orl wrote: Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i + 1}$ and $ A_iA_{i + 2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i + 1}$ and $ C_{i + 2}.$ Prove that the circumcentres of the triangles $ A1B_1I,A_2B_2I,A_3B_3I$ are collinear.

Denote $ L_i (i=1,2,3)$ the intersections of $ A_iI$ and $ (O)$. We have $ L_i$ is the circumcenter of triangle $ IA_{i+1}A_{i+2}$ so $ L_iL_{i+1}$ is the perpendicular bisector of $ [IA_{i+2}]$ Since $ L_1O_1, L_2O_2, L_3O_3$ concur at $ I$ so $ L_1L_2L_3$ and $ O_1O_2O_3$ are perspective. By Desargues's theorem, we get $ (L_1L_2\cap O_1O_2), (L_2L_3\cap O_2O_3), (L_1L_3\cap O_1O_3)$ are collinear. They are circumcenters of triangles $ IA_1B_1, IA_2B_2, IA_3B_3$ Remark: The end still works when $ O_i$ lies on $ IA_i$ and we don't need $ (O_i)$ tangent to $ A_{i+1}A_{i+2}$

The centers of three circles passing through the same point $ I$ and not touching each other are collinear if and only if they have another common point. Hence it is enough to show that the circles $ A_iB_iI$ have a common point other than $ I$. Now apply inversion at center $ I$ and with an arbitrary power. We shall denote by $ X`$ the image of $ X$ under this inversion. In our case, the image of the circle $ C_i$ is the line $ B`_{i + 1}B` _{i + 2}$ while the image of the line $ A_{i + 1}A_{i + 2}$ is the circle $ IA`_{i + 1}A`_{i + 2}$ that is tangent to$ B`_{i}B`_{i + 2}$, and $ B`_{i}B`_{i + 2}$. These three circles have equal radii, so their centers$ P_1,P_2,P_3$ form a triangle also homothetic to$ B`_{1}B`_{2}B`_{3}$. Consequently, points $ A`_1,A`_2,A`_3$ that are the reflections of $ I$ across the sides of $ P_1P_2P_3$, are vertices of a triangle alsohomothetic to$ B`_1B`_2B`_3$. It follows that$ A`_1B`_1,A`_2B`,A`3B`3$are concurrent at some point $ J`$, that the circles $ A_iB_iI$ all pass through $ J$. Probem 10.

Original diagram: Let the incircle be tangent to $A_1A_2A_3$ at points D, E, F on sides $A_2A_3, A_3A_1, A_1A_2$ respectively. Also let L, M, N be the intersection of lines $A_1I, A_2I, A_3I$ with the incircle respectively. Let points $B_1, B_2, B_3$ be defined as stated in the problem. The condition of the circumcenters of $(A_1B_1I), (A_2B_2I), (A_3B_3I)$ being collinear combined with the fact that I lies on all three circles implies that the three circles are coaxial. Hence, there exists another point, J, that lies on all three circles. Inverted diagram: Let us now invert the diagram about the incircle. It follows that circles $C_1, C_2, C_3$ invert to triangle $B_1^*B_2^*B_3^*$. Sides $A_2A_3, A_3A_1, A_1A_2$ invert to circles of diameter $DI, EI, FI$ respectively. Let the centers of these circles be $O_1, O_2, O_3$ respectively. Since L, M, N lie on the incircle, $L=L^*$, $M=M^*$ and $N=N^*$. Note that $(A_1B_1I)$ inverts to the line formed by points $A_1^*$ and $B_1^*$. Hence it suffices to show that lines $A_1^*B_1^*$, $A_2^*B_2^*$, and $A_3^*B_3^*$ are concurrent. This statement is equivalent to proving triangle $A_1^*A_2^*A_3^*$ is similar to triangle $B_1^*B_2^*B_3^*$. Doing so would imply the existence of a homothecy mapping $A_1^* \rightarrow B_1^*$ and likewise for the other 4 points. This would imply collinearity. To prove this, we note that $O_1O_2 \parallel B_1^*B_2^*$ since IM is perpendicular to both lines. (The first perpendicularity follows from radical axis and the second from the fact that IM is the perpendicular bisector of the segment formed by the inverted tangency points on $B_1^*B_2^*$.) Hence, $O_1O_2O_3$ is similar to $B_1^*B_2^*B_3^*$. However, the medial triangle of $O_1O_2O_3$ is similar to $A_1^*A_2^*A_3^*$ by a scale factor of $\frac{1}{2}$. Hence, $O_1O_2O_3$ is congruent to $A_1^*A_2^*A_3^*$. From this, it follows that $B_1^*B_2^*B_3^*$ is similar to $A_1^*A_2^*A_3^*$, as desired.

