Suppose I have a right triangle $ABC$ where the midpoint of the hypotenuse $AB$ is $M$, and the side $AC$ is longer than $BC$. How to determine the point $X$ within the triangle such that the distance from $X$ to any of $A$, $B$, $C$, $M$ is as great as possible? That is, the minimum of $XA$, $XB$, $XC$, $XM$ is maximal? I'm speculating it could be the centroid of $AMC$, but I'm really not sure. Thanks.
First let's solve a simpler problem. Given an isosceles triangle $ABC$ with vertex $C$, the point $X$ inside the triangle that maximizes the minimal distance from the three vertices is:
(For right isosceles triangles, take the limit in either case - we get the center of the hypotenuse $AB$.)
Now for the problem given: the point $X$ might be inside the acute isosceles triangle $BCM$, or it might be inside the obtuse isosceles triangle $ACM$.
These are the only two choices for $X$, and I believe one can show that the latter choice is always optimal.