Reflection Concurrence implies Cyclic

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Reflection Concurrence implies Cyclic

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Source: 2020-2021 Winter SDPC #3

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mira74

#1 h Jan 13, 2021, 11:13 PM

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Let $ABCD$ be a quadrilateral, let $P$ be the intersection of $AB$ and $CD$ , and let $O$ be the intersection of the perpendicular bisectors of $AB$ and $CD$ . Suppose that $O$ does not lie on line $AB$ and $O$ does not lie on line $CD$ . Let $B'$ and $D'$ be the reflections of $B$ and $D$ across $OP$ . Show that if $AB'$ and $CD'$ meet on $OP$ , then $ABCD$ is cyclic.

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#2 h Jan 13, 2021, 11:50 PM • 2 Y

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Let $X=AB'\cap CD'\cap OP$ . Since $O$ lies on the axis of $AB'$ and on the angle bisector of $\angle APB'$ , $O$ is the midpoint of the arc $AB$ in $(APB')$ , and $X$ the foot of the angle bisector. Therefore, by inversion in $P$ with radius $\sqrt{PA\cdot PB'}$ and simmetry across $OP$ , which swaps $X$ and $O$ , $PX\cdot PO=PA\cdot PB'=PA\cdot PB$ , so $PA\cdot PB=PX\cdot PO=PC\cdot PD$ and thus $ABCD$ is cyclic.

This post has been edited 1 time. Last edited by cadaeibf, Jan 13, 2021, 11:52 PM

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#3 h Jun 14, 2021, 5:59 PM

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I am sorry for restarting this thread (I just realized SDPC threads exist). Here was a solution I got after the contest ended (my in-contest solution was three pages long and mostly trig bash).

Solution sketch:
We use the following lemma (which follows from an angle chase):
Let $AB$ be a line that intersects line $\ell$ . Let $A’$ be the reflection of $A$ over $\ell$ . $O, Q \in \ell$ so that $O=BO$ , $Q = A’B \cap \ell$ . Then $ABOQ$ is cyclic.

Let $Q = AB'\cap CD'\cap OP$ . Note that by the lemma, $ABOQ$ and $CDOQ$ are cyclic, and $OQ$ is the radical axis of the circumcircles of these two quadrilaterals. So, $PA*PB= PO*PQ=PC*PD$ , and $ABCD$ is cyclic.

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