Problem 1598

Suppose that $(d)$ is tangent to $(I)$ at $K$. Denote $D$, $E$, $F$ as the tangent points of $(I)$ with the sides $BC$, $CA$, $AB$. Let $V$ be the projection of $K$ onto $A_{1}I$ and $U = AI \cap EF$. It is easy to see that
$$IU \cdot IA = IV \cdot IA_{1} = r^{2}$$
with $r$ is the radius of $(I)$.

Consider the inversion $\Omega$ of center $I$, and power $r^{2}$ then $A \leftrightarrow U$, $A_{1} \leftrightarrow V$ so $AA_{1} \leftrightarrow$ $(IUV)$.

Define $B'$, $C'$ similarly, then to prove the original problem, we just need to show that the circles of diameter $IA'$, $IB'$, $IC'$ have a common point differs from $I$.

Note that $A'$, $B'$, $C'$ belongs to the Simson's line of the point $K$ respect to triangle $DEF$ so by taking the homothety of center $I$, ratio $\frac{1}{2}$ then the center of above circles are collinear. Thus they share some other common point differs from $I$; the image of this point through the inversion is the concurrent point of three line $AA_{1}$, $BB_{1}$, $CC_{1}$.

Continue, consider the figure as below, the other case of position of points can be processed similarly. Denote $S$ as the reflection of $I$ through $K$. Then we have $YZ \parallel AI$, $YX \parallel IC$ so
$$\angle MYP = \angle XYZ = (YX, YZ) = (IA, IC) = 90^{\circ} - \frac{\angle B}{2} .$$
On the other hand, we have
$\angle MSP = \angle MIP = \angle KIM - \angle KIP = \frac{\angle KID - \angle KIE}{2} = \frac{1}{2} \angle DIF = 90^{\circ} - \frac{\angle B}{2}$, so these points $M$, $P$, $Y$, $S$ are concyclic which implies that $S \in (MPY)$.

Similarly, $S \in (NPX)$, $S \in (MNZ)$. From this, we can conclude that $S$ is the Miquel's point of the completed quadrilateral of 6 vertices $M$, $N$, $P$, $X$, $Y$, $Z$. Thus $S \in (XYZ)$.

Continue, note that by doing the similar angle chasing, we can see that two triangles $XYZ$, $DEF$ are similar (with the same direction). Hence, there exist a spiral similarity $\Omega$ transforms $XYZ \rightarrow DEF$ with the angle $90^{\circ}$ (since their sides are perpendicular pairwise). Draw the diameter $KG$ of circle $(I)$, then we have
$$\angle GDE = \angle GKE = \angle IKE = \angle KNI = \angle KNS = \angle PXS = \angle SXY .$$
Similarly, $\angle GED = \angle SYX$. Thus $\Omega: S \rightarrow G$ (two corresponding points). Denote $J$ as the circumcenter of triangle $XYZ$ then $\Omega: J \rightarrow I$ so we have $JS \perp IG$. But these points $I$, $G$, $K$, $S$ are collinear so $\angle JSK = 90^{\circ}$ which implies that $IK$ is tangent to the circle $(XYZ)$.