Concurrency from isogonal Mittenpunkt configuration

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Source: Fake USAMO 2020 P3

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#1 h Apr 28, 2020, 7:07 AM • 5 Y

Y by amar_04, magicarrow, Purple_Planet, CaptainLevi16, Funcshun840

Let $\triangle ABC$ be a scalene triangle with circumcenter $O$ , incenter $I$ , and incircle $\omega$ . Let $\omega$ touch the sides $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ at points $D$ , $E$ , and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$ . The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$ . The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $\text{[math]}$ , $FP$ , and $OI$ are concurrent.

Proposed by MarkBcc168.

This post has been edited 1 time. Last edited by MarkBcc168, Apr 28, 2020, 7:08 AM

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#2 h Apr 28, 2020, 7:07 AM • 10 Y

Y by amar_04, Pluto04, MarkBcc168, Kagebaka, samuel, ValidName, Purple_Planet, mathlogician, JG666, Funcshun840

All poles and polars will be taken with respect to $\omega$ .

Let $M = EF \cap PQ$ and $N = EP \cap FQ \cap AX$ be the radical center of $\odot(AEX), \odot(AFX), \odot(ABC)$ . By La Hire's theorem, $A$ lies on the polar of $M$ . By Brokard's theorem, $N$ also lies on the polar of $M$ . It follows that $AT$ is the polar of $M$ . By Brokard's theorem $IS \perp MN$ where $S = EQ \cap FP$ so it is enough to prove $OI \perp MN$ . We claim that $MN$ is the radical axis $\tau$ of $\odot(ABC)$ and $\omega$ . We have $\text{Pow}(N, \omega) = NP \cdot NE = NX \cdot NA = \text{Pow}(N, \odot(ABC))$ so $N \in \tau$ . Since $T$ lies on the polar of $M$ we have $-1 = (E, F; M, T)$ . Let $R$ be the second intersection of $\odot(ABC)$ and $\odot(AEIF)$ . Then through inverting about $\omega$ we can see that $R, I, T$ are collinear whence $-1 = (A, I; E, F) \stackrel{R}{=} (RA \cap EF, T; E, F)$ so $R, A, M$ are collinear. Hence, $\text{Pow}(M, \odot(ABC)) = MQ \cdot MA = ME \cdot MF = \text{Pow}(M, \omega)$ so $M \in \tau$ also and we are done.

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#3 h Apr 28, 2020, 7:08 AM • 2 Y

Y by amar_04, Purple_Planet

Let $\Gamma$ be the circumcircle of $\triangle ABC$ , and let $\ell$ be the radical axis of $\omega$ and $\Gamma$ . By Radical Axes Theorem on $\odot (AEX),\omega,\Gamma$ , we get that $EP$ and $AX$ meet on $\ell$ . Similarly, applying Radical Axes Theorem to $\odot (AFX),\omega,\Gamma$ , we have that $FQ,AX$ concur on $\ell$ . Combining the above two observations, we can say that $EP,FQ,AX$ meet on $\ell$ at some point $K$ . Let $L$ be the point on $EF$ such that $(L,T;E,F)=-1$ . From now on, all poles/polars are taken wrt $\omega$ . Note that, since $L$ lies on the polar of $A$ , so by La Hire, $A$ must lie on the polar of $L$ . As $T$ lies on the polar of $L$ by definition, so we get that $AT$ is the polar of $L$ . Using Pascal Theorem on $EEQFFP$ , we have that $Z=EQ \cap FP$ lies on line $AU$ , which is nothing but the polar of $L$ . By La Hire's Theorem, we also get that $L$ lies on the polar of $Z$ . But, from Brokard's Theorem on $EFPQ$ , we know that $U$ is on the polar of $Z$ . So $Z$ must be the pole of line $LU$ . In particular, this gives $IZ \perp LU$ . So showing $Z \in OI$ is equivalent to proving that $OI \perp LU$ . As $U \in \ell$ , and $OI \perp \ell$ , so it suffices to to show that $L$ also lies on the radical axis of $\omega$ and $\Gamma$ . For this we require the following Lemma.

