The ever-challenging Adam Atkinson, having noticed my attention to the “impossible” New Zealand exams, pointed me at a tricky question from an Italian exam which asked students to verify that, to give a smooth ride on a bike with square wheels (of side length 2), the height of the floor would need to be of the form $h(x)=\sqrt{2}-\frac{e^x + e^{-x}}{2}$.

I quite like that as a puzzle - it’s a nice mix of geometry and calculus, but I’d be surprised to see it as an A-level question, at least without a good deal of scaffolding.

Below the line be spoilers.

Verification

Verifying that it works isn’t too difficult. For a smooth ride, the height of the centre of the square must stay constant, and the point of contact must be directly below the centre.

Suppose the square rests at a point on the surface with $x$-coordinate, uh, $x$. The gradient there is $h’(x)=-\frac{e^x - e^{-x}}{2}$, which is $\tan(\theta)$, with $\theta$ being the angle between the bottom of the square and the horizontal.

Let $C$ be the centre of the square, $T$ the point of contact and $P$ the point on the base of the square such that $CPT$ is a right angle. Then, angle $TCP=\theta$ and length $CP=1$.

The distance $PT$ is then $\sec(\theta)$, or $\sqrt{1+\tan^2(\theta)}$.

Pretending we haven’t spotted the hyperbolic function, we can square $\tan(\theta)$ to get $\frac{e^{2x}- 2 + e^{-2x}}{4}$, add 1 to get $\frac{e^{2x}+ 2 + e^{-2x}}{4} = \frac{\br{e^{x}+ e^{-x}}^2}{4}$, and square root to get $PT = \frac{e^x + e^{-x}}{2}$.

Adding this on to $h(x)$ gives $h(x)+\sec(\theta)=\sqrt{2}$, and the centre of the wheel remains at a constant height as the wheel moves.

But wait… we could have derived that, and it would have been way more satisfying.

Deriving

The derivation is roughly the same thing, but backwards (with a *really* nice bit of trig towards the end).

Start with the triangle $CPT$ as before, and define $h(x)$ to be the function that keeps $PT + h(x) = \sqrt{2}$ (the height of the centre when the wheel is standing diagonally on its corner).

Now, as before, let $\theta$ be the angle $TCP$; the bottom of the square forms an angle of $\theta$ with the horizontal, which means:

  • $h(x) = \sqrt{2} - \sec(\theta)$
  • $h’(x) = \tan(\theta)$

Differentiating $h(x)$ gives $h’(x) = - \sec(\theta)\tan(\theta) \diff {\theta}{x}$, which means $\diff \theta x = - \cos(\theta)$.

That’s a differential equation we can solve.

$\int \sec(\theta) \d \theta = \int - \d x$

$\ln \left| \tan(\theta)+ \sec(\theta) \right| = -x + C$

Knowing that $\theta = 0$ when $x=0$ means $C=0$.

$\left| \tan(\theta) + \sec(\theta) \right| = e^{-x}$.

For $-\piby 4 < \theta < \piby 4$, which is the interval we’re working on, the thing inside the absolute values is positive, so we can remove them.

$\tan(\theta) + \sec(\theta) = e^{-x}$

We can rewrite the left-hand side as $\frac{\sin(\theta)+1}{\cos(\theta)}$, so we have $\frac{\sin(\theta)+1}{\cos(\theta)} = e^{-x}$ and (taking reciprocals) $\frac{\cos(\theta)}{\sin(\theta)+1} = e^x$.

Adding these together gives $ \frac{\br{\sin(\theta)+1}^2 + \cos^2 (\theta)}{\cos(\theta)\br{1+\sin(\theta)}} = e^x + e^{-x}$

Expanding and simplifying the top gives $\frac{ \sin^2(\theta) + 2 \sin(\theta) + 1 + \cos^2(\theta)}{\cos(\theta)\br{1 + \sin(\theta)}}$

But wait! That’s $\frac{2\br{1+\sin(\theta)}}{\cos(\theta)\br{1+\sin(\theta)}}$, or simply $2\sec(\theta)$!

Therefore, $\sec(\theta)= \frac{e^x + e^{-x}}{2}$, and, going all the way back, $h(x) = \sqrt{2} - \frac{e^x + e^{-x}}{2}$.

I reckon that deserves a $\blacksquare$.