Prove that A, X, O, Y are concyclic

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Prove that A, X, O, Y are concyclic

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Source: Taiwan 2014 TST2 Quiz 2, P1

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#1 h Jul 18, 2014, 7:05 PM • 10 Y

Y by buratinogigle, doxuanlong15052000, anantmudgal09, Davi-8191, Gaussian_cyber, HamstPan38825, LoloChen, Adventure10, Mango247, GeoKing

Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$ . A straight line $L$ is parallel to $BC$ and tangent to the incircle. Suppose $L$ intersects $IO$ at $X$ , and select $Y$ on $L$ such that $YI$ is perpendicular to $IO$ . Prove that $A$ , $X$ , $O$ , $Y$ are cyclic.

Proposed by Telv Cohl

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#2 h Jul 18, 2014, 9:27 PM • 3 Y

Y by Ya_pank, Adventure10, Mango247

I wish there were such nice problems on polish TST (unfortunately, we don't have TST

)

Sketch of proof:
Click to reveal hidden text

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#3 h Jul 20, 2014, 5:14 AM • 2 Y

Y by Adventure10, Mango247

Oooooh any sokuyion not with complex???

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mathuz

#4 h Jul 20, 2014, 9:42 AM • 1 Y

Y by Adventure10

is it hard problem?

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mathuz

#5 h Jul 20, 2014, 6:48 PM • 3 Y

Y by buratinogigle, Adventure10, Mango247

see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=598779&p=3553812#p3553812

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XmL

#6 h Jul 21, 2014, 3:58 AM • 10 Y

Y by nima1376, 61plus, Panoz93, buratinogigle, livetolove212, anantmudgal09, Phie11, Adventure10, and 2 other users

My solution:

Let $BI,CI\cap (O)=E,F, IY\cap BC=Y', L\cap (I)=D$ , $M$ is the midpoint of $BC$ . Since $OM\perp BC$ , therefore $I,O,M,Y'$ are concyclic $\Rightarrow \angle YOI=\angle IOY'=\angle IMY'$ ( $Y,Y'$ are reflexive about $IO$ . Since it's well known that $IM\parallel AD$ , therefore $A,X,Y,O$ are concyclic $\iff \angle XAY=180-\angle YOI=$ $180-\angle IMY'=\angle ADY$ $\iff AY^2=DY*YX=YI\iff AY=YI$ .

Note that If $IY\cap FE=Y''$ , then by butterfly theorem $IY'=IY''\Rightarrow Y=Y''$ , hence $Y$ lies on $EF$ or the perpendicular bisector of $AI$ so $AY=YI$ and we are done.

BTW mathuz's way also works

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#7 h Oct 18, 2014, 7:38 AM • 12 Y

Y by buratinogigle, v_Enhance, anantmudgal09, Davi-8191, amar_04, AmirKhusrau, Siddharth03, LoloChen, Adventure10, GeoKing, and 2 other users

Thanks for all of you like this problem and here is my solution

Lemma :

Given a $\triangle ABC$ with incenter $I$ and circmcenter $O$ .
Let $L$ be a line passing through $I$ and perpendicular to $IO$ .
Let $L$ intersects the external bisector of $\angle BAC , BC$ at $X, Y$ , respectively .

Then $XI=2YI$

Proof of the lemma :

Let $A', B', C', Y', O'$ be the reflection of $I$ with respect to $A, B, C, Y, O$ , respectively.
Let $I_a, I_b, I_c$ be the $A$ -excenter, $B$ -excenter, $C$ -excenter of $\triangle ABC$ , respectively.

Since $O'$ is the circumcenter of $\triangle I_aI_bI_c$ and the circumcenter of $\triangle A'B'C'$ ,
so $I_a, I_b, I_c, A', B', C'$ lie on a circle with center $O'$ and radius $2R$ ( $R$ is the radius of $\odot (ABC)$ ).
From Butterfly theorem (for quadrilateral $I_cI_bB'C'$ ) we get $IX=IY'=2IY$ .
____________________________________________________________
Back to the main problem :

Let $Q$ be the intersection of $IY$ and $BC$ .
Let $S, R$ be the intersection of $AI$ and $BC, L$ , respectively.
Let $P$ be the intersection of the external bisector of $\angle BAC$ and $IY$ .

