[Moderator edit: I have merged Grobber's posts from two different threads containing the same solution to this problem.] Solution, first version: A point is the incenter of a triangle iff it's the orthocenter of the triangle formed by the images of the three points through an inversion having the point as its pole. Consider the inversion with pole $I$ which invariates $\Omega$. For any object $X$, let $X'$ be its image. $d_1=\Omega_1',d_2=\Omega_2'$ are two parallel lines tangent to $\Omega'=\Omega$. We also have $(IA)'=IA\|d_1\|d_2$. Let $\omega=(BC)'$. We can see that $\omega$ is a ircle passing through $I$ and tangent to $d_1,d_2$, so its radius is the same as that of $\Omega$. We have $\{B',C'\}=\omega\cap \Omega$. I think it's pretty clear now why $A$ is the symmetric of $I$ wrt $B'C'$, so $A'=$ the other point of intersection of $IA,\Omega$ is the orthocenter of $IB'C'$ (or equivalently, $I$ is the orthocenter of $A'B'C'$, Q.E.D.). Solution, second version: Let's first observe that if I is the incenter of ABC then, by an inversion of pole I, A, B, C have images A', B', C' s.t. I is the orthocenter of A'B'C' and vice-versa (just look at the case of an inversion of pole H and power HA*HD where D is the foot of the height HA: the images of A, B, C are D, E, F, the feet of the heights of ABC and H is the incenter of DEF). Now, in our original problem, let's consider an inversion of center I and power -k^2 (k is a length; negative means that the image of a point and that point are on opposite sides of I). Our new configuration is this: 2 parallel lines d1 and d2 (the images of the 2 circles tangent to each-other at I) which are also parallel to IA' (X' is, generally, the image of X); a circle C1 through I which is tangent to d1 and d2 (the image of the line BC);another circle C2, also tangent to d1 and d2 (the image of the larger circle ). Now let's observe that A' is on C2 (because A was on the large circle) and B', C' are the intersections between C1 and C2. We can also see that I and A' are on the same side of the line B'C'. It's now fairly easy to see that I is the orthocenter of A'B'C' (angle-chasing and stuff like that) and, by using the first observation in this proof we get that I is the incenter of ABC. Sorry it's not exactly complete and a bit unclear, but I think it can be understood.

Your solution Grobber is very beautiful especially last step was a bit unclear for me at the beginning but finally great !!! I have another solution: Let $BC\cap\Omega_1=X$, $BC\cap\Omega_2=Y$, $\Omega\cap\Omega_1=U$, $\Omega\cap\Omega_2=V$. It is a well known fact that UX goes through the midpoint of the arc BC, same as VY, let $UX\cap VY=D$. Using inversion I with pole D with power $DX\cdot DU$ (which invariates $\Omega_1$) We get that I(U) = X and $\Omega$ transforms in a line tangent to $\Omega_1$ – BC. As a result I(V) = ($\Omega_2\cap BC$) = Y. Hence $DX\cdot DU = DY\cdot DV$. It means D lies on a radical axe of $\Omega_1$ and $\Omega_2$ or UX, VY, AI goes through D. Let $AI\cap BC=T$ then I(A) = T and $DT\cdot DA=DI^2$. Using fact that D is midpoint of arc BC we get AI is a bisector of BAC. Also angle BAD = angle CAD = angle DBC means BD is tangent to ABT or $BD^2=DT\cdot DA=DI^2$. BD=DI Angle DBI = angle BID or Angle DBC + angle CBI = angle BAD + angle ABI or angle CBI = angle ABI. Hence BI is the bisector of the ABC.

Circle $ S,$ with segment $ BC.$ Draw two circles such that they are tangent to each other at a point $ I,$ tangent to $ S,$ and tangent to $ BC.$ Draw the common tangent and $ I,$ which cuts the curve $ BC$ on the same side of the two circles, at point $ A.$ Prove $ I$ is the incentre of $ ABC.$ Solution by Myth: I have just proved it pretty quick using an inversion. Namely, using an inversion with center at point $ B$, we imediately obtain that $ AI$ passes through a fixed point on the circle (the most distant point from $ BC$ in other halfplane than $ A$). Moreover, now, using inversion's properties, we can make the easiest calculations to obtain $ AC: KC = AI: IK$, where $ K = AI\cap BC$. This implies that $ I$ lies on a bisector of $ BAC$. No trigonometry, it is a pure geometry.

Dear Mathlinkers, a way to prove this result is to think to the Thebault-Sawayama theorem on which I wrote an article in Forum Geometricorum. Sincerely Jean-Louis

This problem, proposed by India, was probably the most beautiful Geometry problem in the 92 IMO. Unfortunately, it was not selected for the exam.... a situation which repeat in several instances at IMO.