Let $D_1,D_2,D_3$ be the contact points of the incircle. Under incircle inversion (we'll be lazy with the stars), we see that $B_1^*B_2^*B_3^*$ is a triangle externally tangent to the three circles $(ID_1),(ID_2),(ID_3)$, which all have equal radius since $ID_1=ID_2=ID_3$. Since the radii are all equal, we see that $B_iB_{i+1}\parallel D_iD_{i+1}$, so $B_1B_2B_3$ is homothetic to $D_1D_2D_3$, say $B_i=(1-k)J+kD_i$ for some fixed point $J$ and number $k$. We also see that $A_i^*$ is the midpoint of $D_{i+1}D_{i+2}$, so it's not hard to check that \[\frac{2k}{2k+1}A_i+\frac{1}{2k+1}B_i=\frac{k}{2k+1}(D_i+D_{i+1}+D_{i+2})+\frac{1-k}{2k+1}J,\]which is constant. Thus, all the $A_iB_i$ concur, so pre-inversion, we have that the circles $(A_iB_iI)$ are coaxial, as desired.

IMO Shortlist 1997, Q9 wrote: Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i+1}$ and $ A_iA_{i+2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i+1}$ and $ C_{i+2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear. Let $DEF$ be the contact triangle of $\triangle ABC$, let the midpoints of $DE,EF,FD$ be $A_3^*,A_1^*,A_2^*$ respectively. Let the circumcenters of $\triangle A_1IB_1,A_2IB_2,A_3IB_3$ be $O_1,O_2,O_3$ respectively. Now we will perform an inversion($\Psi$) around the incircle $$\begin{cases}\Psi:A_1\mapsto A_1^* \\ \Psi:A_2\mapsto A_2^* \\ \Psi:A_3\mapsto A_3^* \\ \Psi:B_1\mapsto \text{Radical Axis} (C_2,\odot(I))\cap \text{Radical Axis} (C_3,\odot(I))=B_1^* \\ \Psi:B_2\mapsto \text{Radical Axis} (C_1,\odot(I)) \cap \text{Radical Axis} (C_3,\odot(I))=B_2^* \\ \Psi:B_3\mapsto \text{Radical Axis} (C_2,\odot(I))\cap \text{Radical Axis} (C_1,\odot(I))=B_3^*\end{cases}$$ Since, $\triangle B_1^*B_2^*B_3^*$ is homothetic to $\triangle DEF$ which in turn is homothetic to its medial triangle of $\triangle A_1^*A_2^*A_3^*$. So, $\triangle B_1^*B_2^*B_3^*$ is homothetic to $\triangle A_1^*A_2^*A_3^*$. This implies that $A_1^*B_1^*,B_1^*C_1^*,C_1^*A_1^*$ are concurrent. Now the images of their circumcenters will be reflection of $I$ over $A_1^*B_1^*,A_2^*B_2^*$ and $A_3^*B_3^*$ respectively. So, by a Homothety with a scale factor of $2$ we get that $I,O_1^*O_2^*B_2^*$ are concyclic. Hence, inverting back we get that $O_1,O_2,O_3$ are collinear.

Notice that $C_1=(IB_2B_3)$. Clearly the center of $(IB_2B_3)$ lies on $A_1I$. By Fact 5, the center of $(IA_2A_3)$ lies on $A_1I$, so $(IB_2B_3)$ is tangent to $(IA_2A_3)$. We now invert around $I$ with arbitrary radius. We know that $A_2^*A_3^*\parallel B_2^*B_3^*$, and similarly for the other pairs of sides, and thus $\triangle A_1^*A_2^*A_3^*$ is homothetic to $\triangle B_1^*B_2^*B_3^*$. Therefore, $A_i^*B_i^*$ for $i=1,2,3$ concur, which means $(A_iB_iI)$ are coaxial, and we are done.

