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### National BDMO 2016 : Junior 8

Posted: 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\$.

### Re: National BDMO 2016 : Junior 8

Posted: 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\$.
Q.E.D

### Re: National BDMO 2016 : Junior 8

Posted: Thu Jan 19, 2017 2:13 am
Solution:
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\$.

### Re: National BDMO 2016 : Junior 8

Posted: 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.