Pinko wrote: The points $A_1$,$B_1$,$C_1$ are middle points of the arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ of the circumscribed circle of $\Delta ABC$, respectively. The points $I_a,I_b,I_c$ are the reflections in the middle points of $BC,CA,AB$ of the center $I$ of the inscribed circle in the triangle. Prove that $I_a A_1,I_b B_1$, and $I_c C_1$ are concurrent. Use complex numbers with $a^2,b^2,c^2$ as reference triangle.Then trivially we get $K=\dfrac{abc(a+b+c)}{ab+bc+ca}=I_aA_1\cap I_bB_1\cap I_cC_1$.Furthermore this lies on $\odot{ABC}$

Here goes another synthetic solution Pinko wrote: The points $D$,$E$,$F$ are middle points of the arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ of the circumscribed circle of $\Delta ABC$, respectively. The points $P,Q,R$ are the reflections in the middle points of $BC,CA,AB$ of the center $I$ of the inscribed circle in the triangle. Prove that $DP,EQ$, and $FR$ are concurrent. Let $\triangle A_1B_1C_1$ be the medial triangle.Let $A_2,B_2,C_2$ be midpoints of $IA,IB,IC$ respectively.Clearly $C_2B_1A_1\sim CED$ via mid-point theorm.Hence after a Homothety at $I$ with power $2$ we have $\triangle CQP\sim \triangle CED\implies C : QP\to ED\implies DP\cap EQ\in \odot{ABC}$.Similarly we obtain the other concurrency also.$\square$

A different approach, use the same notation like the above post Pinko wrote: The points $D$,$E$,$F$ are middle points of the arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ of the circumscribed circle of $\Delta ABC$, respectively. The points $P,Q,R$ are the reflections in the middle points of $BC,CA,AB$ of the center $I$ of the inscribed circle in the triangle. Prove that $DP,EQ$, and $FR$ are concurrent. Let $O$ be the circumcenter of $\triangle ABC,$ line through $D$ parallel to $OI$ meets $(O)$ again at $X$, we will prove that $DP$ and $DX$ are isogonal wrt $\angle EDF$ or $\angle ADO,$ then $DP, EQ, FR$ will concurrent at a point on $(O)$ Let $M$ be the midpoint of $BC$ and $D'$ symmetry with $D$ wrt $M$ $DI^2=DB^2=2DM.DO=DD'.DO \Rightarrow \triangle DIO \sim \triangle DD'I \Rightarrow \angle ADX=\angle DIO = \angle DD'I = \angle ODP$ so done By the way, at first I misreaded the problem to $A_1$ is the reflection of midpoint of $BC$ in $I$ and those 3 lines are still concurrent