LEMMA $\omega,\Gamma$ and the nine point circle of $\triangle DEF$ are coaxial. In other words, $\ell$ is also the radical axis of $\omega$ and the nine-point circle of $\triangle DEF$ .

(Proof) Let $M$ be the midpoint of $EF$ . From Coaxality Lemma, and the fact that what we have to show is symmetric with respect to $D,E,F$ , we just wish to prove that $\frac{ET \cdot EM}{EA \cdot EC}=\frac{FT \cdot FM}{FA \cdot FB} \iff \frac{ET}{FT}=\frac{EC}{FB}$ where we use $EM=FM$ and $EA=FA$ . But, this follows from $\frac{ET}{FT}=\frac{DE \cos \angle DEF}{DF \cos \angle DFE}=\frac{2EC \cos \angle CED \cdot \cos \angle BFD}{2FB \cos \angle BFD \cdot \cos \angle CED}=\frac{EC}{FB}$ This completes the proof of the Lemma. $\Box$

Return to the problem. Let $R,S$ be the foot of perpendicular from $E,F$ to $DF,DE$ respectively. Since $(L,T;E,F)=-1$ , so line $RS$ passes through $L$ (This follows from an easy application of Ceva and Menelaus Theorem). But, $EFRS$ is cyclic, and hence, by Power of Point, we get $LE \cdot LF=LR \cdot LS$ . Thus, $L$ lies on the radical axis of $\omega$ and the nine point circle of $\triangle DEF$ . From our Lemma, we have $L \in \ell$ , as desired. $\blacksquare$

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#4 h Apr 28, 2020, 7:08 AM • 12 Y

Y by Orestis_Lignos, teomihai, GeoMetrix, Pluto04, AmirKhusrau, A-Thought-Of-God, BinomialMoriarty, Purple_Planet, Bumblebee60, Abhaysingh2003, CaptainLevi16, iamnotgentle

Wonderful! But easy IMHO.

USAMO Mock D1P3 wrote:

Let $\triangle ABC$ be a scalene triangle with circumcenter $O$ , incenter $I$ , and incircle $\omega$ . Let $\omega$ touch the sides $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ at points $D$ , $E$ , and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$ . The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$ . The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $\text{[math]}$ , $FP$ , and $OI$ are concurrent.

Proposed by MarkBcc168.

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By taking Pairwise Radical Axis of $\{\odot(AEX),\odot(I),\odot(AFX)\}$ we get that $\{FQ,XA,PE\}$ concurs at a point $\{U\}$ . Let $EF\cap PQ=\{Z\}$ . So, by Brocard's Theorem we get that $U\in\text{ Polar of } Z$ WRT $\odot(I)$ and as $Z\in\text{ Polar of } A$ WRT $\odot(I)$ . So, by La Hire's Theorem we get that $A\in\text{ Polar of }Z$ WRT $\odot(I)$ . Hence, $\{UA\}$ is the Polar of $Z$ WRT $\odot(I)$ . But by Brocard's Theorem we again get that $\{V\}=EQ\cap FP\in\text{ Polar of }Z$ WRT $\odot(I)$ . So, $\{EQ,FP,AX\}$ are concurrent at $\{V\}$ . Now if $\odot(X_5)$ is the Nine Point Circle of $\triangle DEF$ then $\{T\}=EF\cap\odot(X_5)$ and $T\ne$ Midpoint of $EF$ . Now consider an Inversion $(\Psi)$ around $\odot(I)$ . $\Psi:\{T\}=EF\cap\odot(X_5)\leftrightarrow\{\Psi(EF)\cap\Psi(\odot(X_5))\}=\{\odot(AI)\cap\odot(ABC)\}$ . So $\{Y\}=IT\cap\odot(ABC)\in\odot(AI)$ . So, if $AY\cap EF=Z'$ then $\angle Z'YT=90^\circ$ and as $YFIE$ is a cyclic quadrilateral and $IF=IE$ . So, $YT$ bisects $\angle FYE$ . Hence, $(Z',T;F,E)=-1$ . But notice that $T\in\text{ Polar of }Z$ WRT $\odot(I)\implies (Z,T;F,E)=(Z',T;F,E)=-1\implies Z'\equiv Z\implies \{AY,EF,QP\}$ are concurrent at $\{Z\}$ . So, $ZY\cdot ZA=ZQ\cdot ZP=ZF\cdot ZE\implies Z$ has equal Power WRT $\{\odot(I),\odot(ABC)\}$ and also $UF\cdot UQ=UE\cdot UP=UA\cdot AX\implies U$ has equal Power WRT $\{\odot(I),\odot(ABC)\}$ . So, $\{UZ\}$ is the Radical Axis of $\{\odot(I),\odot(ABC)\}\implies OI\perp UZ$ . But by Brocard's Theorem we get that $IV\perp UZ\implies \{\overline{O-I-V}\}$ . Hence we get that $\{EQ,FP,OI\}$ are concurrent. $\blacksquare$

