I just deleted and replaced by this updated &nbsp;CORRECT &nbsp;version.


So, &nbsp;Edwin's notes do not relate to this last solution and are not relevant to it anymore.