From the lemma we get $IP=2IQ$ ,
so $Y$ is the midpoint of $IP$ (Since $I$ is the midpoint of $YQ$ ),
hence we get $YA=YP=YI$ . i.e. $\triangle YAI$ is a isosceles triangle

From $\angle OAY=\angle IAY - \angle IAO =\angle YIA - \angle IAO =\angle ACB + \tfrac{1}{2} \angle BAC - \angle AIX =\angle ASB - \angle AIX =\angle IRY - \angle AIX =\angle OXY$ we conclude that $A, X, O, Y$ are concyclic.

Q.E.D
____________________________________________________________
Generalization : Geometry Marathon Problem 29 , Generalization of 2014 Taiwan TST2 Quiz2 P1

This post has been edited 3 times. Last edited by TelvCohl, Apr 3, 2020, 10:26 PM

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#8 h Nov 21, 2014, 1:02 PM • 3 Y

Y by I_m_vanu1996, Adventure10, Mango247

Dear XmL and Mathlinkers,
your proof is very nice because it is based on this interesting idea that IM // AD.
Sincerely
Jean-Louis

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#9 h Nov 21, 2014, 1:55 PM • 1 Y

Y by Adventure10

@jayme,,,agree n thnx for ur post!!!!!

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#10 h Nov 28, 2014, 3:50 PM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

Thanks for all of you like this problem

My solution:

Lemma:

Given a $\triangle ABC$ with incenter $I$ and circmcenter $O$ .
$L$ is a line pass through $I$ and perpendicular to $IO$ .
$L$ intersects the external bisector of $\angle BAC , BC$ at $X, Y$ .

Then $XI=2YI$

==============================
Proof of the lemma:

Let $A', B', C', Y', O'$ be the reflection of $I$ with respect to $A, B, C, Y, O$ , respectively.
Let $Ia, Ib, Ic$ be the $A$ -excenter, $B$ -excenter, $C$ -excenter of $\triangle ABC$ , respectively.

Since $O'$ is the circumcenter of $\triangle IaIbIc$ and the circumcenter of $\triangle A'B'C'$ ,
so $Ia, Ib, Ic, A', B', C'$ lie on a circle with center $O'$ and radius $2R$ ( $R$ is the radius of $(ABC)$ ).
From Butterfly theorem (for quadrilateral $IcIbB'C'$ ) we get $IX=IY'=2IY$ , so we are done.

==============================
Back to the main problem

Let $Q$ be the intersection of $IY$ and $BC$ .
Let $S, R$ be the intersection of $AI$ and $BC, L$ , respectively.
Let $P$ be the intersection of the external bisector of $\angle BAC$ and $IY$ .

From the lemma we get $IP=2IQ$ ,
so we get $Y$ is the midpoint of $IP$ (Since $I$ is the midpoint of $YQ$ ),
hence we get $YA=YP=YI$ . ie. $\triangle YAI$ is a isosceles triangle,
so
$\angle OAY=\angle IAY - \angle IAO =\angle YIA - \angle IAO =\angle ACB + ( \frac { \angle BAC}{2}) - \angle AIX =\angle ASB - \angle AIX =\angle IRY - \angle AIX =\angle OXY$
hence we get $A, X, O, Y$ are concyclic.

Q.E.D

Same with mine

And can you post the other versions of this problem that you made, too ?

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#11 h Jan 23, 2015, 10:00 AM • 2 Y

Y by Adventure10, Mango247

I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$ . $D,E,F$ are projections of $P$ on $BC,CA,AB$ . $PD$ cuts $(DEF)$ again at $G$ . $d$ is the line passing through $G$ and parallel to $BC$ . $O$ is circumcenter of $ABC$ . $Q$ is isogonal conjugate of $P$ . $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$ . Prove that $A,X,Y,O$ lie on a circle.

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#12 h Jan 23, 2015, 5:30 PM • 2 Y

Y by Adventure10, Mango247

A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$ . $IY$ cuts $(\omega_a)$ again at $D$ . Define cyclically $E,F$ . Prove that $AD,BE,CF$ are concurrent on $OI$ .

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#13 h Jan 23, 2015, 7:42 PM • 5 Y

Y by buratinogigle, amar_04, Adventure10, Mango247, and 1 other user

buratinogigle wrote:

I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$ . $D,E,F$ are projections of $P$ on $BC,CA,AB$ . $PD$ cuts $(DEF)$ again at $G$ . $d$ is the line passing through $G$ and parallel to $BC$ . $O$ is circumcenter of $ABC$ . $Q$ is isogonal conjugate of $P$ . $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$ . Prove that $A,X,Y,O$ lie on a circle.

Dear buratinogigle, thank you for your interest and nice generalization

.

