62nd Putnam 2001

Problem A6

A parabola intersects a disk radius 1. Can the arc length of the parabola inside the disk exceed 4?



Answer: yes (just).

Without loss of generality, we may take the circle to be x2 + (y - 1)2 = 1 and the parabola to be y = kx2. The parabola touches the circle at the origin and intersects it at ( ±X, Y), where X = √(2k - 1) )/k, Y=(2k - 1)/k. The arc length within the disk is L = 2 ∫0X √(1 + 4k2x2) dx. Substituting t = 2kx, and putting T = 2√(2k - 1), we have L = 1/k ∫0T √(1 + t2) dt.

Clearly k small is a poor choice - as k tends to zero the arc length obviously tends to zero. The extreme case k = infinity gives a degenerate parabola comprising two half-lines along the y-axis. That gives arc length exactly 4. So we might expect that for k large the gain from the extra width of the parabola (in the x-direction) would outweigh the loss from the loss of length in the y-direction (compared to k infinite).

So far the question is easy, but it is not clear what to do next. The integral can be evaluated exactly but it is a complicated: 1/2 t √(1 + t2) + 1/2 ln(t + √(1 + t2) ). That is not promising. The next obvious approach is to approximate the integral. The simplest approximation is √(1 + t2) > t. That evaluates to 4 - 2/k, so it is not quite good enough.

If we expand √(1 + t2) we get t + 1/(2t) - 1/(8t3) + ... . So that suggests that t + 0.4/t might be suitable. This type of approximation will not work for t near zero. However, we can easily check that √(1 + t2) > t + 0.4/t for t ≥ 1. Integrating the extra term from 1 to T gives 0.4/k ln T. Evidently ln T increases without limit (as k increases), so (0.4 ln T)/k will exceed 2/k for k sufficiently large.

A slightly slicker approximation (which avoids the need to approximate the integrand by zero for 0 < t < 1) is t + 1/(2(1 + t)).  


62nd Putnam 2001

© John Scholes
16 Dec 2001