Pretty much a circle 1

For convenience, I'll refer to both the region and its area as $A(r)$ . Let $C(r)$ be the boundary of $A(r)$ , ie the curve defined by $\cos x + \cos y = r$

We'll work out the limit by sandwiching $A(r)$ between two circles.

Since for all $\theta$ , $1-\frac12 \theta^2 \le \cos \theta$ (with equality only at $\theta=0$ ), $A(r)$ contains the region $1-\frac12 x^2 + 1-\frac12 y^2 \ge r$ which is the interior of a circle centred at the origin with radius $\sqrt{4-2r}$ . Hence $A(r) > \pi(4-2r)$ for all $r$ .

Now, we'll find a region that contains $A(r)$ .

Claim: the circle through the intersection points of $C$ with the axes is always bigger than $A(r)$ .

Another way to put this is that for a point $P$ on $C$ , the distance $OP$ is maximised when $x=0$ or $y=0$ . (Slightly long-winded proof below.)

Now we have $\pi(4-2r) < A(r) < \pi \left(\cos^{-1} (r-1) \right)^2$

Dividing by $2-r$ , this is

$2\pi < \frac{A(r)}{2-r} < \frac{\pi \left(\cos^{-1} (r-1) \right)^2}{2-r}$

Two applications of L'Hôpital's rule show that as $r \to 2$ , the expression on the right also tends to $2\pi$ .

Hence by the squeeze theorem, in the limit, we have $\frac{A(r)}{2-r} \to \boxed{2\pi}$

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I'm sure there must be a simpler proof using a clever inequality. Any ideas?

Proof of claim: we want to maximise $x^2+y^2$ subject to $\cos x + \cos y = r$ . We can do this with Lagrangian multipliers: define $F(x,y,L)=x^2+y^2+L(\cos x + \cos y-r)$

The stationary points of $F$ solve the optimisation problem. Taking partial derivatives, we need all three of the following to hold $\begin{aligned} \frac{\partial F}{\partial x} &=2x-L\sin x=0 \\ \frac{\partial F}{\partial x} &=2y-L\sin y=0 \\ \frac{\partial F}{\partial L} &=\cos x + \cos y-r=0 \end{aligned}$

If neither $x$ nor $y$ is zero, we get the solutions $x=\pm y$ . In this case, $x^2+y^2=\left(\cos^{-1} \frac{r}{2} \right)^2$

But we can also make all three equations hold if (exactly) one of $x$ or $y$ is zero; we find $x^2+y^2=\left(\cos^{-1} (r-1) \right)^2$

It's (fairly) easy to show that $\cos^{-1} \frac{r}{2} < \cos^{-1} (r-1)$ for $0<r<2$ , so the first case corresponds to a minimum and the second to a maximum, and the claim is proved.