My solution:

Lemma:

Let $P$ be a point on the bisector of $\angle BAC$ .
Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ .
Let $R$ be the second intersection of $AQ$ and $\odot (QBC)$ .
Let $\ell$ be a line passing through $R$ and perpendicular to $OP$ .

Then the intersection $K$ of $BC$ and the perpendicular bisector of $AR$ lie on $\ell$ .
__________________________________________________
Proof of the lemma:

Let $X=\odot(ABC) \cap \odot (P, PA)$ .
Let $\Psi$ be the composition of Inversion $\mathcal{I}(A, \sqrt{AB \cdot AC})$ and reflection $\mathcal{R}(AP)$ .

Easy to see $\odot (ABC)$ is the image of the line $BC$ under $\Psi$ .

From $\angle PBA=\angle CBQ=\angle CRA \Longrightarrow \triangle ABP \sim \triangle ARC$ ,
so we get $AP \cdot AR=AB \cdot AC \Longrightarrow P$ is the image of $R$ under $\Psi$ ,
hence $\odot(P, PA)$ is the image of the perpendicular bisector of $AR$ under $\Psi$ ,
so $X\equiv \odot(ABC) \cap \odot (P, PA)$ is the image of $K$ under $\Psi$ .

Since $OP$ is the perpendicular bisector of $AX$ ,
so $OP$ pass through the center of $\odot (APX)$ ,
hence the tangent of $\odot (APX)$ at $P$ is perpendicular to $OP$ ,
so from $\angle (PA, OP)=90^{\circ}-\angle (\ell, PA )$ we get $\ell$ is the image of $\odot (APX)$ under $\Psi$ .
i.e. $K \in \ell$
____________________________________________________________
Back to the main problem:

Let $R=AP \cap \odot (PBC), B'=XY \cap AB, C'=XY \cap AC$ .
Let $D', E', F'$ be the projection of $Q$ on $BC, CA, AB$ , respectively .

Since $\angle CRB=180^{\circ}-\angle BPC=180^{\circ}-\angle BPD-\angle DPC$
$=180^{\circ}- \angle F'FD-\angle DEE'=180^{\circ}-\angle D'F'E'-\angle D'E'F'$
$=180^{\circ}-\angle PEQ-\angle PFQ =180^{\circ}-\angle PC'B'-\angle PB'C' =\angle C'PB'$ ,
so we get $\triangle RBC$ and $\triangle PB'C'$ are homothetic with center $A$ .

From lemma we get the perpendicular bisector of $AR$ and the perpendicular from $R$ to $OQ$ are concurrent on $BC$ ,
so after doing honothety with center $A$ which send $R \mapsto P$ and $BC \mapsto B'C'$ we get $YA=YP$ ,
hence $\angle YAO=\angle YAP+\angle PAO=\angle APY+\angle PAO =90^{\circ}-\angle OQA+\angle PAO$
$=90^{\circ}-\angle OQA+\tfrac{1}{2} \angle BAC+\angle CBA-90^{\circ}=\angle(BC, PA)-\angle OQA=\angle YXO$ .
i.e. $A, X, O, Y$ are concyclic

Q.E.D

Attachments:

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#14 h Jan 24, 2015, 1:19 AM • 2 Y

Y by Adventure10, Mango247

You are welcome dear Telv. Thanks for nice solution. Actually, the original problem is very nice. I try to find some more properties of circle $(\omega_a)=(AXYO)$ . But It seems this circle is quite specially.

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2312 posts

#15 h May 16, 2015, 4:11 PM • 5 Y

Y by THVSH, buratinogigle, amar_04, enhanced, Adventure10

buratinogigle wrote:

A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$ . $IY$ cuts $(\omega_a)$ again at $D$ . Define cyclically $E,F$ . Prove that $AD,BE,CF$ are concurrent on $OI$ .

My solution :

Let $A'$ be the reflection of $A$ in $L$ and $T=L \cap \odot (I), H=L \cap AA'$ .

From the solution in post #6 or post #7 we get $YI=YA \Longrightarrow Y$ is the center of $\odot (AA'I)$ .

