## IMO 2020 #1

Discussion on International Mathematical Olympiad (IMO)
FuadAlAlam
Posts: 23
Joined: Wed Sep 16, 2020 11:10 am
Location: Dhaka, Bangladesh
Contact:

### IMO 2020 #1

Consider the convex quadrilateral \$ABCD\$. The point \$P\$ is in the interior of \$ABCD\$. The following ratio equalities hold:
\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]Prove that the following three lines meet in a point: the internal bisectors of angles \$\angle ADP\$ and \$\angle PCB\$ and the perpendicular bisector of segment \$AB\$.

Proposed by Dominik Burek, Poland

FuadAlAlam
Posts: 23
Joined: Wed Sep 16, 2020 11:10 am
Location: Dhaka, Bangladesh
Contact:

### Re: IMO 2020 #1

Let the circumcircle of \$\triangle PAB\$ intersects line \$DA\$ and \$BC\$ at points \$P_1\$ and \$P_2\$, respectively. Now, \$\measuredangle P_1PA = \measuredangle P_1BA = \measuredangle PBA - \measuredangle PBP_1 = \measuredangle PBA - \measuredangle PAP_1 = \measuredangle PBA - \measuredangle PAD = \measuredangle PAD\$. Again, \$\measuredangle DPP_1 = \measuredangle DPA - \measuredangle P_1PA = \measuredangle DPA - \measuredangle PAD = \measuredangle PBA = \measuredangle PP_1A = \measuredangle PP_1D\$. Thus, \$DP = P_1D\$. Therefore, the internal bisector of \$\angle ADP\$ is the perpendicular bisector of segment \$PP_1\$. Similarly, the internal bisector of \$\angle PCB\$ is the perpendicular bisector of segment \$PP_2\$. Clearly, the perpendicular bisectors of segments \$PP_1, PP_2\$ and \$AB\$ passes through the circumcenter of \$\triangle PAB\$, as desired.