This post has been edited 2 times. Last edited by amar_04, Apr 28, 2020, 8:05 AM

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#5 h Apr 28, 2020, 2:25 PM • 1 Y

Y by Purple_Planet

not as good as the above solutions but posting anyway :whistling:

:whistling:

Let $L_A$ be the midpoint of arc $BC$ not containing $A$ on $(ABC)$ , and $L_{A_0}$ its antipode. Let $A_0$ be the antipode of $A$ on $(ABC)$ and $R$ the intersection of $(AI)$ with $(ABC)$ other than $A$ . It is well-known that $R,T,I,A_0$ are collinear and that $R,D,L_A$ are collinear. Let $T_A$ be the $A$ -mixtilinear touch point of $\triangle ABC$ and $U:= AR \cap BC$ satisfies $UI \perp AI$ by radical axis on $(AEF),(BIC),(ABC)$ . Moreover, $U\in L_AT_A$ is a well-known mixtilinear fact. Now define $V = L_{A_0}R \cap BC$ , we claim that $X\in VL_A$ . Projecting onto $(ABC)$ from $L_A$ it is equivalent that $(BC;RX) = -1$ . Projecting this from $A$ onto line $EF$ it's equivalent that $(T_1T;EF) = -1$ where $T_1 := AR \cap EF$ , but this is true because $AR$ is the pole of $T$ with respect to the incircle ( $R$ is the inverse of $T$ wrt incircle and $RA \perp RI$ ). Moreover since $UD$ is tangent to the incircle and $D_2 := DT \cap (DEF) \setminus \{D\}$ it follows that $U$ is the pole of $DD_2$ wrt incircle. Moreover, $A,D_2,T_A$ are collinear because $AT_A$ and $AD_1$ are isogonal where $D_1$ is the reflection of $D$ over $I$ .

We now claim that $VRTDX$ is cyclic with diameter $VD$ . Angle chase reveals that $\frac{\widehat{BR} + \widehat{CL_A}}{2} = \frac{\widehat{RL_A}}{2} \implies (RDXY)$ is cyclic. Moreover
$\angle IDD_2 =|\angle E - \angle F| = \frac{|\angle B - \angle C|}{2} = \angle L_AAA_0 = \angle L_ARA_0 \implies ID$ is tangent to $(RD_2D)$ , but $\measuredangle VRD = \measuredangle VD_2D = 90^\circ \implies RVD_2D$ cyclic $\implies X \in (RD_2DV)$ , as desired.
This implies that $L_{A_0},D,X$ are collinear.

Now we note that $O,I$ and the center of the image of $(ABC)$ under inversion wrt $(DEF)$ are collinear, hence $OI$ is the Euler line of $\triangle DEF$ .

Let $X'$ be the image of $X$ after inversion around $(DEF)$ , then $X'$ lies on the ninepoint circle $\gamma$ of $\triangle DEF$ and $IX$ ; moreover it lies on $(DVR)$ as well since $(DVR)$ is orthogonal to $(DEF)$ . Therefore $\{X',T\}$ are the intersections of $\gamma$ and $(VRTX)$ . By radical axis it follwos that $EP, FQ, AX, A'X'$ concur where $A'$ is the midpoint of $EF$ .

Moreover we note that $AIDX$ is cyclic since $\measuredangle DXA = \measuredangle DVT = \measuredangle DWA = \measuredangle DIA$ where $W:= AL_{A_0}\cap BC$ , hence $D,X',A'$ are collinear.