From $OI= AI \cdot \frac{\sin \angle IAO}{\sin \angle AOI} \Longrightarrow \sin \angle AOI=\frac{AI \cdot \sin \angle IAO}{OI}= \frac{AI \cdot \sin \angle A'AI}{OI}=\frac{TH}{OI}$ ,

so combine $\triangle AA'I \sim \triangle AIO$ $\Longrightarrow \frac{AH}{YI}=\frac{AH}{YA}=\cos \angle YAH=\sin \angle AIA'=\sin \angle AOI=\frac{TH}{OI}$ ,

hence $\frac{AH}{TH}=\frac{YI}{OI} \Longrightarrow \triangle AHT \sim \triangle YIO \Longrightarrow \angle HAT=\angle IYO=\angle DAO \Longrightarrow \angle BAD=\angle TAC$ ,

so $AD$ is the isogonal conjugate of A-Nagel line WRT $\angle A \Longrightarrow AD$ pass through the exsimilicenter $Z$ of $\odot (I) \sim \odot (O)$ .

Similarly, we can prove $Z \in BE , Z \in CF$ $\Longrightarrow AD, BE, CF, OI$ are concurrent at the exsimilicenter of $\odot (I) \sim \odot (O)$ .

Q.E.D

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tranquanghuy7198

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tranquanghuy7198

#16 h May 16, 2015, 4:54 PM • 2 Y

Y by Adventure10, Mango247

My solution for the problem in post #11:
Lemma.
Let $ABC$ be a triangle, $P, Q$ are isogonal conjugates on the bisector of $\angle{BAC}$ and $S$ is the midpoint of arc $BC$ not containing $A$ . We have: $SP.SQ = SB^{2} = SC^{2}$

Back to our main problem:
We’ll prove that $YA = YP$ . Indeed:
$I$ is the midpoint of $PQ$
Let the points $K, D, M, S, J, G, R$ be constructed as in the figure.
$\angle{PJQ}$ = 90 $\Rightarrow$ $IP = IQ = IJ$
Now:
$YA = YP$ $\Longleftrightarrow$ $PJ.PY = PI.PA$ $\Longleftrightarrow$ $PR.PG = PI.PA$ $\Longleftrightarrow$ $\frac{PR}{SO}.QK.SO = PI.PA$ $\Longleftrightarrow$ $\frac{QP}{QS}.\frac{QK}{SM}.SM.SO = \frac{PQ}{2}.PA$ $\Longleftrightarrow$ $\frac{1}{QS}.\frac{TQ}{TS}.(2.SM.SO) = PA$ $\Longleftrightarrow$ $\frac{1}{QS}.\frac{TQ}{TS}.(SP.SQ) = PA$ (because $2.SM.SO = SM.SZ = SP.SQ$ ) $\Longleftrightarrow$ $\frac{TQ}{TS} = \frac{PA}{PS}$ $\Longleftrightarrow$ $\frac{SQ}{ST} = \frac{SA}{SP}$ $\Longleftrightarrow$ $SP.SQ = SA.ST$ (this is right)
Now we have: $\angle{OAY} = \angle{PAY}-\angle{PAO} = \angle{APY}-\angle{APG} = \angle{YPG} = \angle{OXY}$ and the conclusion follows.

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THVSH

#17 h May 16, 2015, 5:22 PM • 3 Y

Y by buratinogigle, Adventure10, Mango247

TelvCohl wrote:

My solution:

Lemma:

Let $P$ be a point on the bisector of $\angle BAC$ .
Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ .
Let $R$ be the second intersection of $AQ$ and $\odot (QBC)$ .
Let $\ell$ be a line passing through $R$ and perpendicular to $OP$ .

Then the intersection $K$ of $BC$ and the perpendicular bisector of $AR$ lie on $\ell$ .

(This lemma was mentioned by buratinogigle at here (post # 8).
I did not read Xml's solution but I think my proof is similar to Xml's proof

.)
__________________________________________________
Proof of the lemma:

Let $X=\odot(ABC) \cap \odot (P, PA)$ .
Let $\Psi$ be the composition of Inversion $\mathcal{I}(A, \sqrt{AB \cdot AC})$ and reflection $\mathcal{R}(AP)$ .

Easy to see $\odot (ABC)$ is the image of the line $BC$ under $\Psi$ .

From $\angle PBA=\angle CBQ=\angle CRA \Longrightarrow \triangle ABP \sim \triangle ARC$ ,
so we get $AP \cdot AR=AB \cdot AC \Longrightarrow P$ is the image of $R$ under $\Psi$ ,
hence $\odot(P, PA)$ is the image of the perpendicular bisector of $AR$ under $\Psi$ ,
so $X\equiv \odot(ABC) \cap \odot (P, PA)$ is the image of $K$ under $\Psi$ .

