Problem:
Tetrahedron ABCD has AD=BC=28,AC=BD=44, and AB=CD=52. For any point X in space, define f(X)=AX+BX+CX+DX. The least possible value of f(X) can be expressed as mn, where m and n are positive integers, and n is not divisible by the square of any prime. Find m+n.
Solution:
Let M and N be midpoints of AB and CD, respectively. The given conditions imply that △ABD≅△BAC and △CDA≅△DCB, and therefore MC=MD and NA=NB. It follows that M and N both lie on the common perpendicular bisector of AB and CD, and thus line MN is that common perpendicular bisector. Points B and C are symmetric to A and D with respect to line MN. If X is a point in space and X′ is the point symmetric to X with respect to line MN, then BX=AX′ and CX=DX′, so f(X)=AX+AX′+DX+DX′.
Let Q be the intersection of XX′ and MN. Then AX+AX′≥2AQ and, similarly, DX+DX′≥2DQ, from which it follows that f(X)≥2(AQ+DQ)=f(Q). It remains to minimize f(Q) as Q moves along MN.
Allow D to rotate around line MN to point D′ in the plane AMN on the side of line MN opposite to A. Because N is on the axis of rotation, D′N=DN. It then follows that f(Q)=2(AQ+D′Q)≥2AD′, and equality occurs when Q is the intersection of AD′ and MN. Thus min f(Q)=2AD′. Because MD is the median of △ADB, the formula for the length of a median shows that 4MD2=2AD2+2BD2−AB2=2⋅282+2⋅442−522 and MD2=684. By the Pythagorean Theorem MN2=MD2−ND2=8.
Because ∠AMN and ∠D′NM are right angles, (AD′)2=(AM+D′N)2+MN2=(2AM)2+MN2=522+8=4⋅678. It follows that minf(Q)=2AD′=4678. The requested sum is 4+678=682.
Note that more is true. In any tetrahedron ABCD, the three lines passing through the pairs of midpoints of non-intersecting sides are concurrent. That is, the point Q above is the midpoint of MN as well as the midpoint of the segment connecting the midpoints of AC and BD and the midpoint of the segment connecting the midpoints of AD and BC. An easy way to show this is to represent the sides of the tetrahedron using vectors.