On the circumference of a unit radius circle, we mark the vertices of a regular hexagon on one hand, and the vertices of a regular heptagon on the other. Between what limits does the perimeter of the convex hull of the marked points vary?
In the case of a unit radius circle, the marking of a single vertex on its circumference completely determines the desired regular hexagon or heptagon. The shape of the convex hull, since it is influenced only by the relative positions of the regular polygons, is determined by a single angle parameter. A fortunate choice of this parameter can simplify the necessary calculations.

Let us consider the initial position where the regular polygons share a common axis of symmetry but do not have a common vertex (see figure; the vertices of the regular hexagon are marked with solid dots, those of the regular heptagon with open circles). The convex hull is now a symmetric 13-sided polygon, whose vertices are formed by the vertices of the regular polygons.

$1984-10-298-1$. eps

We examine the central angles of this 13-sided polygon. The vertices of the regular hexagon fall into different central angle sectors of the regular heptagon, since $\frac{2 \pi}{6}>\frac{2 \pi}{7}$. Therefore, the regular heptagon has exactly one central angle sector that does not contain a hexagon vertex. This $\frac{2 \pi}{7}$ sized angle sector is necessarily symmetric with respect to the axis of symmetry of the convex hull. The adjacent central angles in the 13-sided polygon are equal due to symmetry, and are specifically $\frac{\pi}{42}$ in size, since their sum is $\frac{2 \pi}{6}-\frac{2 \pi}{7}$. Continuing in both directions, the sides of the central angles alternately coincide with the vertices of the hexagon and heptagon.

Let us call the direction of the central angles the direction of the smallest rotation less than $180^{\circ}$ that maps the side of the angle coinciding with a heptagon vertex (initial side) to the side coinciding with a hexagon vertex.

Except for the $\frac{2 \pi}{7}$ sized angle, every central angle has a symmetric pair, which is therefore equal in size but opposite in direction. If we start from the $\frac{\pi}{42}$ sized angles and proceed in the direction of the angle, each angle is $\frac{2 \pi}{42}$ larger than the previous one, since one angle can also be obtained from the previous one by rotating its initial side by $\frac{2 \pi}{7}$ and its other side by $\frac{2 \pi}{6}$ in the direction of the angle.

In the initial position, therefore, the convex hull has a central angle of $\frac{2 \pi}{7}$, and two pairs of central angles of $\frac{\pi}{42}, \frac{3 \pi}{42}, \frac{5 \pi}{42}, \frac{7 \pi}{42}, \frac{9 \pi}{42}, \frac{1}{4}$ in size.

Let us fix the regular hexagon and rotate the regular heptagon around the center in the positive direction, with the angle of rotation being $2 x$. Our goal is to have the convex hull take on all possible shapes. We claim that for this, it is sufficient for $2 x$ to vary between 0 and $\frac{\pi}{42}$. Indeed, when $2 x=\frac{2 \pi}{42}$, the initial position is restored. For $\frac{\pi}{42}<2 x<\frac{2 \pi}{42}$, we only get the mirror images of the convex hulls that were formed for $0<2 x<\frac{\pi}{42}$. The mirror axis is the common axis of symmetry that the regular polygons have when $2 x=\frac{\pi}{42}$. (In this case, there is also a common vertex, so the convex hull is a 12-sided polygon.)

Let us write down the perimeter of the convex hull as a function of $x$! As a result of the rotation, the negative direction central angles increase by $2 x$, the positive direction ones decrease by $2 x$, and the $\frac{2 \pi}{7}$ sized central angle does not change. Thus, we have the following 13 (or 12 if $2 x=\pi / 42$) central angles:

$$
\begin{gathered}
\frac{2 \pi}{7}, \frac{\pi}{42}-2 x, \frac{\pi}{42}+2 x, \frac{3 \pi}{42}-2 x, \frac{3 \pi}{42}+2 x, \frac{5 \pi}{42}-2 x, \frac{5 \pi}{42}+2 x, \frac{7 \pi}{42}-2 x, \frac{7 \pi}{42}+2 x \\
\frac{9 \pi}{42}-2 x, \frac{9 \pi}{42}+2 x, \frac{11 \pi}{42}-2 x, \frac{11 \pi}{42}+2 x
\end{gathered}
$$

In a unit radius circle, the length of the chord corresponding to the central angle $\alpha$ is $2 \sin \frac{\alpha}{2}$, so the perimeter $k(x)$ of the convex polygon is

$$
k(x)=2 \sin \frac{\pi}{7}+2 \cdot \sum_{i=1}^{6}\left(\sin \left(\frac{2 i-1}{84} \pi-x\right)+\sin \left(\frac{2 i-1}{84} \pi+x\right)\right)
$$

Based on the identity $\sin (\alpha+x)+\sin (\alpha-x)=2 \sin \alpha \cos x$,

$$
\begin{aligned}
& k(x)=2 \sin \frac{\pi}{7}+2 \cdot \sum_{i=1}^{6}\left(2 \sin \frac{(2 i-1) \pi}{84} \cdot \cos x\right)=A+B \cos x \\
& \text { where } A=2 \sin \frac{\pi}{7}=0.868, \text { and } B=4 \cdot \sum_{i=1}^{6} \sin \frac{(2 i-1) \pi}{84}=5.297
\end{aligned}
$$

Since $\cos x$ is strictly decreasing on the interval $0 \leqq x \leqq \pi / 84$,

$$
A+B \geqq k(x) \geqq A+B \cos \frac{\pi}{84}
$$

and equality holds only for $x=0$ and $x=\pi / 84$. This solves the problem.

Remark. The expression denoted by $B$ in (1) can also be written in closed form based on the following identity:

$$
\sin t+\sin 2 t+\cdots+\sin n t=\frac{\sin \frac{(n+1) t}{2} \sin \frac{n t}{2}}{\sin \frac{t}{2}}
$$

Then

$$
\begin{gathered}
\sum_{i=1}^{6} \sin \frac{(2 i-1) \pi}{84}=\sum_{i=1}^{11} \sin \frac{i \pi}{84}-\sum_{i=1}^{5} \sin \frac{2 i \pi}{84}=\frac{\sin \frac{\pi}{14} \cdot \sin \frac{11 \pi}{168}}{\sin \frac{\pi}{168}}-\frac{\sin \frac{\pi}{14} \sin \frac{5 \pi}{84}}{\sin \frac{\pi}{84}}= \\
=\frac{\sin \pi / 14}{\sin \pi / 84}\left(2 \sin \frac{11 \pi}{168} \cos \frac{\pi}{168}-\sin \frac{5 \pi}{84}\right)=\frac{\sin ^{2} \pi / 14}{\sin \pi / 84}
\end{gathered}
$$

In the last step, we used the following identity for the first term in the parentheses:

$$
2 \sin u \cos v=\sin (u+v)+\sin (u-v)
$$

Based on these,

$$
k(x)=2 \sin \frac{\pi}{7}+4 \frac{\sin ^{2} \pi / 14}{\sin \pi / 84} \cos x
$$

The value of $k(x)$ can be calculated more accurately with a pocket calculator:

$$
k\left(\frac{\pi}{84}\right)=6.1610929, \text { and } k(0)=6.1647971
$$

The perimeter of the convex hull is always between these two limits.