Since $OP$ is the perpendicular bisector of $AX$ ,
so $OP$ pass through the center of $\odot (APX)$ ,
hence the tangent of $\odot (APX)$ at $P$ is perpendicular to $OP$ ,
so from $\angle (PA, OP)=90^{\circ}-\angle (\ell, PA )$ we get $\ell$ is the image of $\odot (APX)$ under $\Psi$ .
i.e. $K \in \ell$

See the proof without inversion in this link i don't think it's...

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#18 h Mar 24, 2016, 4:34 PM • 2 Y

Y by baopbc, Adventure10

Here is a consequence of problem in #11, I get a solution base on solution of Telv Cohl in #13.

Problem. Let $ABC$ be a triangle with circumcenter $O$ and $P$ is a point on bisector of $\angle BAC$ . $Q$ is isogonal conjugate of $P$ . $QB,QC$ cut $CA,AB$ at $E,F$ . The line passes through $Q$ and is perpendicular to $OP$ cut $BC,EF$ at $M,N$ . Prove that $\frac{QM}{QN}=1+\frac{d(Q,BC)}{d(P,BC)}$ .

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doxuanlong15052000

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doxuanlong15052000

#19 h Jul 2, 2016, 3:52 PM • 1 Y

Y by Adventure10

buratinogigle wrote:

I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$ . $D,E,F$ are projections of $P$ on $BC,CA,AB$ . $PD$ cuts $(DEF)$ again at $G$ . $d$ is the line passing through $G$ and parallel to $BC$ . $O$ is circumcenter of $ABC$ . $Q$ is isogonal conjugate of $P$ . $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$ . Prove that $A,X,Y,O$ lie on a circle.

My solution:
Let $PY$ cut $OQ$ at $L$ , $PD$ cut $d$ at $G$ , $OQ$ cut $PD$ at $R$ , $U$ and $V$ be the midpoint of arc $BC$ and $BAC$ , $M$ is the midpoint of $BC$ , $K$ be the midpoint of $PQ, AP$ cut $BC$ at $T$ , $QR\perp BC$ , $OQ$ cut $PD$ at $H$ .
We have $\frac {PG}{UM}=\frac {QR}{UM}=\frac {QT}{TU}$ (well-know) $=\frac {PA}{PU}$ $\Longrightarrow$ $\triangle APG\sim \triangle PUM$ . Since $UP.UQ=UM.UV$ $\Longrightarrow$ $\triangle UQV\sim \triangle PGA$ $\Longrightarrow$ $\frac {PQ}{PH}=\frac {UQ}{2UO}=\frac {PG}{PA}$ $\Longrightarrow$ $PH.PK=PG.PA$ . Let $O_1$ be the midpoint of $AP$ $\Longrightarrow$ $PO_1.PQ=PG.PH=PY.PL$ $\Longrightarrow$ $YO_1\perp AP$ $\Longrightarrow$ $YA=YP$ . We have $\angle YXO=\angle HPQ+\angle QPL=\angle YPA+\angle OUA=\angle YAO$ $\Longrightarrow$ $A,X,Y,O$ lie on a circle

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#20 h Dec 13, 2016, 9:42 PM • 3 Y

Y by amar_04, Adventure10, Mango247

Beautiful Problem! Congratulations to Telv Cohl

v_Enhance wrote:

Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$ . A straight line $L$ is parallel to $BC$ and tangent to the incircle. Suppose $L$ intersects $IO$ at $X$ , and select $Y$ on $L$ such that $YI$ is perpendicular to $IO$ . Prove that $A$ , $X$ , $O$ , $Y$ are cyclic.

Proposed by Telv Cohl

Let $Y'$ be the reflection of $I$ wrt $Y$ . Time Travelling to Sharygin Olympiad 2016, we get $\angle IAY'=90^{\circ}$ so $YA=YI$ .

Suppose line $L$ touches the incircle at $D$ and let the $A$ -excircle touch $BC$ at point $E$ . It is well-known that $A, D, E$ are collinear. Let $K$ be the midpoint of $DE$ . $K$ lies on the perpendicular bisector of $BC$ .

Observe that $\angle IKO=\angle IDY=90^{\circ}$ and $\angle YIO=\angle KID=90^{\circ} \Longrightarrow \angle YID=\angle OIK.$ So, $\triangle IDY \sim \triangle IKO$ . By spiral similarity, $\triangle YIO \sim \triangle DIK$ . Thus, $\angle ADY=\angle IKD=\angle IOY$ .

