National BDMO 2016 : Junior 8

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National BDMO 2016 : Junior 8

Unread post by dshasan » Tue Jan 10, 2017 11:39 pm

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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Re: National BDMO 2016 : Junior 8

Unread post by ahmedittihad » Wed Jan 11, 2017 4:55 pm

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, $\angle CPQ= \angle BPC+\angle BPQ=\angle BPC+BC'Q=90+30=120$.
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Re: National BDMO 2016 : Junior 8

Unread post by Kazi_Zareer » Thu Jan 19, 2017 2:13 am

Take $X,Y$ reflections of $B$ about $P,Q$ respectively and $Z$ reflection of $C$ about $P$.Now see that $\triangle ZYC$ is equilateral triangle, so $ZY=ZC=ZD$, $Z$ is circumcenter of $\triangle CXY$, thus $\angle CXY=30^\circ$, but $PQ$ is midline of $\triangle CXY$, so $PQ\parallel XY$, that's $\angle CPQ=120^\circ$.
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Re: National BDMO 2016 : Junior 8

Unread post by Thamim Zahin » Thu Feb 02, 2017 7:11 pm

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

Unread post by prottoy das » Fri Feb 23, 2018 7:53 pm

[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

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