IMO 1986 SL 19

Origin: BUL

Problem

A tetrahedron ABCD is given such that AD = BC = a; AC = BD = b; AB \cdot CD = c2. Let f(P) = AP + BP + CP + DP, where P is an arbitrary point in space. Compute the least value of f(P).

Solution

Let M and N be the midpoints of segments AB and CD, respectively. The given conditions imply that \triangleABD ∼= \triangleBAC and \triangleCDA ∼= \triangleDCB; hence MC = MD and NA = NB. It follows that M and N both lie on the perpendicular bisectors of AB and CD, and consequently MN is the common perpendicular bisector of AB and CD. Points B and C are symmetric to A and D with respect to MN. Now if P is a point in space and P ′ the point symmetric to P with respect to MN, we have BP = AP ′, CP = DP ′, and thus f(P) = AP + AP ′ + DP + DP ′. Let PP ′ intersect MN in Q. Then AP + AP ′ \geq2AQ and DP + DP ′ \geq2DQ, from which it follows that f(P) \geq2(AQ + DQ) = f(Q). It remains to minimize f(Q) with Q moving along the line MN. Let us rotate point D around MN to a point D′ that belongs to the plane AMN, on the side of MN opposite to A. Then f(Q) = 2(AQ + D′Q) \geq AD′, and equality occurs when Q is the intersection of AD′ and MN. Thus min f(Q) = AD′. We note that 4MD2 = 2AD2 + 2BD2 −AB2 = 2a2 + 2b2 −AB2 and 4MN 2 = 4MD2 −CD2 = 2a2 + 2b2 −AB2 −CD2. Now, AD′2 = (AM + D′N)2 + MN 2, which together with AM + D′N = (a + b)/2 gives us AD′2 = a2 + b2 + AB \cdot CD = a2 + b2 + c2 . We conclude that min f(Q) =

(a2 + b2 + c2)/2.