Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.

TheDarkPrince wrote: Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem. Just a simple question to this nice solution: Why $S$ goes to $R$ and not to another point $Q$ ,of the circle $(AI)$ ? Thank you ! ( Ok ! The circle $(A,B,C)$ has invers the Euler circle of triangle $DEF$ and $S$ belongs to this circle.Allmost obvious, but ....)

Dear Matlinkers, http://www.artofproblemsolving.com/community/c6h614584 Sincerely Jean-Louis

Lets complex bash Set the incircle as the unit circle with incentre as the origin and let $d = 1$. Then $a = \frac{2ef}{e + f} , b = \frac{2f}{1 + f} , c = \frac{2e}{1 + e}$ $\therefore \frac{a - b}{\overline{a - b}} = -f^2$. $ nb \perp ba \implies -\frac{a - b}{\overline{a - b}} = \frac{c - n}{\overline{c - n}} = f^2$ $\therefore \overline{n} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$ Similarly $\overline{n} = \frac{(1+e)n-2e+2e^2}{e^2(1+e)}$ $\therefore \frac{(1+e)n-2e+2e^2}{e^2(1+e)} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$ After simplifying we obtain that $n = (1 + e + f - ef)k$ where $k = \frac{2ef}{(e + 1)(f + 1)(e + f)}$ Now we also obtain that $\overline{k} = \frac{2ef}{(e + 1)(f + 1)(e + f)}$ So $\overline{k} = k \implies k \in\mathbb{R}$ Now also note that $s = \frac{1}{2}(1 + e + f - ef)$ So $\frac{n}{s} = 2k \in \mathbb{R}$ So $N$, $I$ and $S$ are collinear.

Let $L$ be the midpoint of arc $BAC$ in $\Gamma$, and let $M$ be the antipode of $L$ in $\Gamma$. Let $K$ be the second intersection point of $MD$ with $\Gamma$. Claim 1. $K$ is the Miquel point of $BFEC$. Let $\Psi$ be the inversion around $(BIC)$. By the Incenter Lemma, we get that such inversion is centered in $M$. Clearly, $$ \left\{\begin{array}{lll} \Psi ((ABC))=BC \\ \Psi (KM)=KM \\ \end{array} \right. \Longrightarrow \Psi (K)=D $$Therefore, $$ MD\cdot MK=MI^2 \Longrightarrow (KID) \thickspace \text{tangent to} \thickspace IM $$Hence $$ \measuredangle DKI=\measuredangle DIM=\measuredangle LMA=\measuredangle LKA $$But then $$ 90^o=\measuredangle MKL=\measuredangle DKL=\measuredangle DKI+\measuredangle IKL=\measuredangle LKA+\measuredangle IKL=\measuredangle IKA $$This way $K\in (AEIF)$ and so Claim 1 is proved. Claim 2. $K,S,I$ are collinear. Right triangles $\triangle FSD$ and $\triangle IDC$ are similar because $$ \measuredangle DFS=\measuredangle DFI+\measuredangle IFS=\measuredangle DBI+\measuredangle IAE=90^o-\measuredangle ICD $$Analogously, right triangles $\triangle ESD$ and $\triangle IDB$ are similar. From these two similarities, we get $$ \left\{\begin{array}{lll} \frac{FS}{SD}=\frac{ID}{DC} \\ \\ \frac{SE}{SD}=\frac{ID}{BD} \\ \end{array} \right. \Longrightarrow \frac{FS}{SE}=\frac{BD}{DC} \qquad (\star) $$From Claim 1, we get that $K$ is the center of the spiral similarity sending $BC$ to $FE$. But from $(\star)$, we get that such spiral similarity also sends $D$ to $S$. Hence $\triangle KFS$ and $\triangle KBD$ are similar. Consequently, $$ \measuredangle FKS=\measuredangle BKD=\measuredangle BKM=\measuredangle BAM=\measuredangle FAI=\measuredangle FKI $$Therefore, $K,S,I$ are collinear and so Claim 2 is proved. Finally, from Claim 1 we have $K\in (AEFI) \Longrightarrow IK\perp KA$. But, in light of Claim 2, this implies that lines $ISK$ and $AK$, are perpendicular, so ray $SI$ will intersect $\Gamma$ at $N$, the antipode of $A$.

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