It is a well-known fact (verifiable by barycentric coordinates) that for any triangle $DEF$ , if $X'$ is the second intersection of the nine-point circle with the D-median $DA'$ , $P$ the second intersection of $(A'EX') \cap (DEF)$ and similarly define $Q$ , then $FP \cap EQ$ lies on the Euler line of $\triangle DEF$ , so we are done.

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#6 h Apr 28, 2020, 3:57 PM • 10 Y

Y by cmsgr8er, GeoMetrix, BinomialMoriarty, ayan_mathematics_king, Purple_Planet, Bumblebee60, Mango247, MS_asdfgzxcvb, ehuseyinyigit, Funcshun840

It's well known that the Perspector of the Orthic Triangle of the Contact Triangle of $\Delta ABC$ and $\Delta ABC$ is the Isogonal Mittenpunkt Point $(X_{57})$ (See #11 here ) and $X_{57}$ lies on $\overline{OI}$ (Well known). Hence, the Concurrency Point is the $X_{57}$ from Kimberling's Triangle Center.

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SpecialBeing2017

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#7 h May 1, 2020, 6:06 PM • 1 Y

Y by Purple_Planet

Here is what I submitted to MarkBcc168

Problem 3

Attachments:

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#8 h May 18, 2020, 12:36 AM

Y by

Here is my terrible solution.

Let $A\neq G=(AEF)\cap (ABC)$ .

Lemma

Proof.

Finish

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#9 h Jun 26, 2020, 11:25 PM • 1 Y

Y by JG666

We make the following point definitions:

$T = EF \cap PQ$
$R = FQ \cap EP$
$Z = AX \cap EF$
$K = EQ \cap FP$

Let $\Gamma$ be the nine-point circle of $\triangle DEF$ . To start, notice that by radical axes on $(AEPX), (AFQX), (DEF)$ the points $A,X,R$ are collinear. Moreover, Pascal on $FFQEEP$ yields that $A, R, K$ are collinear. Let $M$ be the midpoint of $EF$ . Note that $(T,Z;E,F) = -1$ , so $TE \cdot TF = TZ \cdot TM$ , so $T$ lies on the radical axis of $(DEF)$ and $\Gamma$ . In addition, it is well-known that the circumcircle $(ABC)$ , the incircle $(DEF)$ , and $\Gamma$ are coaxial, say, by inverting around $I$ , so $T$ lies on the radical axis of $(DEF)$ and $(ABC)$ .

Furthermore, since $(AFQX)$ is cyclic, $RQ \cdot RF = RX \cdot RA$ , so $R$ lies on the radical axis of $(ABC)$ and $(DEF)$ . Now $TR$ is the radical axis of $(ABC)$ and $(DEF)$ , so $OI \perp TR$ . But by Brocard on $EFQP$ , note that $IK \perp TR$ , so $I,O,K$ is collinear, as desired.

Remark: Nice problem! Felt easy for a P3, but still spent quite a long time on it, oops :blush:

:blush:

This post has been edited 1 time. Last edited by mathlogician, Jun 26, 2020, 11:33 PM

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#11 h Oct 26, 2020, 11:37 AM • 2 Y

Y by amar_04, Nathanisme

Nice but rather easy since the configuration of $\odot(AEF) \cup T$ is well known.
Name $U= \overline{EP} \cap \overline{FQ}$ , $S= \overline{EQ} \cap \overline{FP}$
$R = \overline{EF} \cap \overline{PQ}$ , $M = \frac{E + F}{2}$
Claim 1: $\overline{S,A,T,U,X}$ are collinear.
Proof: Notice $S$ is concurrency of radax of $\odot (EQA) , \omega$ and $\odot(PFA) , \omega$ so $S =$ radical center of $\omega , \odot(AEQ),\odot(AFP) \implies S \in \overline{AX}$
So $\overline{S,A,T,X}$ are collinear.
$\textrm{All pole polars taken wrt}$ $\omega$ .
Notice $\overline{SU} = \mathcal{P}(R)$
$R \in \overline{EF} = \mathcal{P}(A) \implies A \in \mathcal{P}(R)$
So $A \in \overline{SU} \implies \overline{S,A,T,U,X}$ are collinear.