For any two non-intersecting circles the incenters of the triangle(s) in question (there are two internal tangents) are the limiting points. When these two circles touch, the two limiting points degenerate into one, and it remains to be the incenter (of one triangle now, for there is one internal tangent). The result of Thebault's theorem happens and holds just because the incenter is a limiting point. M.T.

This problem can be solved using Casey's theorem.

Since D is the midpoint of the arc BC, it's easy to see that $ \angle UVD = \angle DXY$, hence XUVY is cyclic and $ DX \cdot DU = DY \cdot DV = DI^{2}$, that is, XY is the inverse of the circle $ \Omega$ through the inversion centered D and power $ DX \cdot DU$, hence B and C are their own images through the said inversion, that is, $ DB = DI = DC$, i.e. I is the incenter of $ \Delta ABC$ (this is a well known property of incenter, namely the point of intersection of the bisector of $ \angle BAC$ with the circle (D, DB), point placed inside the $ \Delta ABC$). Best regards, sunken rock

Another solution without using inversion: Let the two circles $\Omega_{1}$ with center $O_{1}$ and $\Omega_{2}$ with center $O_{2}$ tangent to $BC$ at $E$ and $F$. The two circles are tangent to circle $\Omega$ with center $O$ at $N$ and $S$, respectively. Let $NE$ cut $\Omega$ again at Q, then $EO_{1}\parallel QO$, so $O_{2}F\parallel QO$. As $\Delta SFO_{2}$ and $\Delta SQO$ are both isosceles and $FO_{2}\parallel QO$, points $S, F, Q$ are collinear. Let the two tangents to $\Omega$ at $N$ and $S$ meet at $T$. $\angle NSQ = 90^{\circ} - \angle TSN - \angle QSO = 90^{\circ} - \angle NQS - \angle SQO = 90^{\circ} - \angle NQO$ $\angle NEF = 90^{\circ} + \angle NEO_{1} = 90^{\circ} + \angle NQO$ Thus $\angle NSF + \angle NEF = 180^{\circ}, NEFS$ is cyclic. Therefore, $QE. QN = QF. QS$, this means points $Q, A, I$ collinear, and $QE. QN = QF. QS = QI^{2}$. As $OQ\parallel O_{1}E, OQ\perp BC$, this means that $Q$ is the midpoint of, let say, the lower arc $BC$, and $AQ$ is the angle bisector of $\angle BAC$ (*). Since $Q$ is the midpoint of the lower arc $BC, \angle BNQ = \angle BCQ = \angle QBE$, thus $\Delta BEQ\sim \Delta NBQ$, therefore $QB^{2} = QE. QN = QI^2$, which means $QB = QI$. Now $\angle ABI = \angle BIQ - \angle BAQ = \angle IBQ - \angle BCQ = \angle IBQ - \angle QBC = \angle IBC$, so BI is the angle bisector of $\angle ABC$ (**). With (*) and (**) we are done.

let intersection of (O1, BC)=E (O2, BC)=F, (O1,O)=X, (O2,O)=Y, arc BC's midpoint that not contain A=M EX and FY intersect at M(by harmonic, and similiailty.), and (E,X,T,Y)=concyclic. so M is on the radical axis of (O1, O2), so, A,I,M is collinear. then MI^2=ME*MX=MB^2, then mention's thereom.

Dear Mathlinkers, 1. P, Q the resp. centers of (1), (2) 2. U, V the resp. points of contact of BC with (1), (2) 3. J the incenter of ABC 4. according to the Sawayama Thebault theorem, p, Q and J are collinear 5. According to the Sawayama lemma, U, I and J are collinear. 6. I and J are the same… and we are done. See : Sawayama and Thébault's theorem, Forum Geometricorum vol. 3 (2003) 225-229 ; http://forumgeom.fau.edu/ Sincerely Jean-Louis

My solution: We start off with the following lemma. LEMMA Let $I$ be the incenter of $\triangle ABC$, having circumcircle $\Gamma$, and $P$ be a point on $\overline{BC}$. Let $\omega$ be a circle tangent to $AP$ at $X$, $BC$ at $Y$ and $\Gamma$ at $T$. Then $I$ lies on $XY$. PROOF Let $TX \cap \Gamma = K, TY \cap \Gamma = M, AM \cap XY = Z$. As $T$ is the center of homothety that takes $\omega$ to $\Gamma$, we get that $M$ is the midpoint of $\overarc{BC}$ (not containing $A$) of $\Gamma$. Also, By Reim's Theorem, $XY \parallel KM$. Now, $\measuredangle TXZ = \measuredangle TKM = \measuredangle TAZ \Rightarrow T,A,X,Z$ are concyclic. And, $\measuredangle MZT = \measuredangle AZT = \measuredangle AXT = \measuredangle XYT = \measuredangle ZYM \Rightarrow \triangle MTZ \sim \triangle MZY \Rightarrow MY \cdot MT = MZ^2$ Also, By Shooting Lemma, $MY \cdot MT = MB^2 = MC^2 \Rightarrow MB = MC = MZ \Rightarrow$ By Fact 5, $Z = I$ $\Box$ Return to the problem at hand. Let $BC \cap \Omega_1 = D, BC \cap \Omega_2 = E$. Also, let $I'$ be the incenter of $\triangle ABC$. By our LEMMA, $D, I, I'$ and $E, I, I'$ are collinear $\Rightarrow I = I'$ $\blacksquare$

