The lines which connect the vertices of a triangle with the tangent points between the Spieker circle and the medial triangle are concurrent. Which kimberling center does this point correspond to?
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It's the Nagel point $X_8$.
Name the incircle of $\triangle ABC$ $\gamma$, the incircle of the median triangle $\omega$. If we enlarge $\omega$ by a factor of $2$ from the center $A$, we get the incircle $\omega_A$ of $\triangle BCA'$, where $A'$ is the fourth vertex of the parallelogram $ABA'C$. Thus the point at which $\omega_A$ is tangent to $BC$ and the point at which $\gamma$ is tangent to $BC$ are symmetric about $BC$'s midpoint. By the well-known property of excircle tangent points, the point at which $\omega_A$ is tangent to $BC$ is the point at which the $A$-excircle is tangent to $BC$. This point is also the image of the point at which $\omega$ is tangent to the $A$-median under enlarging by $2$ from $A$. Thus (using the point labels in the diagram) the intersection of $AI$ and $BC$ is the point at which the $A$-excircle is tangent to $BC$. We thus recover the definition of the Nagel point.
