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Let $M,N$ be the midpoints of the arcs $CA,CB$ respectively. We have $MO_{1}=MC,\ NO_{2}=NC$, so $\frac{MO_{1}}{NO_{2}}=\frac{MC}{NC}$, which is constant (doesn't depend on $X$). Let $P$ be a point on the arc $AB$ s.t. $\frac{MP}{NP}=\frac{MC}{NC}$. Since we alos have $\angle PMO_{1}=\angle PNO_{2}$, it means that $PMO_{1},PNO_{2}$ are similar, so $\angle O_{1}PO_{2}=\angle MPN=\angle MXN=\angle O_{1}XO_{2}$, so $P$ always belongs to $\Omega$. [Moderator edit: This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=555 .]

OFFICIAL SOLUTIONS: Let $M$ be the midpoint of the arc BC. It is a known result (otherwise easy to prove) that triangles $MO_{2}B$ is isosceles with $MO_{2} = MB$. Therefore $MO_{2} = MB = MC = MO$, where $O$ is the incenter of the triangle $ABC$. In the same way $A,O_{1},O,B$ are on a circle with center $N$, the midpoint of the arc $AC$. Let $P$ be on $\Omega$ such that $CP||MN$ and $T$ be the second intersection of $OP$ with $\Omega$. Then $MO = MC = NP$ and $MP = NC = NO$, therefore $MPNO$ is a parallelogram. It follows that $S_{MPT} = S_{NPT}$, whence $MP \cdot MT = NP \cdot NT$. Thus $NC \cdot MT = MC \cdot NT$, which shows that $NT : NO_{1} = MT : MO_{2}$. But then $\triangle NO_{1}T \sim \triangle MO_{2}T$ (because $\widehat{O_{1}NT} = \widehat{XNT} = \widehat{XMT} = \widehat{O_{2}MT}$), $\widehat{NTO_{1}} = \widehat{MTO_{2}}, \widehat{O_{1}XO_{2}} = \widehat{MTN} = \widehat{O_{1}TO_{2}}$ therefore the quadrilateral $XO_{1}O_{2}T$ is cyclic. This proves that the circle $(XO_{1}O_{2})$ passes through the fixed point $T$ or every position of $X$. Another Solution (Figure 2): It is easy to prove that if $A,B$ are fixed points and $C$ is a obile point on one of the arcs $\overset{\frown}{AB}$ of a fixed circle then the locus of the incenter of the triangle $ABC$ is an arc with endpoints $A,B$. Therefore $O_{1}$ and $O_{2}$ are on the circles with centers $N$ and $M$ which pass through $A,C$ and $B,C$ respectively. This suggests a transformation of the figure using an inversion of pole $C$. The circle $(ABC)$ becomes the straight line $AB$, the straight line $CO_{1}$ becomes the bisector of the angle $\widehat{CAX}$, the tangent at $C$ to the circle $(ABC)$ becomes the parallel $(p)$ from $C$ to $AB$, the straight line $CN$ becomes the bisector $q$ of the angle $(\widehat{p,CA})$, the circle $(CAO_{1})$ becomes the perpendicular from $A$ on $(q)$, that is the external bisector of the angle $\widehat{BAC}$. Hence $O_{1}$ becomes an excenter for the triangle $CAX$. In the same way $O_{2}$ becomes an excenter for the triangle $BAX$. It follows that $O_{1}X \bot O_{2}X$, therefore the circle $(XO_{1}O_{2})$ becomes the circle of diameter $O_{1}O_{2}$. For $X = A$ we have $O_{1} = A, O_{2} = I_{c}$ (excenter for $\triangle ABC$), therefore the fixed point $T$ must be the projection of $I_{c}$ on $AB$. This implied by the following facts: - $O_{2}X$ passes through $O^{'}_{1} = $ incenter of $\triangle CAX$; \item $\triangle AO_{1}T \sim \triangle O^{'}_{1}O_{1}O_{2} \Rightarrow \frac{AT}{O^{'}_{1}O_{2}} = \frac{AO_{1}}{O^{'}_{1}O_{1}}$; - $\triangle CAI_{c} \sim \triangle CO^{'}_{1}O_{2} \Rightarrow \frac{O^{'}_{1}O_{2}}{AI_{c}} = \frac{CO^{'}_{1}}{CA}$; \item $AT = AI_{c} \cdot \frac{AO_{1}}{AC} \cdot \frac{CO^{'}_{1}}{O^{'}_{1}O_{1}} = AI_{c} \cdot \frac{4R \sin \frac{C}{2} \cos \frac{X}{2}}{AC} \cdot \frac{4R \sin \frac{X}{2} \sin \frac{A}{2}}{4R \sin \frac{C}{2}} = AI_{c} \sin \frac{A}{2}$ where $R$ and $A,C,X$ are the usual notations meant in $\triangle ACX$. Mentioned figures are coming in due course.