Since $YA^2=YI^2=YD\cdot YX \Longrightarrow \angle YOI=\angle ADY=\angle XAY,$ points $A, X, O, Y$ lie on a circle, as desired. $\square$

This post has been edited 1 time. Last edited by anantmudgal09, Dec 13, 2016, 9:43 PM

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#23 h Apr 15, 2017, 4:08 AM • 1 Y

Y by Adventure10

buratinogigle wrote:

A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$ . $IY$ cuts $(\omega_a)$ again at $D$ . Define cyclically $E,F$ . Prove that $AD,BE,CF$ are concurrent on $OI$ .

Note that by d'Alembert theorem it's suffice to prove:
$\dfrac {Theorem}{}$ : let $A'$ be the tangent point of $A-$ mixtilinear circle and $(ABC)$ , then $A, A', D$ are collinear.

$\dfrac {Proof}{}$ : let $D=AA'\cap IY$ , then it's suffice to prove $A, X, O, D$ are concyclic.

$\dfrac {Lemma}{}$ : Given a triangle $\Delta ABC$ , let $I$ , $O$ be the incenter and circumcenter of $\Delta ABC$ , respectively. Let $A'$ be the tangent point of $A-$ mixtilinear circle and $(ABC)$ . $P$ is the midpoint of $\widehat {BAC}$ . Then $OI$ , the line perpendicular to $AA'$ at $A$ , the line tangent to $(ABC)$ at $P$ is concurrent.

$\dfrac {Proof of Lemma}{}$ : let the antipode of $A'$ in $(ABC)$ be $A''$ , $A_1$ be the midpoint of $\widehat {BC}$ (not contain $A$ ), since it's well-known $A', I, P$ are collinear, then use Pascal's theorem to $PPA_1AA''A'$ complete the proof.

$\dfrac {Back to the main problem}{}$ : let $OI\cap AA'=K$ , $AA'\cap \odot I=S$ , by d'Alembert theorem $K$ is the external homothetic center of $(ABC)$ and $\odot I$ , so use the lemma we have $L$ , $OI$ , the line perpendicular to $AA'$ at $S$ are concurrent $\Rightarrow$ $D, S, I, X$ are concyclic. Since $SI\parallel AO$ $\Rightarrow$ $A, O, D, X$ are concyclic. $\Box$

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#24 h Jun 10, 2017, 10:33 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

Let $U\equiv L\cap \odot(I)$ , $Y'\equiv YI\cap BC$ , and suppose $V$ is the $A$ -mixtilinear touchpoint on $\odot(ABC)$ . By angle chasing and using the fact that $\{AU,AV\}$ are isogonal, we establish that $O$ is the spiral center sending $YY'$ to $AV$ ; consequently, $Y'V=AY$ . On the other hand, we know that $IY'=Y'V$ , so $AY=Y'V=Y'I=YI$ . Let $M$ be the midpoint of $\overline{BC}$ ; then $AY^2=YI^2=YU\cdot YX$ , so $\angle XAY=\angle AUY=\pi-\angle IMY'=\pi-\angle IOY$ as desired.

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#25 h Apr 3, 2020, 2:09 AM

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$[asy] size(7cm); defaultpen(fontsize(10pt)); pair O=(0,0); pair A=dir(121); pair B=dir(195); pair C=dir(345); pair I=incenter(A,B,C); pair D=foot(I,B,C); pair Dp=2I-D; pair X=extension(I,O,Dp,Dp+B-C); pair Y=extension(X,Dp,I,I+rotate(90)*(O-I)); pair Xp=2I-X; pair Yp=2I-Y; draw(Xp--X); draw(Y--Yp); draw(circumcircle(A,X,Y)); draw(incircle(A,B,C)); draw(A--B--C--A); draw(X--Y); dot("$O$",O,dir(210)); dot("$A$",A,N); dot("$B$",B,dir(120)); dot("$C$",C,SE); dot("$I$",I,S); dot("$D$",D,N); dot("$D'$",Dp,S); dot("$X$",X,NW); dot("$Y$",Y,NE); dot("$X'$",Xp,S); dot("$Y'$",Yp,S); [/asy]$

