Four points are concyclic

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Four points are concyclic

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Source: Moldova IMO-BMO TST 2003, day 1, problem 3

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#1 h Aug 14, 2008, 6:31 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a quadrilateral inscribed in a circle of center $O$ . Let M and N be the midpoints of diagonals $AC$ and $BD$ , respectively and let $P$ be the intersection point of the diagonals $AC$ and $BD$ of the given quadrilateral .It is known that the points $O,M,Np$ are distinct. Prove that the points $O,N,A,C$ are concyclic if and only if the points $O,M,B,D$ are concyclic.

Proposer: Dorian Croitoru

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#2 h Feb 19, 2010, 6:18 PM • 2 Y

Y by fystic, Adventure10

Let $\tau_a,\tau_b,\tau_c,\tau_d$ be the tangents of $(O,R)$ through $A,B,C,D,$ respectively. $X \equiv \tau_b \cap \tau_d,$ $Y \equiv \tau_a \cap \tau_c.$ Inversion with respect to $(O)$ transforms $A,B,C,D$ into themselves and $M \mapsto Y,$ $N \mapsto X,$ because of $R^2 = OM \cdot OY = ON \cdot OX.$ If $O,N,A,C$ are concyclic, then $X,A,C$ are collinear, i.e. $X \equiv \tau_b \cap \tau_d \cap AC$ $\Longleftrightarrow$ $ABCD$ is harmonic $\Longleftrightarrow$ $Y \equiv \tau_a \cap \tau_c \cap BD,$ i.e $Y,B,D$ are collinear $\Longleftrightarrow$ $O,M,B,D$ are concyclic.

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#3 h Oct 23, 2012, 4:09 PM • 2 Y

Y by Adventure10, Mango247

Suppose that $OMBD$ is cyclic. Let $\{P\}=OM\cap BD$ we have $PM\cdot OP = PB\cdot PD = OP^2 - R^2$ or
$R^2 = OM\cdot OP$ . Let $\{Q\}=ON\cap AC$ ,it remains to show that $ON\cdot OQ = R^2$ .
We have also that $OQ\perp BD$ and $OP\perp AC$ $\Longrightarrow$ $PMNQ$ is cyclic,
so $ON\cdot OQ = OM\cdot OP= R^2$ .
Similarly we prove the reciprocal.

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#4 h Nov 1, 2013, 12:04 AM • 2 Y

Y by Adventure10, Mango247

A little different

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Sardor

#6 h Nov 26, 2014, 9:58 AM • 1 Y

Y by Adventure10

Let $ON \cap AC=T$ and $OM \cap BD=S$ .We know that the points $M$ and $N$ lies on circle with dimametr $OP$ .Let $Q$ is midpoint of $OP$ ,then we know that $TS \perp QO$ .Also know that the points $B,M,O,D$ are concyclic if and only if the radical axis of $(O),(Q),(BMOD)$ are concurrent at $T$ ,and the radical axis of $(O)$ and $(Q)$ is perpendicular to $(OQ)$ , so is $TS$ , similarly for $ANOC$ .Hence we have $ONAC$ is cyclic if and only if $BMOD$ is cyclic if and only if $TS$ is radical axis of $(O)$ and $(Q)$ , so we are done !

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#7 h Nov 26, 2014, 8:37 PM • 1 Y

Y by Adventure10

Suppose, w.l.o.g. $AB>AD$ .
If $BOMD$ was cyclic, $\angle OBD+\angle OMD=180^\circ\ (\ *\ )$ , but $\angle OBD=90^\circ-\angle BAD$ and, for $(*)$ to be true, since $\angle OMC=90^\circ$ , we need $\angle CMD=\angle BAD$ , or $\angle ADM=\angle BAC=\angle BDC$ , i.e. $BD$ is symmedian of $\triangle ADC$ and $ABCD$ is a harmonic quadrilateral, thus we are done.

Best regards,
sunken rock

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#8 h Nov 26, 2014, 9:28 PM • 1 Y

Y by Adventure10

Can someone give us a diagram?

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#9 h Jun 22, 2016, 1:16 PM • 1 Y

Y by Adventure10

patrickhompe wrote:

Invert about pole $O$ with power $r^2$ . $A,B,C,D$ go to themselves. Call the intersection of $ON$ and $AC$ $N'$ . Call the intersection of $OM$ and $BD$ $M'$ . Clearly, $NN'MM'$ is cyclic since $N,M$ are midpoints of chords. Therefore, $\triangle ONM' \sim \triangle OMN'$ , yielding $ON \cdot ON' = OM \cdot OM'$ . Now, if $ANOC$ is cyclic $I(N)$ must lie on $AC$ , and is therefore $N'$ . From there we obtain $I(M)=M'$ easily, which means $M$ is on the circumcircle of $\triangle OBD$ . It works identically for the other direction.

Can someone explain why the following holds:

Now, if $ANOC$ is cyclic $I(N)$ must lie on $AC$ , and is therefore $N'$ .

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#10 h Apr 27, 2020, 6:08 PM

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https://i.imgur.com/7erk2Th.png

We start off with a little lemma:
Lemma: The quad $ONPM$ is cyclic
Proof: $\angle OMP = \angle ONP = 90 \implies \angle OMP + \angle ONP = 180$

So now we assume that $ONAC$ is cyclic.
We start off by defining a inversion $\psi(O,OC)$
Then by that inversion the circle around $ONAC$ is sent to $\overline{AC}$ , and the circle around the quad $ONPM$ is sent to a straight line.
The inverses of points we denote by adding a $'$ to it.
So now we have since $N$ belongs to two circles we have that $N'$ is the intersection of two lines.
Now we get that the points $C,M,P,A,N'$ are all colinear.
So now we take a look at $M'$ , we want to show that $\angle ONM' = 90$ , if we show that we are done, since then we would get that the points $M',D,P,N,B$ are all colinear and they invert into a circle, hence proving that the quad $OMBD$ is cyclic.

To show this we just take a look at the angles $\angle ONM' = \angle OMN' = 90$ , the last relation is true because of colinearity.
Hence we have shown that the quad $OMBD$ is cyclic . . .

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