Let $D$ and $E$ be the midpoints of arcs $\widehat{CA}$ and $\widehat{CB}$. If $Y$ denotes the second intersection point of $\Omega$, which is the circumcircle of $\triangle{DEX}$, and the circumcircle of $\triangle{XO_1 O_2}$, then $Y$ is the center of spiral similarity sending segment $DE$ to $O_1 O_2$. This implies that $Y$ is also the center of the spiral similarity sending segment $DO_1$ to $EO_2$. Hence triangles $\triangle{YDO_1}$ and $\triangle{YEO_2}$ are similar which implies that $YD/YE=DO_1/EO_2$. However, since $O_1$ is the incenter of $\triangle{CAX}$ and $D$ is the midpoint of arc $\widehat{CA}$, it follows that $DO_1 = DA = DC$. Similarly, $EO_2 = EB = EC$ and hence $DY/EY=DC/EC$. Since the locus of all points $P$ such that $DP/EP=DC/EC$ is a fixed circle passing through $C$ and not passing through $D$ and $E$, it intersects $\Omega$ again exactly once. Furthermore, $Y$ cannot coincide with $C$ since $\angle{XO_1 C} + \angle{XO_2 C}=270^\circ$, which implies that $XO_1 O_2 C$ is not cyclic. Therefore $Y$ is the unique point satisfying $DY/EY=DC/EC$ and $Y \neq C$ and therefore is fixed.

Dear MLs, Maybe knowledge of the fact that this fixed point is the tangency point of mixtilinear incircle, will provide some fresh ideas for a solution . M.T.

Dear Armpist and Mathlinkers, the synthetic proof can be seen on http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure I p. 41-43. Sincerely Jean-Louis

Dear Jean-Louis In your paper A new mixtilinear incircle adventure I on p 42 you use 'Un quadrilatere harmonique' property to prove that the fixed point of intersection of two circles in on mixtilinear circle tangency point, but the earlier proof of this 'harmonique' property on p. 38 is based on the very fact that point P is the mixtilinear tangency point. I find it a bit confusing. BTW, who named this point after Longchamps and when? M.T.

Dear Armpist and Mathlinkers, yes, I have to read again this part.... Sincerely Jean-Louis

Let the midpoint of $\widehat{BC}=G$ and $\widehat{AC}=F$, and the intersection of the two circles as $P$. Note that the $F,O_1,X$ and $G,O_2,X$ are collinear respectively. Then we have $\angle O_1FP=\angle PGO_2$ and $\angle O_2PO_1=\angle O_2XO_1=\angle GBF\rightarrow\angle FPO_1=\angle GPO_2$, thus $\triangle PFO_1\sim\triangle PGO_2$. We note that $GO_2$ and $FO_1$ are constant lengths as $X$ varies since $F$ is always and the circumcenter $BO_2C$ and likewise for $G$, thus we have the lengths a constant (the radius of the circle). Let $FO_1/GO_2=c$ a constant. But since $PF/PG=FO_1/GO_2=c$, and $P$ lies on the circle we have that $P$ is fixed since the ratio $PF/PG$ is fixed.