The center of $(IA_2A_3)$ is the midpoint of $\widehat{BC}$ and the center of $(IB_2B_3)$ also lies on $\overline{A_1I}$ so $\overline{A_2'A_3'}\parallel\overline{B_2'B_3'}$ where we are inverting about $I.$ Hence, $\overline{A_iB_i}$ concur and so $(A_iB_iI)$ are coaxial. $\square$

Apply the inversion with the respect to $I$. We leave to the reader to draw the inverse picture. Notice that the condition that $I$ is the incentar now reads that the circumcircles $A^*_i A^*_{i+1}I$ are of the same radii. Indeed if $R$ is the radius of the circle of inversion and $r$ the distance between $I$ and $XY$ then the radius of the circumcircle of $\triangle IX^*Y^*$ is equal to $R^2/r$. Now we use the following statement that is very easy to prove: "Let $k_1, k_2, k_3$ be three circles such that all pass through the same point $I$, but no two of them are mutually tangent. Then the centers of these circles are colinear if and only if there exists another common point $J \neq I$ of these three circles." In the inverse picture this transforms into proving that the lines $A^*_1B^*_1, A^*_2B^*_2,$ and $A^*_3B^*_3$ intersect at a point. In order to prove this it is enough to show that the corresponding sides of the triangles $A^*_1A^*_2A^*_3$ and $B^*_1B^*_2B^*_3$ are parallel (then these triangles would be perspective with respect to the infinitely far line). Afterwards the Desargue's theorem would imply that the triangles are perspective with respect to a center. Let $P^*_i$ be the incenter of $A^*_{i+1}A^*_{i+2}I$, and let $Q^*_i$ be the foot of the perpendicular from $I$ to $P^*_{i+1}P^*_{i+2}$. It is easy to prove that $$\overrightarrow{A_1^*A_2^*}= 2\overrightarrow{Q_1^*Q_2^*}= -\overrightarrow{P_1^*P_2^*}.$$Also since the circles $A^*_iA^*_{i+1}I$ are of the same radii, we have, $P^*_1P^*_2 || B^*_1B^*_2$, hence $A^*_1A^*_2||B^*_1B^*_2$. As desired!

Let $D_i$ be touch point of incircle with $A_{i+1}A_{i+2}$ and $\gamma$ is inversion wrt this circle. $\gamma (A_i)$ is midpoint of $D_{i+1}D_{i+2}$ and all circles $\odot (I\gamma(A_i)\gamma(D_{i+2})\gamma(A_{i+1}))$ are congruent. Therefore $\gamma(C_i)\parallel D_{i+1}D_{i+2}$ and since $\overline{\gamma(A_i)\gamma(B_i)}$ concur (by homothety) $\odot (IA_iB_i)$ are coaxial as desired.

It suffices to prove the circles are coaxial. Let $DEF$ be the intouch triangle. Note thatof $\Gamma_1^\ast$ is exactly the circle with diameter $\overline{ID}$, etc. We proceed by inversion around $I$. Claim: The triangle $A_1^\ast A_2^\ast A_3^\ast$ is the medial triangle of $DEF$. Proof. Circles $\Gamma_2$ and $\Gamma_3$ are mapped to the circles with diameter $\overline{IE}$ and $\overline{IF}$, hence their second intersection $A_1^\ast$ is exactly the midpoint of $\overline{EF}$. $\blacksquare$ Claim: The triangle $B_1^\ast B_2^\ast B_3^\ast$ is homothetic to triangle $DEF$. Proof. This is the triangle determined by the lines $\Gamma_1^\ast$, $\Gamma_2^\ast$, $\Gamma_3^\ast$. Since $\Gamma_1^\ast$ is clearly perpendicular to $\overline{A_1 I}$, it is parallel to $\overline{EF}$, and similarly. $\blacksquare$ This means $A_1^\ast B_1^\ast$, $A_2^\ast B_2^\ast$, $A_3^\ast B_3^\ast$ are indeed concurrent as needed.

It suffices to show that the three circles are coaxial, i.e. intersect twice. Let $DEF$ be the intouch triangle. Invert around the incircle, denoting images with $\bullet'$. Then $\overline{A_2A_3}$ gets sent to the circle $\omega_1$ with diameter $\overline{ID}$, etc., and the circle $C_i$ gets sent to the common external tangent to $\omega_{i+1}$ and $\omega_{i+2}$ which is closer to $A_1'$ than $I$ for $1 \leq i \leq 3$ (note that $\triangle DEF$ is acute). We wish to prove that $\overline{B_1'A_1'},\overline{B_2'A_2'},\overline{B_3'A_3'}$ concur. Observe that $A_1'$ is the midpoint of $\overline{EF}$, etc., hence $\triangle A_1'A_2'A_3'$ is the medial triangle of $\triangle DEF$. Furthermore, it is clear that $\overline{DE} \parallel \overline{B_1'B_2'}$ since $\omega_1$ and $\omega_2$ have equal radius, so triangles $DEF$ and $B_1'B_2'B_3'$ are homothetic, hence there is a homothety sending $\triangle A_1'A_2'A_3'$ to $\triangle B_1'B_2'B_3'$ which implies the desired concurrence. $\blacksquare$

