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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.

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If the hypotenuse is $AB$ then $AC$ can't be the longest side. The hypotenuse is always longest. –  John Moeller Jan 19 '13 at 3:46
    
@John, perhaps one is meant to understand "side" as "leg". –  Gerry Myerson Jan 19 '13 at 4:08
    
@GerryMyerson ...yes, perhaps. It would be nice to have that clarified, however. –  John Moeller Jan 19 '13 at 4:09
    
Sorry, Gerry is right. That was slackness on my part. –  Joshua Ciappara Jan 19 '13 at 4:10
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@JoshuaCiappara : I gave this some thought -- if $X$ is a centroid of either $AMC$ or $BMC$, then it will be the centroid of $BMC$. To show this, let $AC$ become really long, almost as long as $AB$. Then $AMC$ is very skinny, and you can't find a point far from all three of $A$, $M$, and $C$ without choosing a point on one of the sides. When you do that though, you may as well choose a point inside $BMC$, which will be close to the midpoint of $MC$, but a little farther away (i.e., the centroid of $BMC$). –  John Moeller Jan 19 '13 at 4:54

1 Answer 1

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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.

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You might be right in this answer, but your claims in the simpler problem (the dot points) aren't completely obvious to me. Could you elaborate on why they are true? –  Joshua Ciappara Jan 19 '13 at 7:05
    
I guess it's just a matter of if you're NOT at the circumcentre, then your distance to some vertex is less than the radius of the circumcircle. What about dot point two? –  Joshua Ciappara Jan 19 '13 at 7:11
    
Without loss of generality we're closer to $B$ than to $A$ (in the "bottom half" of the triangle), by symmetry. Then we're just trying to maximize the minimum of the distances to $B$ and $C$. Both distances are increasing as we go away from $BC$, so the optimum must be on the side $AB$ somewhere. But as we move along that side (between $B$ and the midpoint of $AB$), one distance increases while the other decreases. Therefore the highest minimum occurs when they are equal. And that means it must lie on the perpendicular bisector of $BC$. –  Greg Martin Jan 19 '13 at 7:58

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