2nd Putnam 1939

Problem A7

Do either (1) or (2):

(1)   Let Ca be the curve (y - a2)2 = x2(a2 - x2). Find the curve which touches all Ca for a > 0. Sketch the solution and at least two of the Ca.

(2)   Given that (1 - hx)-1(1 - kx)-1 = ∑i≥0  ai xi, prove that (1 + hkx)(1 - hkx)-1(1 - h2x)-1(1 - k2x)-1 = ∑i≥0  ai2 xi.



(1)   It is not hard to see that C1 is a figure 8 lying on its side with the double point at (1,0), vertical tangents at (1,1) and (-1,1) and horizontal tangents at (1/√2, 1/2), (1/√2, 3/2), (-1/√2,1/2), (-1/√2,3/2). Ca is obtained from C1 by transforming (x,y) to (ax,a2y) (so stretching by a factor a in the x-direction and a factor a2 in the y-direction). So we get a sequence of ever-larger horizontal 8s centered ever-further up the y-axis. This suggests that the envelope is something like a parabola y = k x2.

At this point it is distinctly helpful to know more classical differential geometry than today's undergraduate. The classical method for finding the envelope for a 1-parameter family of curves (which usually works) is to differentiate wrt the parameter. Thus we eliminate a from:

  (y - a2)2 - x2(a2 - x2) = 0, and   4a(y - a2) + 2a x2 = 0

giving y = 3/4 x2 (or the y-axis, which is presumably a spurious solution arising from the double points).

Alternatively, we might guess that that the envelope is a curve y = k x2 for some k. The intersection of this with Ca is given by:
  (k2 + 1)x4 - (2k + 1) a2x2 + a4 = 0, or x2 = (2k + 1)a2/(2k2 + 2) +/- a2(4k - 3)1/2/(2k2 + 2).

For this to be a tangent we need double roots and hence k = 3/4. It is now easy to check that this parabola meets Ca at (2a/√5,3a2/5), (-2a/√5,3a2/5) and to check that the gradients match.

(2)   Fairly easy.

This is obviously not a general result (whereby we derive ∑ ai2xi from ∑ aixi), so we need to evaluate the ai. In fact, it is easily seen that ai = (hi+1 - ki+1)/(h - k). [For example, multiply the expansions of (1 - hx)-1 and (1 - kx)-1 to get ai = hi + hi-1k + ... + h ki-1 + ki = (hi+1 - ki+1)/(h - k).]

Multiplying across to try to show that higher powers of (1 - hkx)(1 - h2x)(1 - k2x) ∑ ai2xi   have zero coefficients is a mistake (doable, but much algebra). It is better to evaluate ∑ ai2xi directly. After substituting for ai, we are going to get terms of the form   ∑ zi   which evaluates immediately to (1 - z)-1, giving the expression in the question in partial fraction form.  


2nd Putnam 1939

© John Scholes
4 Sep 1999