IMO 2020 #1

Discussion on International Mathematical Olympiad (IMO)
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FuadAlAlam
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IMO 2020 #1

Unread post by FuadAlAlam » Fri Dec 04, 2020 12:39 pm

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

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FuadAlAlam
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Re: IMO 2020 #1

Unread post by FuadAlAlam » Mon Mar 01, 2021 1:03 pm

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.

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