SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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 $$IU\cdot IA=IV\cdot I{A}_1=r^2$$ 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 $A{A}_1\leftrightarrow$ $(IUV)$.

Define $B^{\prime }$, $C^{\prime }$ similarly, then to prove the original problem, we just need to show that the circles of diameter $I{A}^{\prime }$, $I{B}^{\prime }$, $I{C}^{\prime }$ have a common point differs from $I$.

Note that $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ 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 $A{A}_1$, $B{B}_1$, $C{C}_1$.

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 $XYZ\to DEF$ 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 $\Omega :S\to G$ (two corresponding points). Denote $J$ as the circumcenter of triangle $XYZ$ then $\Omega :J\to I$ 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 $(XYZ)$.