My favourite from LMAO this year. Suppose $\Delta_0=ABC$, $\Delta_1=P_AP_BP_C$, and $\Delta_2=Q_AQ_BQ_C$. Invert at $P$, and we'll denote the inverted image of any point $X$ by $X'$. Since $P$ is the Fermat point of $P_AP_BP_C$, it remains the Fermat point of $P_A'P_B'P_C'$ (angles at $P$ are preserved). Further $P_A'$ becomes center of circle $(B'C'P)$ and so on. Also being a reflection originally, $Q_A'$ becomes center of circle $(P_C'P_B'P)$ and so on, which are just centers of the three equilateral triangles constructed externally. Finally, since the sides $BC,CA,AB$ become circles passing through $P$ with centers $P_A,P_B,P_C$, the incircle becomes a common tangent to them, say $\ell$. We have to prove that $(Q_A'Q_B'Q_C')$ is also tangent to $\ell$. Let the radii of the above three circle be $a,b,c$ respectively. Let $G$ be centroid of $P_A'P_B'P_C'$, and suppose point $D$ is outside $P_A'P_B'P_C'$ such that $DP_B'P_C'$ is equilateral. By properties of the Fermat point, $P_A'D=a+b+c$. Then by basic proportionality from midpoint of $P_B'P_C'$, $GQ_A'= \frac{a+b+c}{3}$. Since this is symmetric in $a,b,c$, $G$ is the center of $(Q_A'Q_B'Q_C')$ and has radius $\frac{a+b+c}{3}$. But, since $P_A,P_B,P_C$ are a distance of $a,b,c$ respectively away from $\ell$ (and on the same side of $\ell$), and $G$ is their centroid, $G$ is a distance of $\frac{a+b+c}{3}$ away from $\ell$, so $(Q_A'Q_B'Q_C')$ is tangent to $\ell$ as well, as required.

Let $\Delta_0=\triangle ABC$, $\Delta_1=\triangle A_1B_1C_1$ and $\Delta_2=\triangle A_2B_2C_2$ where symmetry is preserved in naming. Also, $\triangle A_0B_0C_0$ is the intouch triangle of $\Delta_0$. Let $P_A$ be the reflection of $P$ in $B_0C_0$ and define $P_B$ and $P_C$ analogously. Finally, let $M_A$, $M_B$ and $M_C$ be the small arc midpoints of $\triangle A_0B_0C_0$. Let $Q$ be the Anti-Steiner point of $P$, which is known to be on $P_A M_A$ and $\omega$. We claim $Q$ is the point of tangency. Firstly, let's show: Lemma wrote: Let $\angle BAC=60^\circ$ and $\gamma=(ABC)$. If $X$ and $Y$ are the reflections of $A$ in the tangents at $B$ and $C$ to $\gamma$, then $XY$, $BC$ and the $A-$angle bisector concur.

Back to the problem. We get $PM_C$, $A_1B_1$ and $A_0B_0$ concur at, say, $K$. Claim: $PQP_CC_2$ is cyclic. Proof. \[\angle PC_2P_C=\angle PKA_0=\angle M_CKA_0=\angle PA_0M_C=\angle PQM_C=\angle PQP_C.\]Similarly, we get $PQP_BB_2$ and $PQP_AA_2$ are cyclic. Claim: $QA_2B_2C_2$ is cyclic. Proof. \[\angle A_2B_2C_2=\angle A_2B_2P+\angle PB_2C_2=\angle B_1C_1P+\angle PA_1B_1\]and \[\angle C_2QP=\angle C_2P_CP=\angle A_1KP=\angle B_1A_1P+\angle A_1PM_C\]hence

ZVFrozel wrote:

And even if $P$ doesn't lie on the incircle of $\Delta_0$, we still have that $(\Delta_0)$ is tangent to $(\Delta_2)$. The original problem can also be generalized by taking $P$ on any circle $\omega$ concentric with $(\Delta_0)$ intersecting the sides of $\Delta_0$. Then, we can prove that $\omega$ makes the same angle with both the sides of $\Delta_0$ and the circle $(\Delta_2)$.