MHB Problem of the Week # 212 - April 19, 2016

[Ackbach]

Here is this week's POTW:

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An ellipse, whose semi-axes have lengths $a$ and $b$, rolls without slipping on the curve $y=c\sin\left(\dfrac{x}{a}\right).$ How are $a,b,c$ related, given that the ellipse completes one revolution when it traverses one period of the curve?

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[Ackbach]

Re: Problem Of The Week # 212 - April 19, 2016

This was Problem B-2 in the 1995 William Lowell Putnam Mathematical Competition.

No one solved this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

For those who haven't taken enough physics, ``rolling without
slipping'' means that the perimeter of the ellipse and the curve pass
at the same rate, so all we're saying is that the perimeter of the
ellipse equals the length of one period of the sine curve. So set up
the integrals:
$$
\int_{0}^{2\pi} \sqrt{(-a \sin \theta)^{2} + (b \cos \theta)^{2}}\,
d\theta = \int_{0}^{2\pi a} \sqrt{1 + (c/a \cos x/a)^{2}}\,dx.
$$
Let $\theta = x/a$ in the second integral and write 1 as $\sin^{2}
\theta + \cos^{2} \theta$ and you get
$$
\int_{0}^{2\pi} \sqrt{a^{2} \sin^{2} \theta + b^{2} \cos^{2}
\theta}\,d\theta
= \int_{0}^{2\pi} \sqrt{a^{2} \sin^{2} \theta +
(a^{2} + c^{2}) \cos^{2} \theta}\,d\theta.
$$
Since the left side is increasing as a function of $b$, we have
equality if and only if $b^{2} = a^{2} + c^{2}$.