Point within triangle maximally distant from four points 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. 
 A: 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:


*

*the circumcenter, if it lies within the triangle (that is, if the triangle is acute);

*the intersection of $AB$ with the perpendicular bisector of $BC$, if the triangle is obtuse (that is, if angle $ACB$ is obtuse). We can also use the perpendicular bisector of $AC$ instead, by symmetry.


(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 $ACM$.


*

*In the first case, any point inside $BCM$ is farther from $A$ than it is from $M$. So the best we can do is take $X$ to be the circumcenter of $BCM$.

*In the second case, let $N$ be the midpoint of $AC$. Any point in triangle $ANM$ has an analogous point in triangle $CNM$ (its reflection in the line $NM$0; but the point in $ANM$ is farther from $B$ than the analogous point in $CNM$. Therefore the best we can do is to take $X$ to be the intersection of $AN$ with the perpendicular bisector of $AM$.


These are the only two choices for $X$, and I believe one can show that the latter choice is always optimal.