Let $L= \odot (AEF) \cap \odot(ABC)$ and $R' = \overline{AL} \cap \overline{EF}$
Claim 2: $R \equiv R'$
Notice by SDL , $\angle ALT = 90 \implies \odot(ALTM)$ is cyclic.
So By PoP $R'F \cdot R'E =R'L \cdot R'A = R'T \cdot R'M \implies (R',T,F,E)=-1$
But we knew $\overline{S,T,U}= \mathcal{P}(R) \implies (R,T,F,E)=-1$
So $R \equiv R'$ $\square$

Claim 3: $S,R$ are on radax of $\omega , \odot(ABC)$
Proof: Notice by PoP $RF \cdot RE = RL \cdot RA$ and $SF \cdot SP = SA \cdot SX$ $\square$

Ending: $O,I,U$ are collinear.
Proof:
By Brocard Theorem on $(FEQP) \cup \{S\} \cup \{R\}$ , we get $\overline{IU} \perp \overline{SR}$
By $\textbf{Claim 3}$ $R,S$ are on radax of $\omega , \odot (ABC) \implies$ $\overline{RS}$ is the radax so $\overline{OI} \perp \overline{RS}$
So $O,I,U$ are collinear.

$\textbf{Diagram}$

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#12 h Aug 15, 2024, 12:12 AM

Y by

Let $N$ be the major arc midpoint. It is well known that $NX$ passes through $D$ ; for a proof of this, note that $T$ is the $BC$ -harmonic conjugate of the $A$ -sharkydevil point by say, projection through $A$ onto $(AEF)$ .

Now by 2002 G7 it suffices to prove that $(ABP)$ is tangent to the incircle as that implies that $FP$ passes through $X_{57}$ . Perform a $\sqrt{bc}$ inversion.

We see $D'$ gets sent to the $A$ -mixtilinear extouch point. $N'$ is then $AN \cap BC$ . We have $X' = (D'AN') \cap \overline{BC}$ , from which we angle chase $\measuredangle ND'A = \measuredangle ND'B + \measuredangle BD'A = \measuredangle CBN + \measuredangle BNA = \measuredangle N'BN + \measuredangle BNN' = \measuredangle X'N'A = \measuredangle X'D'A$ , whence $D'X'N$ collinear.

Now $P'$ is mapped to $F'X' \cap (D'E'F')$ .

Let $M$ be the minor arc midpoint, $I_a$ the $A$ -excenter. We observe that $(MD'I_a)$ is tangent to $(BIC)$ , so radical axis gives $MD'$ , $E'F'$ (mixtlinear excircle properties) and $BC$ concur at a point $Y$ . Perspectivity at $X'$ reveals that $(B, C; X', Y) = -1$ . Perspectivity at $F'$ gives that the tangents at $E'$ and $P'$ concur on line $F'B$ , whence $BP'$ is tangent, and so after undoing the inversion we are done.

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Batsuh

#13 h Oct 15, 2024, 11:25 AM

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First we need a lemma.
\textbf{lemma:} Building on top of the main diagram, let $I_{A}, I_{B}, I_{C}$ be the excenters of $\triangle ABC$ . Then, the lines $I_{A}D, I_{B}E,I_{C}F$
and $IO$ are concurrent.
\textbf{proof:} Let $D' = EF \cap BC$ , $G=I_{A}D \cap \omega \ne D$ and $L$ be the reflection of $D$ over $I$ .
Note that $-1 = D(A,AI\cap BC;I,I_{A}) \overset{\omega}{=} (AD \cap \omega,D;L,G)$ which implies that $GL$ passes through the intersection of $BC$ and the tangent to $\omega$ at $AD\cap \omega$ , which by an easy pole-polar argument is
just the point $D'$ . Therefore, since $\angle D'GD = 90$ , we have $KG = KD$ which means that $KG$ is tangent to $\omega$ .
In particuler, the polar of $K$ wrt $\omega$ is line $I_AD$ .
On the other hand, since $KD^{2} = KB\cdot KC$ , the point $K$ must line on the radical axis of $\omega$ and $(ABC)$ . Therefore, by La-Hire, the pole
of the radical axis wrt $\omega$ must lie on the polar of $K$ , or $I_AD$ . Doing this for all three sides, we get the concurrency of $I_AD, I_BE, I_CF$ .
Call this concurrency point $V$ . Then, the radical axis $\ell$ of $\omega$ and $(ABC)$ is the polar of $V$ wrt $\omega$ which means that $IV \perp \ell$ .
Also, we have $IO \perp \ell$ as well, so $I,O$ and $V$ are collinear, as desired.

