National BDMO 2016 : Junior 8
National BDMO 2016 : Junior 8
In $\bigtriangleup ABC$ , $\angle A = 20$, $\angle B = 80$, $\angle C = 80$, $BC = 12$ units. Perpendicular $BP$ is drawn on $AC$ from from $B$ which intersects $AC$ at the point $P$. $Q$ is a point on $AB$ in such a way that $QB = 6$ units. Find the value of $\angle CPQ$.
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 Charles Caleb Colton
 Charles Caleb Colton
 ahmedittihad
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Re: National BDMO 2016 : Junior 8
This was a nice problem. Let $C'$ be the reflection of $C$ w.r.t $BP$. Now, $\angle C'BC = 2*10$. So, $\angle QBC'=8020=60$. As $QB=6, BC'=12$ and $\angle QBC' =60$ we see that $\triangle QBC'$ is a $306090$ triangle. So, $\angle BQC'=90$. We get, $BQC'P$ is cyclic. So, $\angle CPQ= \angle BPC+\angle BPQ=\angle BPC+BC'Q=90+30=120$.
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 Kazi_Zareer
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Re: National BDMO 2016 : Junior 8
Solution:
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 Thamim Zahin
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Re: National BDMO 2016 : Junior 8
I had made the reflection but didn't get that it was a right triangle. Have to draw diagram as scale from now on.
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Re: National BDMO 2016 : Junior 8
[I'm solving this problem with the geometry of class 910.]
Take a point D in AC such that PD=CP. Join B and D.Now two triangle BCP and BPD are congruent.So BD=BC=12.Here angle PBQ=60.Draw a perpendicular from D to AB. Let it intersect AB at E.Now cos60=12/BE.Hence BE=6.But we know BQ=6.So E and Q are the same point.Now quadrilateral BPDQ is cyclic because the sum of the opposite angle of the quadrilateral is equal to 180.S0 angle BDQ=angle BPQ.In the right angled triangle BDQ=30.So BPQ=30 and angle CPQ=120
Take a point D in AC such that PD=CP. Join B and D.Now two triangle BCP and BPD are congruent.So BD=BC=12.Here angle PBQ=60.Draw a perpendicular from D to AB. Let it intersect AB at E.Now cos60=12/BE.Hence BE=6.But we know BQ=6.So E and Q are the same point.Now quadrilateral BPDQ is cyclic because the sum of the opposite angle of the quadrilateral is equal to 180.S0 angle BDQ=angle BPQ.In the right angled triangle BDQ=30.So BPQ=30 and angle CPQ=120