Another interesting solution is to apply an inversion centered at C with arbitrary power. Let the image of $P$ be $P$ for simplicity. Then our problem can be phrased: Let $ABC$ be a triangle, and let $X$ be a variable point on $AB$. Let $O_1, O_2$ be the C-excenters of $ACX$ and $BCX$ respectively. Prove that the circumcircle of $O_1XO_2$ passes through a fixed point on $AB$. Proof: Let this fixed point be $H$. We in fact prove that $H$ is the projection of $O_C$ onto $AB$, where $O_C$ is the C-excenter of $ABC$. Since $\angle O_1XO_2 = 90^\circ$ (easy angle chasing), we must prove that $O_1HO_2 = 90^\circ$, which is equivalent to $O_1H^2+O_2H^2 = O_1X^2 + O_2X^2$. Let $H_1, H_2$ be the projections of $O_1, O_2$ onto $AB$. Then $O_1H^2+O_2H^2 = O_1X^2 + O_2X^2$ is equivalent to $O_1H_1^2+H_1H^2+O_2H_2^2+H_2H^2 = O_1H_1^2+H_1X^2+O_2H_2^2+H_2X^2$ $H_1H^2+H_2H^2 = H_1X^2+H_2X^2$. So proving that $H_1H = H_2X$ is sufficient. (then $H_2H = H_1X$ would follow, so the equality would be obvious) But $H_1H = \left\vert AH - AH_1 \right\vert =$ $ \left\vert \frac{AB+BC-AC}{2} - \frac{AX+CX-AC}{2} \right\vert = $ $\left\vert \frac{(AB-AX)+BC-CX}{2} \right\vert =$ $\left\vert \frac{BX+BC-CX}{2} \right\vert = H_2X$ (all equalities are just from simple properties of excircles tangencies).

SnowEverywhere wrote: Since the locus of all points $P$ such that $DP/EP=DC/EC$ is a fixed circle passing through $C$ and not passing through $D$ and $E$ Can someone provide a proof for why this is true? EDIT: ok i got it just let $X$ be the point on $DE$ such that $DX/XE = DC/EC$. Then let $Y$ be the point such that $(DXEY)$ is harmonic. Then $DP/EP = DX/XE$ implies $PX$ is the angle bisector of angle $DPE$. Thus, by a well-known lemma, angle $XPY=90$ and $P$ lies on the circle of diameter $XY$.

Consider $N,M$ the midpoints of arcs $CB, CA$ and $\Omega_1, \Omega_2$ the circles centered at $M,N$ with radii $r_1=MC, r_2=NC$, respectively. Then by continuity, there exists a point $P \in \odot (ABC)$ such that $ \frac{MP}{PN} = \frac{r_1}{r_2} := r$. This is true since with $P=B$ the ratio equals $\frac{BM}{r_2} > r$ and with $P=A$ the ratio equals $\frac{r_1}{AN} < r$. Then notice $\angle PNX = \angle PMX$ and $\frac{PN}{MO_2}=\frac{PM}{MO_1}$, so $\triangle PNO_2 \equiv \triangle PMO_1$. Thus, by Miquel's Theorem, $P=\odot (NXM) \cap \odot (XO_1O_2)$. Since $P$ is constant, we're done.

This was pretty tricky but a bit dumb/silly in motivation. Here is my solution:- Let $M,N$ be the mid-points of arcs $CA,CB$ and let $T$ be the touch-point of the $C$-mixitlinear incircle with the circumcircle and $E$ be the touch-point of the $C$ excircle with $BC$. Let $M',N'$ be points on line $AB$ such that $CM.CM'=CN.CN'=CB.CA$ and $\angle BCM = \angle ACM'$ and $\angle ACN = \angle BCN'$. (That is, $M',N'$ are the images of $M,N$ in the inversion about $C$ of radius $\sqrt{CA.CB}$ composed with the reflection in the $C$ angle bisector.) Now, we see that $\angle \frac{A}{2} = \angle ACM'$ and so $AC=AM'$ Similarly, $BC=BN'$ And so we get that $E$ is the mid-point of $M'N'$. Therefore, $CMTN$ is a harmonic quadrilateral and hence we get that $\frac{TM}{TN}=\frac{MA}{NA}=\frac{MO_1}{NO_2}$ and so we get that $\triangle TO_1M$ is similar to $\triangle TO_2N$ and hence $T$ is the center of a spiral similarity sending $MO_1$ to $NO_2$ and so points $X,O_1,O_2,T$ are on a circle.

$\blacksquare \text{ } CMTN \text{ is harmonic where M,N are midpoints of arcs CA , CB and T is point of contact of C-mixtilinear incircle with (CAB) .} $ Proof -- Let $TM \cap AC = M' $ , $TN \cap CB=N' $ and $TA \cap \omega =A' $ where $\omega$ is C-mixtilinear incircle. It is well known that $CM'$ and $CN'$ are tangents to $\omega$ . So , we have $CM'TN'$ is harmonic .But as $CM'TN'$ and $CMTN$ are homothetic with center $T$ ,so $CMTN$ is harmonic . So , now , $\frac{TM}{TN}=\frac{MC}{NC}=\frac{MO_1}{NO_2}$ . This yields $ \triangle TO_1M \sim \triangle TO_2N \implies$ $T$ is unique center of spiral similarity sending $MO_1 \rightarrow NO_2 \implies T \in (XO_1O_2)$.

