Farthest Possible Points in a Rectangular Room A rectangular room has dimensions 12x12x24. That is, the floor and ceiling and both the side walls are 12x24 and the 2 end walls are 12x12. A point M corresponds to one of the vertices on the floor. If we define the distance between any 2 points on the surface area of the room to be strictly along the sides of the room, what is the location of point N, which is the farthest possible point away from M?
Hint: The location of N is not the diametrically opposite vertex of M.
So I've flattened the sides of the room on the 2 dimensional plane and considered simpler cases like cubes but I still can't figure it out. Any help?
 A: If we take the geodetic ball of radius $r$ centered in a vertex $0$, we can easily convince ourselves that its boundary is composed by a finite number of circular arcs. We increase the radius up to a value $R$ so that the whole space is reached. It means that such circular arcs collapse to a point $P$ (in general there can be many points). For this to happen it is necessary to have at least three circular arcs which correspond to three different geodesics from $O$ to $P$. So if we fix the point $P$ we need to find all possible developments of the box, find all possible positions of the images of point $O$ and identify the shortest geodesics.
Here is a picture where you can see that the point $P$ can be reached by three different paths of equal length from the point $O$ and its images $O'$ and $O''$. 
The coordinates of the points $O$, and its images $O'$ and $O''$ can be easily found. The point $P$ is the circumcenter of the triangle $OO'O''$. For example one can compute the equation of the axis of the two segments $OO''$ and $O'O''$ and make the intersection.
Then one finds that
$$
OP = O'P = O''P = \frac{\sqrt{130}}{4} AB.
$$
