11th Putnam 1951

Problem B5

T is a torus, center O. The plane P contains O and touches T. Prove that P ∪ T is two circles.

Solution

We may take the torus to be ( √(x2 + y2) - a )2 + z2 = b2, where a > b. Viewed in the x-z plane the torus appears as two circles each radius b and centred at x = ±a, z = 0. The plane P appears as the line through the origin tangent to both circles and hence having equation c z = b x, where c = √(a2 - b2). P also contains the y-axis which is perpendicular to the line z = k x. Hence the equation of the plane P is z = k x.

It is far from obvious that P meets T in two circles. If we draw the plane P we see that T contains the four points on the y-axis, y = ±a ± b, and also the points a distance c either side of the origin. The obvious way to put these onto two circles is to take circles radius c, centred at (x, y, z) = (0, ±b, 0). These can be written as (x, y, z) = (c cos t, ±b + a sin t, b cos t). Hence √(x2 + y2) = a ± b sin t. So ( √(x2 + y2) - a )2 = b2sin2t and ( √(x2 + y2) - a )2 + z2 = b2sin2t + b2cos2t = b2. Thus these circles do indeed lie on the torus.

We may write the equation of T as 2a √(x2 + y2) = x2 + y2 + z2+ c2. Hence 4a2(x2 + y2) = ( x2 + y2 + z2+ c2 )2. So (x2 + y2 + z2 - c2)2 = 4a2x2 + 4a2y2 - 4c2(x2 + y2 + z2) = 4b2x2 + 4b2y2 - 4c2z2. Hence (x2 + y2 + z2 - c2)2 - (2by)2 = 4b2x2 - 4b2c2. So (x2 + (y + b)2 + z2 - a2)(x2 + (y - b)2 + z2 - a2) = 4(bx - cz)(bx + cz).

So if a point lies on the plane P, which has equation bx - cz = 0 and on the torus T, then it also lies on one of the two spheres (x2 + (y + b)2 + z2 - a2) = 0 and (x2 + (y - b)2 + z2 - a2) = 0. But a plane always intersects a sphere in a circle (or not at all) so the intersection of P and T is at most two circles. We have already established that it contains two circles, so it cannot contain any other points.