Use complex numbers with circumcircle as the unit circle. Let $x'=2i-x$ be $\overline{BC}\cap\overline{OI}$ . Then
$\begin{align*} x'&=\frac{(\overline bc-b\overline c)(i-o)-(\overline io-i\overline o)(b-c)}{(\overline b-\overline c)(i-o)-(\overline i-\overline o)(b-c)}\\ &=\frac{i\left(\frac cb-\frac bc\right)}{i\left(\frac1b-\frac1c\right)-\overline i(b-c)}=\frac{i\cdot\frac{c^2-b^2}{bc}}{i\cdot\frac{c-b}{bc}+\overline i(c-b)}\\ &=\frac{i(c+b)}{i+\overline ibc}. \end{align*}$ Hence we compute
$\begin{align*} x&=2i-x'=i\left(2-\frac{b+c}{i+\overline ibc}\right)\\ &=i\left(\frac{2i+2\overline ibc-b-c}{i+\overline ibc}\right). \end{align*}$
Now let $y'=2i-y$ lie on $\overline{BC}$ . The intersections $R$ , $S$ of $\overline{IY}$ with circumcircle are the roots of the quadratic in $t$ :
$\begin{align*} \frac{t-i}i\in\mathbb I&\iff\frac{t-i}i+\frac{\frac1t-\overline i}{\overline i}=0\\ &\iff t^2-ti+i/\overline i-ti=0\\ &\iff t^2-2i\cdot t+i/\overline i=0. \end{align*}$ Hence $r+s=2i$ , $rs=i/\overline i$ , and
$\begin{align*} y'&=\frac{bc(r+s)-rs(b+c)}{bc-rs}\\ &=\frac{2bci-\frac i{\overline i}(b+c)}{bc-\frac i{\overline i}} =\frac{2bci\overline i-i(b+c)}{bc\overline i-i}. \end{align*}$ Hence we compute
$\begin{align*} y&=2i-y'=2i-\frac{2bci\overline i-i(b+c)}{bc\overline i-i}\\ &=\frac{-2i^2+i(b+c)}{bc\overline i-i}=\frac{i(b+c-2i)}{bc\overline i-i}. \end{align*}$ Let $P=\frac{a-y}a$ , $Q=\frac{x-y}x$ , $F=\frac PQ$ . It suffices to show $F\in\mathbb R$ . Let $a=u^2$ , $b=v^2$ , $c=w^2$ , so that $i=-(uv+vw+wu)$ .

We have
$\begin{align*} P&=\frac{a-y}a=1-\frac ya =1-\frac{i(b+c-2i)}{a(bc\overline i-i)}=\frac{abc\overline i-i(a+b+c-2i)}{a(bc\overline i-i)}\\ &=\frac{-u^2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)+(uv+vw+wu)(u+v+w)^2}{u^2\left(-v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)+uv+vw+wu\right)}\\ &=\frac{(u+v+w)\left[-uvw+(uv+vw+wu)(u+v+w)\right]}{u\left(u^2-vw\right)(v+w)}\\ &=\frac{u+v+w}{u\left(u^2-vw\right)(v+w)}\left[2uvw+\sum_\mathrm{sym}u^2v\right]\\ &=\frac{(u+v+w)(u+v)(v+w)(w+u)}{u\left(u^2-vw\right)(v+w)}\\ &=\frac{(u+v+w)(u+v)(u+w)}{u\left(u^2-vw\right)}. \end{align*}$ The main computation:
$\begin{align*} Q&=\frac{x-y}x=1-\frac yx=1-\left(\frac{i(b+c-2i)}{bc\overline i-i}\bigg/\frac{i(2i+2\overline ibc-b-c)}{i+\overline ibc}\right)\\ &=\frac{(bc\overline i-i)(2i+2\overline ibc-b-c)-(b+c-2i)(i+\overline ibc)}{(bc\overline i-i)(2i+2\overline i-b-c)}\\ &=\frac{2(i+\overline ibc)(bc\overline i-i)-(b+c)(bc\overline i-i)-(b+c)(bc\overline i+i)+2i(i+\overline ibc)}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\ &=\frac{2(i+\overline ibc)bc\overline i-2(b+c)bc\overline i}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\ &=\frac{2bc\overline i(i+\overline ibc-b-c)}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\ &=\frac{2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\left(v^2+w^2+uv+vw+wu+v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\right)}{\left(v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)-uv-vw-wu\right)\left(v^2+w^2+2(uv+vw+wu)+2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\right)}\\ &=\frac{\left[2\frac{vw}u(u+v+w)\right]\cdot\left[\frac1u\left(2uvw+\sum_\mathrm{sym}u^2v\right)\right]}{\left[\frac1u\left(u^2-vw\right)(v+w)\right]\cdot\left[\frac1u\left(2uvw(u+v+w)+u(u+v+w)^2-u^3\right)\right]}\\ &=\frac{2vw(u+v+w)(u+v)(u+w)}{\left(u^2-vw\right)\big(2vw(u+v+w)+u(v+w)(2u+v+w)\big)}. \end{align*}$
From here, $F=\frac PQ=\frac{2uvw(u+v+w)+u(v+w)(2u+v+w)}{2uvw},$ and we can check
$\begin{align*} \overline F&=\frac{\frac2{vw}\left(\frac1u+\frac1v+\frac1w\right)+\frac1u\left(\frac1v+\frac1w\right)\left(\frac2u+\frac1v+\frac1w\right)}{\frac2{uvw}}\\ &=\frac{2u(uv+vw+wu)+(v+w)(2vw+wu+uv)}{2uvw}\\ &=\frac{2vw(u+v+w)+2\left(u^2v+u^2w\right)+(v+w)(uw+uv)}{2uvw}\\ &=\frac{2vw(u+v+w)+u(v+w)(2u+v+w)}{2uvw}=F. \end{align*}$ This implies $F\in\mathbb R$ , and we are done.