This my first hardest SL on IMO, i solved this on 11th of november in 2020 . Here we will denote $X \Omega_K$ as the distance of tangent from $X$ to the circle $\Omega_K$, and we will consider a point $Y$ as a degenrate circle with rad $0$. And we will fix our watch on the $A$-side. Let $AI \cap BC=D$, $BI \cap AC=E$ and $CI \cap AB=F$ We will apply Casey's at $(B, \Omega_1, A, \Omega_2)$ to get: $B \Omega_1 \cdot A \Omega_2+B \Omega_2 \cdot A \Omega_1=AB \cdot \Omega_1 \Omega_2$ We will also use Casey's at $(C, \Omega_1, A, \Omega_2)$ to get: $C \Omega_1 \cdot A \Omega_2+C \Omega_2 \cdot A \Omega_1=AC \cdot \Omega_1 \Omega_2$ We will choose $A \Omega_1=A \Omega_2=AI$ and note that $D \Omega_1=DI=D \Omega_2$ thus $\Omega_1 \Omega_2=2 \cdot ID$ Now use the fact's on the last line, and adding both casey's results we have: $2 \cdot BC \cdot AI =2(AB+AC) \cdot ID$. From the last equation $\frac{AI}{ID}=\frac{AB+AC}{BC}$ and similarily we get $\frac{BI}{IE}=\frac{BA+BC}{AC}$ and $\frac{CI}{IF}=\frac{CA+CB}{AB}$ Thus we have that $I$ is the incenter of $\triangle ABC$ and we are done

Solution without any computations. Let $S$ be the midpoint of arc $BC$, not containing $A$. We would like to show that $S$ is the center of $(BIC)$, which would directly give us that $I$ is the incenter of $\triangle ABC$ by the Incenter-Excenter Lemma. Let $O_1,O_2$ be the centers of $\Omega_1,\Omega_2$, respectively. Let $P,Q$ be touchpoints of $\Omega,\Omega_1$ and $\Omega,\Omega_2$, respectively. Let $E,F$ be touchpoints of $\Omega_1,\Omega_2$ with $\overline{BC}$, respectively. Firstly, $\overline{OS}\perp\overline{BC}\perp\overline{O_1E}\parallel \overline{O_2F}$. Hence, obviously by similarities, we get that $P,E,S$ and $Q,F,S$ are collinear. By Monge's theorem, $\overline{BC},\overline{PQ},\overline{O_1O_2}$ are concurrent at $X$. Next, $(IPQ)$ is tangent to $\overline{O_1O_2}$ being $O$-excircle of $\triangle OO_1O_2$ by just suitable length conditions. Hence, inversion at $X$ with radius $XI^2$ yields that $\Omega_1$ and $\Omega_2$ are getting swapped, which moreover yields that $PQEF$ is cyclic. On top of that, the inversion at $X$ gives us that $(BIC)$ is tangent to $\overline{O_1O_2}$. Hence, by radical axis on $(PQEF),\Omega_1,\Omega_2$, we get that $S$ lies on the common tangent to $\Omega_1,\Omega_2$ from $I$. Therefore, $S$ must be the center of $(BIC)$, hence done.

Fairly straightforward with shooting lemma. Let $K$ and $L$ be the tangency pints of $\Omega_1$ and $\Omega_2$ with $\Omega$, and $T_1$ and $T_2$ be their tangency points respectively with $\overline{BC}$. First, note that $\overline{KT_1}$ and $\overline{LT_2}$ both pass through the midpoint $M$ of arc $\widehat{BC}$ (not containing $A$). So $KT_1T_2L$ is cyclic and $\overline{MT}$ is the common tangent by radical axis. Then by shooting lemma $MB=MI=MC$, thus $I$ is the desired incenter.

Let one of the two smaller circles be $\omega_1$ and suppose it touches $(ABC)$ and $BC$ at $X$ and $R$ respectively. Then due to shooting lemma, $XR$ passes through midpoint of arc $BC$ not containing $A$ which we call $M$. Similarly the other analogous chord of the other smaller circle also goes through $M$. Now by shooting lemma also, it is known that $MR.MX=MB^2$ and the analogous version for the other circle so it follows that $M$ lies on the radical axis which is $AI$. Now to finish, notice that inversion with center $M$ and radius $MB$ preserves the two small circles and so $I$ must be preserved and so it must be the incenter as it is the unique point lying on the $A$ angle bisector on the same side of $BC$ as $A$ which is preserved after the above inversion.