Let $O_1X \cap \Omega =M_1$ and $O_2X \cap \Omega =M_2$. Also let $T$ be the $C$-mixtillinear intouch point. We claim that $T$ is the desired fixed point. From here, this is equivalent to proving that $$\frac{TM_1}{TM_2}=\frac{M_1O_1}{M_2O_2}= \frac{b \cos \frac{A}{2}}{a \cos \frac{B}{2}}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}$$Now it is well known that $CM_1M_2T$ is harmonic, which gives $$\frac{TM_1}{TM_2}=\frac{CM_1}{CM_2}=\frac{\sin \angle CM_2M_1}{\sin \angle CM_1M_2}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}} \text{ } \blacksquare$$

(probably a different solution). We claim $T$ is the fixed point where $T$ is $C-$ mixtillinear intouch point. Let $M$ and $N$ be points where $M = XO_1 \cap \odot(ABC)$ and $N = XO_2 \cap \odot(ABC)$. By fact 5, $CM = MA$ and $NC = NB$. Let $K$ be the second intersection of $\odot(ABC)$ and $\odot(XO_1O_2)$. Let $\Psi$ be an inversion with center $X$ and arbitrary radius. Now $A^*,B^*,M^*,N^*$ are collinear and $O_1$ and $O_2$ are excenters of $\triangle A^*C^*X$ and $\triangle B^*C^*X$ respectively. Now $K^* = O_1^*O_2^* \cap A^*B^*$. Let $Y = O_2^*C^*\cap O_2^*X$. We have \[-1 = (M^*,X;Z,O_1^*) \stackrel{O_2^*}{=} (M^*,N^*;C^*,K^*).\]We need to show that $K^* \equiv T^*$ or \[\frac{N^*C}{M^*C} = \frac{N^*T}{M^*T}\Leftrightarrow \frac{NC}{MC} = \frac{NT}{MT}.\] Adding a proof to : $CMNT$ is harmonic (because I didn't know it was true till today ) Note: The $A^*, B^*$, etc. used below are different as the ones used above. Consider an inversion $\mathcal{I}$ centered at $C$ with radius $\sqrt{CB\cdot CA}$ and reflection over $C-$ internal angle bisector. Now $M^*A^* = A^*C$ and $N^*B^* = B^*C$. Also $\mathcal{I}$ maps $T$ to $C-$ excircle touchpoint of $\triangle A^*B^*C$. Therefore $T^*M^* = T^*N^*$. This gives $\frac{MT}{CM} = \frac{TN}{CN}$ showing the desired result and we are done $\qquad \square$ Also my first G8

Let $M_1,M_2$ denote midpoints of arcs $AC,BC$ respectively, so $X-O_1-M_1,X-O_2-M_2.$ Clearly $\odot (XO_1O_2)$ meet $\Omega$ again at Miquel point $Y$ of $M_1O_1O_2M_2$ and by Trillium theorem $$|YM_1|:|YM_2|=|O_1M_1|:|O_2M_2|=|CM_1|:|CM_2|$$which is fixed and so defines $Y$ uniquely.

Let $I$ be incenter. Let $E$ be midpoint of arc $AC$, $F$ midpoint of arc $BC.$ We have $AE=O_1E=CE=IE,BF=O_2F=CF=IF.$ Let $G$ on $(ABC)$ such that $CG$ and $EF$ parallel, and $GI$ intersect $(ABC)$ again at $D$ where $I.$ $EI=EC=GF,IF=CF=EG$ implies $EGFI$ parallelogram. We claim that $\triangle O_1DE\sim \triangle O_2DF.$ Since $\angle DEO_1=\angle DFO_2$ we only need to show \[ED/EO_1=FD/FO_2\]\[\iff ED/EC=FD/FC\]\[\iff ED\cdot EG=FP\cdot FG\]\[\iff [EDG]=[FDG]\]True by $EGFI$ parallelogram. Thus, our claim is true and so $\angle O_1DO_2=\angle EDF=\angle O_1XO_2.$ Thus, $D$ is the fixed point on $(XO_1O_2)$

Let $(O_1XO_2) \cap (ABC)=T_C$, $M,N$ the midpoints of minor arcs $AC,BC$ respectivily, now by I-E lemma we have that $MC=MO_1=MA$ and $NC=NO_2=NB$ so now note that $M,O_1,X$ and $N, O_2, X$ are colinear so $T_C$ is miquel point of $MNO_1O_2$ so: $$\frac{MT_C}{T_CN}=\frac{MO_1}{O_2N}=\frac{MC}{CN} \implies -1=(M, N; C, T_C) \implies T_C \; \text{is unique and fixed}$$Fun fact: If u project through the incenter of $\triangle ABC$ the harmonic and use $\sqrt{ab}$ inversion u get that $T_C$ is in fact the C-mixtilinear intouch point which is very hilarious tbh.