who made these point labels :sob: solved with popop614 We relabel the diagram one time. Let the triangle be $ABC$, still with incenter $I$. The $\Gamma_i$ closest to each $A,B,C$ will be $\omega_A, \omega_B,\omega_C$. The circles will intersect at $X_A, X_B, X_C$ instead of $B_1, B_2, B_3$. Let $DEF$ be the contact triangle. We invert around the incircle. $A'B'C'$ becomes the medial triangle of $DEF$, and the sides of triangle $X_A'X_B'X_C'$ are the common tangents of the three pairs of circles from $(A'IB'F), (A'IC'E), (B'IC'D)$ that lie strictly outside of $DEF$, as these three circles are the inverses of $AB, AC, BC$. Let the circles have centers $O_D$, $O_E$, and $O_F$. Note that these circles have the same diameter, which is also the circumradius of $DEF$. This implies that the common tangent of $(A'IB'F)$ and $(A'IC'E)$ must be parallel to $O_EO_F$, and similar for the other two. As a result, $O_DO_EO_F$ is similar to $X_A'X_B'X_C'$. However, a homothety of factor $2$ at $I$ sends $O_DO_EO_F$ to $DEF$, and another homothety of factor $-\frac12$ at the centroid of $DEF$ sends $DEF$ to $A'B'C'$. Thus, $A'B'C'$ is similar to $X_A'X_B'X_C'$, implying that there must be a center of homothety. As a result, $A'X_A', B'X_B', C'X_C'$ concur. Therefore, $(AX_AI),(BX_BI),(CX_CI)$ are coaxial, so their centers are collinear.

Geez bruh;

Let the contact triangle be $\triangle DEF$. Invert about the incircle: The sides of $\triangle A_1A_2A_3$ invert to congruent circles with diameters $ID$, $IE$, and $IF$. Circles $C_i$ map to pairwise external tangents of the above circles. The desired circles map to $A_i^*B_i^*$. Thus the condition is equivalent the desired circles being coaxial, so we need to show the three lines $A_i^*B_i^*$ being concurrent. This simply results from $\triangle A_1^*A_2^*A_3^*$ and $\triangle B_1^*B_2^*B_3^*$ being homothetic with parallel sides. $\blacksquare$

All indices are taken modulo $3$. Also rename $C_i$ to $\Gamma_i$. Let the projection of $I$ onto $A_iA_{i+1}$ be $D_{i+2}$. Then we do inversion wrt the incircle. Note that $\overline{A_iA_{i+1}}^* = (ID_{i+2})$ and $\Gamma_i^*$ is the common external tangent of $(ID_{i+1})$ and $(ID_{i+2})$. $(A_iB_iI)^* = \overline{A_iB_i}$. Our condition is equivalent to $(A_iB_iI)$ being coaxial which is then equivalent to $A_iB_i^*$ concurring. So then $A_iB_i$ must also concur. We can take an affine transformation sending $A_1A_2A_3$ to an equilateral triangle and by symmetry, $A_iB_i$ must concur. the affine transfomation is a bit dubious sorry

Invert about the incircle, let $P_i$ be the inverse of $A_i.$ By Fact 5, $A_iI$ is a diameter of $(A_{i+1}A_{i+2}I),$ so $I$ is the orthocenter of $P_1P_2P_3.$ Now let $Q_i$ be the inverse of $B_i.$ Since the $(IP_iP_{i+1})$ have the same radius we see $Q_{i+1}Q_{i+2}\perp IP_i\perp P_{i+1}P_{i+2}$ so $P_1P_2P_3,Q_1Q_2Q_3$ are homothetic so $P_1Q_1,P_2Q_2,P_3Q_3$ concur. Thus inverting back the circumcircles of $A_1B_1I,A_2B_2I,A_3B_3I$ are coaxial which finishes.

I think my solution is same as #17. I am also lazy to just do affine homography. It is equivalent to prove that $(A_1B_1I)$, $(A_2B_2I)$, $(A_3B_3I)$ are coaxial. Obviously we are going to invert about the incircle, and hence (with a bit of different notation); the problem transforms to this. Let $\triangle ABC$ be a triangle with $O$ circumcenter. Let the ``external" pairwise common tangent lines to $\overline{AO}$, $\overline{BO}$, $\overline{CO}$ determine a triangle $\triangle XYZ$. If $\triangle DEF$ is the medial triangle of $\triangle ABC$, then prove that $\overline{XD}$, $\overline{YE}$, $\overline{ZF}$ concur. Now obviously $\triangle XYZ$ is homothetic to $\triangle ABC$ and even though there are easier ways to do this probably, we can just take an affine homography sending $\triangle ABC$ to an equilateral triangle, and move on with our lives.