## National BDMO 2016 : Junior 8

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### 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\$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

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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'=80-20=60\$. As \$QB=6, BC'=12\$ and \$\angle QBC' =60\$ we see that \$\triangle QBC'\$ is a \$30-60-90\$ 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\$.
Q.E.D
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Kazi_Zareer
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### Re: National BDMO 2016 : Junior 8

Solution:
We cannot solve our problems with the same thinking we used when we create them.

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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prottoy das
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### Re: National BDMO 2016 : Junior 8

[I'm solving this problem with the geometry of class 9-10.]
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