The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.
Problem
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Tags: invariant, geometry, cyclic quadrilateral, radical axis, power of a point, geometry proposed
Raúl
22.07.2012 16:47
Consider the inversion with center $O$ and circle $\Gamma= \odot ABCD$. That send $P$ to $P'$, $Q$ to $Q'$ and $R$ to $R'$. As $P$ is in the circumferences $\odot ODC$ and $\odot OAB$, $P'$ is in the lines $AB$ and $DC$. As $Q$ and $R$ are in the line $OP$, $R'$ and $Q'$ is still in the line $OP$. Note that $A$, $B$, $C$ and $D$ are on $\Gamma$, so those points are sent to itself.
Using angle chasing e get:
$\angle QAP=\angle P'AQ'$
$\angle PDQ=\angle Q'AP'$
$\angle OBR=\angle OR'B$
$\angle RCO=\angle CR'O$
So, representing by $\odot(P)$ the sentence "The poligon $P$ is cyclic" we get
$\angle QAP=\angle OBR\Leftrightarrow \angle P'AQ'=\angle OR'B \Leftrightarrow \angle P'AQ'=\angle P'R'B\Leftrightarrow \odot([Q'R'BA])$
Similarly
$\angle PDQ=\angle RCO\Leftrightarrow \angle Q'AP'=\angle CR'O \Leftrightarrow \angle Q'AP'=\angle CR'P'\Leftrightarrow \odot([Q'R'CD])$
Concluding with power of the point:
$\odot([Q'R'BA])\Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'A}\times\overline{P'B} \Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'P}\times\overline{P'O}\Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'D}\times\overline{P'C}\Leftrightarrow \odot ([Q'R'CD])$
And we are done
cmtappu96
26.07.2012 14:02
Note that we do not require the fact that points $A,B,C,D$ remain invariant under inversion. So we can have any two circles, one passing through $A,B$ and the other through $C,D$ meeting at $O,P$ and the result would still be true.
subham1729
17.12.2012 08:10
Let $H$ be radical center of $\odot ABCD, \odot ABOP, \odot CDOP$.Now so radical axis of any two of three circles , the lines $AB,CD,OP$ pass through $H$ Since $P$ lies on the shorter arcs $AO,DO$ , it follows that $H$ lies on the extension of $OP$ beyond $P$ so radical center satisfies $HA.HB=HC.HD=HO.HP$. Now since $ABOP$ is cyclic , easy angle chasing gives $\angle{QAB}+\angle{BRQ}=\pi +\angle{QAP}-\angle{OBR}$ so $\angle{QAP}=\angle{OBR}$ iff $ABRQ$ is cyclic same as to $HA.HB=HQ.HR$ similarly $\angle{QAP}=\angle{OBR}$ iff $HC.HD=HQ.HR$. So $\angle{QAP}=\angle{OBR}$ iff $HA.HB=HC.HD$ which is true iff $\angle{PDQ}=\angle {RCO}$.
In hall I did that by trigo bashing of $5$ pages
anantmudgal09
10.09.2015 01:30
wow, one inversion kill. Note that both conditions are equivalent to the fact that $Q$ and $R$ are inverses of one another wrt inversion about $T$ where $T$ is the intersection of $AB$ and $CD$.(possibly at infinity we dont care much)