SL 1999 G8 wrote: Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $CB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $ABX$ and $CAX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point. Nobody thought about complex, lol wutt Let $(ABC)$ be the unit circle , $A=a^2,B=b^2,C=c^2,X=x^2$ So $O_1=-ab-ax-bx,O_2=-ac+ax-cx$, let $M_C,M_B$ be midpoints of minor arcs $AB,AC$. Then $M_C=-ab,M_B=-ac$, since $\overline{X-O_1-M_C},\overline{X-O_2-M_B}$ if $(XO_1O_2) \cap (ABC)=T$, we've $$\triangle TO_1O_2 \sim \triangle TM_BM_C$$So, by similarity formula, $$T=\frac{(-ab-ax-bx)(-ac)-(-ac+ax-cx)(-ab)}{-ac-ab-ax-bx-(-ab-ac+ax-cx)}=\frac{a^2(b+c)}{c-2a-b}$$Which is independent of $x$ , thus $T$ is a fixed point .The End $\blacksquare$

Solved with BVKRB Let $P$ intersection point and$M,N$ be midpoint of arc $CA,AB$ then we know $\overline {N-O_1-X}$ and $\overline{M-O_2-X}$ From incenter lemma $NO_1=NA=NB$ and $MO_2=MA=MC$ Note $X$ is center of spiral similarity send $O_1O_2 \rightarrow NM$ so that $NO_1 \rightarrow MO_2$ $$\frac{NO_1}{MO_2}=\frac{PN}{PM}=\frac{NA}{MA}$$ $\implies (A,P;M,N)=-1$ and as only one such point satisfied we get $P$ is fixed point.

We claim that this fixed point is $C-$mixtilinear touch point. Consider $\sqrt{ab}$ inversion and reflect over the angle bisector of $\measuredangle C$ and rephrase the problem as it would be symmetric to $B,C$ (or $A$ index). New Problem Statement: $ABC$ is a triangle with $A-$excenter $J$ and $X$ is an arbitrary point on $BC$. Angle bisectors of $\measuredangle XAB$ and $\measuredangle CAX$ intersect segments $JB,JC$ at $P,Q$. If $D$ is the tangency point of $A-$excenter with $BC$, then show that $X,P,Q,D$ are concyclic. If the circle centered at $J$ with radius $JA$ intersects $BC$ at $K,L$, then $K\in (APX)$ and $L\in (AQX)$. Hence \[\measuredangle QXP-\measuredangle QAP=\measuredangle APX+\measuredangle XQA=\measuredangle AKX+\measuredangle XLA=\frac{\measuredangle B+\measuredangle C}{2}=90-\frac{\measuredangle A}{2}=90-\measuredangle QAP\]Thus, $\measuredangle QXP=90$. Now, we will prove that $\measuredangle QDP=90$ with the method of moving points. Consider the map $f:JB\mapsto JC$ such that $\measuredangle P'AP=\frac{\measuredangle A}{2}$. Also $g: JB\mapsto JC$ such that $\measuredangle P'DP=90$ (note that we use directed angles). Since we used only rotation around fixed points, $deg \ f,deg \ g\leq 2$. For $P=B$, $f(P)=J=g(P)$. For $P=J$, $f(P)=C=g(J)$. Pick $P=JB_{\infty}$. $H$ is the intersection of the perpendicular to $JB$ from $D$ with $JC$. If $T$ is the perpendicular from $A$ to $JC$ and $AT\cap BC=T_1$, then $\measuredangle CAJ=\measuredangle JT_1D=\measuredangle JTD=90-\measuredangle BJC$ hence $H=T'$ which means $AH\perp JC$. $\measuredangle HAP=\measuredangle (HA,JB)=\frac{\measuredangle A}{2}$ as desired.$\blacksquare$