### 15th Putnam 1955

Problem B7

A three-dimensional solid acted on by four constant forces is in equilibrium. No two lines of force are in the same plane. Show that the four lines of force are rulings on a hyperboloid.

Solution

Let the lines be L1, L2, L3, L4. If the line L intersects each of L1, L2, L3, then the moment of the first three forces about L is zero. Hence the moment of the fourth force about L is also zero and so L must also intersect L4. This is sufficient to prove that L1, L2, L3, L4 lie in a ruled quadric.

In fact the union of all the lines which intersect L1, L2, L3 is a ruled quadric. Let P be a point on L4. For a point X and a line L not containing it we take XL to be the unique plane containing X and L. If L and L' are skew lines and X does not lie on either, then XL ∩ XL' is the unique line through X which meets L and L'. So PL1 ∩ PL2 is the unique line through P which meets L1 and L2. Suppose it meets L1 at Q. Now QL2 ∩ QL3 meets L1, L2 and L3 so it must also meet L4. But there is a unique line through Q meeting L2 and L4, so QL2 ∩ QL3 = PL1 ∩ PL2. Hence P lies on the line QL2 ∩ QL3 which is a line of the quadric. P was arbitrary, so we have shown that L4 lies in the quadric.

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002