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Next, it's easy to see by angle chasing that triangles $\triangle DEF$ and $\triangle I_{A}I_{B}I_{C}$ are homothetic with center $V$ .
Then, since $\frac{FT}{TE} = \frac{I_{C}A}{AI_{B}}$ points $A, T$ and $V$ must be collinear. Therefore, all four points $A,T,V,X$ line on one line.

Now, by radical axis, let $S$ be the concurrency point of $FQ, EP$ and $AX$ . Clearly, $S$ lies on $\ell$ (the radical axis of $\omega$ and $(ABC)$ ). This means that $V$
lines on the polar of $S$ wrt $\omega$ . On the other hand, by Brocards theorem, $V'=FP \cap EQ$ is also on the polar of $S$ . Additionally, by pascals theorem on
$FFQEEP$ , we get that $A$ , $V'$ and $S$ are collinear. This is enough to force $V'=V$ , thereby solving the problem.

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684 posts

#14 h Nov 20, 2024, 4:36 PM

Y by

Let $EP\cap FQ=S,EF\cap PQ=R,EW\cap FP=K$ and let $U$ be $A-$ sharky devil point.
Since radical axises of $(PQEF),(AXQF),(AXPE)$ are concurrent, we see that $S$ lies on $AX$ .

Claim: $K$ lies on $AS$ .
Proof: $R$ is on the polar line of $A$ with respect to the incircle hence by La Hire, $A$ lies on the polar line of $R$ . And polar of $R$ is $KS$ thus, $K,S,A$ are collinear. $\square$

Claim: $RS$ is the radical axis of $(I)$ and $(ABC)$ .
Proof: Since $Pow(S,(I))=SP.SE=SX.SA=Pow(S,(ABC))$ , we conclude that $S$ lies on the radical axis of $(I)$ and $(ABC)$ . Set $AT\cap RI=L$ .
Note that $I,T,U$ are collinear. Since $RT\perp AI$ and $RI\perp AT$ by Brocard, we see that $A,T,R,I$ is an orthogonal system which implies $AR\perp IT\perp AU$ thus, $A,R,U$ are collinear. We have $Pow(R,(ABC))=RU.RA=RF.RE=Pow(R,(I))$ . So $R$ also lies on the radical axis of $(I)$ and $(ABC)$ . $\square$

$I,O$ are the circumcenters of $(I)$ and $(ABC)$ hence $IO\perp RS$ . Also by Brocard, $IK\perp RS$ which yields the collinearity of $I,O,K$ as desired. $\blacksquare$

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cursed_tangent1434

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cursed_tangent1434

#15 h Feb 15, 2025, 11:26 AM

Y by

Beautiful problem. Evan is Old Point once again. A rather different solution from most of the ones above. We denote by $I_b$ and $I_c$ the $B-$ excenter and $C-$ excenter of $\triangle ABC$ respectively. Let $L$ denote the $BC$ major arc midpoint. We start off with the following observation.

Claim : Quadrilaterals $FTPX$ and $ETQX$ are cyclic.

Proof : Observe that,
$\measuredangle TXP = \measuredangle AXP = \measuredangle CEP = \measuredangle EFP = \measuredangle TFP$ which implies that $FTPX$ is cyclic. Similarly we can show that $ETQX$ is cyclic, as desired.

Now note that by Radical Center Theorem on circles $(DEF)$ , $(XPF)$ and $(XQE)$ it follows that lines $\overline{TX}$ , $\overline{FP}$ and $\overline{EQ}$ concur at a point. We wish to show that this point also lies on $\overline{OI}$ which we shall now do. The following is the key claim.

Claim : Quadrilaterals $AEXI_c$ and $AFXI_b$ are cyclic.