This post has been edited 1 time. Last edited by Asuboptimal, Apr 3, 2020, 2:10 AM

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#27 h Apr 3, 2020, 9:24 PM • 1 Y

Y by Mango247

For anyone who is bashing with $I=0$ we get : $\bar x=-\frac{2 \sum (xy)}{(y+z)(x^2+yz)+2xyz}$ and $\bar y=\frac {2\sum (xy)}{(y+z)(yz-x^2)}$

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jayme

#28 h Jul 8, 2020, 12:54 PM

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Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/24.%203.%20Cocyclite.pdf p. 6…

Sincerely
Jean-Louis

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#29 h Jul 3, 2023, 11:19 AM • 1 Y

Y by GeoKing

It is a well known lemma that the locus of points such that the sum of their oriented distances to the sides of $\triangle ABC$ is constant, is a line perpendicular to $OI$ . In particular, letting $r$ denote the inradius of $\triangle ABC$ , and using $YI\perp OI$ this implies that
$3r=d(I, BC)+d(I, AC)+d(I, AB)=d(Y, BC)+d(Y, AB)+d(Y, AC)=2r+d(Y, AB)+d(Y, AC),$ so we learn $d(Y, AB)+d(Y, AC)=r$ . Now, I claim this implies the equality $YA=YI$ . Let $K\in AC$ such that $IK=IA$ , whence $IK\parallel AB$ . By the above we find that $d(Y, IK)=d(Y, KA)$ , so $Y$ lies on the interior angle bisector of $\angle AKI$ , and the claim follows.

Finally, we can conclude by a long but straightforward angle chase (directing angles modulo 180). First we compute
$\angle AYX=\angle (AY, BC)=\angle ACB+\angle IAC-\angle IAY=\angle ACB+\angle IAC-\angle YIA.$ Now, by definition $\angle YIA=\angle OIA-90$ , so upon substituting
$\angle AYX=90+\angle ACB+\angle IAC-\angle OIA.$ On the other hand
$\angle AOX=\angle AOI=180-\angle OIA-\angle IAO=180-\angle OIA-(90-\angle ACB-\angle BAI)=90+\angle ACB+\angle BIA-\angle OIA,$ so indeed $\angle AOX=\angle AYX$ , so $AYXO$ is cyclic, as desired.

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#30 h Dec 24, 2024, 9:12 PM

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Very nice config, let $IY \cap BC=Z$ and also $M$ midpoint of $BC$ and also $OY \cap (OIMZ)=J$ (where that circle exists due to $\angle OIZ=90=\angle OMZ$ ), also let $D'$ touchpoint of incircle with $L$ , from I-E Lemma we have that perpendicular bisector of $AI$ is the line connecting midpoints of minor arcs $AB,AC$ in $(ABC)$ and from a little angle chase we notice that $AO,ID'$ are symetric on the perpendicular bisector of $AI$ , so now also note that $IY=IZ$ and $\angle YJZ=90$ which means that $IJ=IZ$ which means from arcs that $\angle XAY=\angle IZJ=\angle IJZ=\angle IMB=\angle AD'Y$ (from $IM \parallel AD'$ by homothety)m which means that $YA^2=YD' \cdot YX=YI^2$ so $YA=YI$ which means $Y$ lies on the perpendicular bisector of $AI$ and therefore $\angle YAO=\angle YID'=\angle YXO$ therefore $AXOY$ is cyclic as desired, thus we are done :cool:

:cool:

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