Proof : We perform an inversion centered at $A$ with radius $\sqrt{AB\cdot AC}$ . Then, $E$ and $F$ maps to the intersections $E'$ and $F'$ of the perpendicular to the $\angle A-$ bisector at $I_a$ with $\overline{AB}$ and $\overline{AC}$ . It is well known that $I$ maps to $I_a$ and points $I_b$ and $I_c$ map to each other. Now note that by Pappus Theorem on collinear points $\overline{I_bAI_c}$ and $\overline{E'I_aF'}$ it follows that $Z=I_bE' \cap I_cF'$ lies on $\overline{BC}$ .

But further note that, $I_bI_c \perp AI_a \perp E'F'$ which implies that $I_bI_c \parallel E'F'$ . This means that $Z$ is in fact the center of homothety mapping $\triangle ZE'F'$ to $\triangle ZI_bI_c$ . But now note that $L$ is the midpoint of $I_bI_c$ (well known) and $I_a$ is the midpoint of $E'F'$ (since $AE'=AF'$ ). Thus, the aforementioned homothety sends $I_a$ to $L$ implying that points $I_a$ , $L$ and $Z$ are collinear. But now inverting back this implies that $(AI_cE)$ and $(AI_bF)$ intersect for the second time at the second intersection of $(ABC)$ and $(ADR)$ where $R = AL \cap BC$ . It is well known that this second intersection is the Evan is Old Point $X$ which implies the claim.

Now note that,
$\measuredangle AI_cP = \measuredangle CEP = \measuredangle EFP$ which since $I_bI_c \parallel EF$ implies that points $I_c$ , $F$ and $P$ are collinear. Similarly we can note that points $I_b$ , $E$ and $Q$ are collinear. This implies that $\overline{EQ}$ and $\overline{FP}$ intersect at the homothety center mapping $\triangle DEF$ to $\triangle I_aI_bI_c$ which is well known to lie on $\overline{IO}$ which finishes the proof of the claim.

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#16 h Feb 23, 2025, 1:53 AM • 1 Y

Y by drmzjoseph

I see you have long solutions, I will give you a simple one

It is enough to prove that $FP$ passes through $X_{57}$ . So it is enough to prove that $(APB)$ is tangent to $\omega$ . Let $Y$ and $Z$ be points on $\omega$ such that $(AZB)$ is tangent to $\omega$ and $(AYC)$ is tangent to $\omega$ , it is well known that $I_C \in (AEZ)$ and $I_B \in (AFY)$ where $I_B$ and $I_C$ are the $B,C-$ excenters of $\triangle ABC$ . Let $M$ be the midpoint of arc $BAC$ of $(ABC)$ .Let $F'$ be the reflection of $F$ in $B$ , By 2002 ISL G7 we know that $(AZB)$ passes through the midpoint of $I_CF$
$\Rightarrow \frac{FI_C.FZ}{2}=FA.FB\Rightarrow FZ.FI_C=FA.FF'\Rightarrow F'\in (I_CAE)\Rightarrow \frac{Pow_{(I_CAE)}B}{Pow_{I_BAF}B}=-1=\frac{Pow_{(I_CAE)}M}{Pow_{I_BAF}M}$ and $A\in (I_CAE)\cap (I_BAF)$ , so, by the forgotten coaxiality lemma we have that $(ABC), (I_CAE)$ and $(I_BAF)$ are coaxial.
Since $X_{57}$ is the homothetic center of the contact triangle and the excentral triangle $\Rightarrow X_{57}I_C.X_{57}E=X_{57}I_B.X_{57}F\Rightarrow X_{57}I_C.X_{57}Z=X_{57}I_B.X_{57}Y\Rightarrow X_{57} \text{ is on the radical axis of these 3 circles}\Rightarrow Z\equiv P_\blacksquare$

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445 posts

#17 h Feb 23, 2025, 3:09 AM • 1 Y

Y by hectorleo123

Another solution using famous ISL2002-G7

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#18 h Feb 23, 2025, 3:34 AM

Y by

Tbh i solved it at the beginning by a reformulation, using metrics:
The inversion(negative) at $X_{57}$ sending the incircle to Bevan circle, also send $\odot (ABC)$ to $\odot (ABC)$
So bellow i proved it, using another notation

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