$A B C$ is an equilateral triangle $(A B=a)$. Around vertex $A$, we draw two arcs, one with radius $r$ beyond the extensions of sides $A B$ and $A C$, and the other with radius $a+r$ between the rays of angle $B A C$. Similarly, we proceed around vertices $B$ and $C$, and the six arcs enclose a so-called Reuleaux $\mathbf{R}$ domain of width $a+2 r$. We rotate this domain within a square frame of side $a+2 r$. What segments of the frame's sides does $\mathbf{R}$ rub against during this process: 11[^0]


[^0]:    ${ }^{1}$ Readers should cut out the shapes from cardboard, or draw one on transparent paper and move it over the other.
During rotation, the shape $\mathbf{R}$ touches the frame $\mathbf{K}$ at four points in any position, one point on each side. It is evident that every point on the perimeter of $\mathbf{R}$ will touch $\mathbf{K}$, but the reverse is not true; for example, the vertices of $\mathbf{K}$ and a certain neighborhood around them do not touch $\mathbf{R}$. However, if points $D$ and $E$ on a side of $\mathbf{K}$ can touch $\mathbf{R}$, then every point on the segment $D E$ can touch. Therefore, the question can also be posed as follows: how long are the segments on the sides of $\mathbf{K}$ that cannot touch $\mathbf{R}$, and what is the shortest distance that the contact point can be from the vertices of $\mathbf{K}$? - It is more convenient to follow the process by rotating $\mathbf{K}$ around the stationary $\mathbf{R}$, as this allows us to more easily utilize the fourfold rotational symmetry of $\mathbf{K}$, rather than the threefold rotational symmetry of $\mathbf{R}$. It is sufficient to observe the motion until the side $S T$ of $\mathbf{K}$ moves from a position parallel to $B C$ to a position perpendicular to $B C$ after a $+90^{\circ}$ rotation, landing on the position of $T U$, and to write the distances of the contact points of $S T$ and $T U$, $S^{*}$ and $U^{*}$, from $T$, as functions of the angle $\varphi$ between $B C$ and the side $S T$, while $0^{\circ} \leq \varphi \leq 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a3b91d9cfcbf77a3baf2g-1.jpg?height=517&width=1241&top_left_y=560&top_left_x=450)

While $0^{\circ} \leq \varphi \leq 30^{\circ}$, with our notation, $S^{*}$ touches $\mathbf{R}$ at a point on the half of the arc $B_{2} C_{1}$ centered at $A$, and for $30^{\circ} \leq \varphi \leq 90^{\circ}$, at a point on the arc $C_{1} C_{2}$ centered at $C$; $U^{*}$, on the other hand, touches $\mathbf{R}$ on the arc $C_{2} A_{1}$ centered at $B$ for $0^{\circ} \leq \varphi \leq 60^{\circ}$, and on the arc $A_{1} A_{2}$ centered at $A$ for $60^{\circ} \leq \varphi \leq 90^{\circ}$. The length of $T S^{*}$ is determined by the position of the side $T U$, and the length of $T U^{*}$ by the position of the side $S T$, so the sought lengths must be determined separately for the intervals $\left(0^{\circ}, 30^{\circ}\right),\left(30^{\circ}, 60^{\circ}\right),\left(60^{\circ}, 90^{\circ}\right)$.
I. For $0^{\circ} \leq \varphi \leq 30^{\circ}$, $S T \perp A S^{*}$ and $T U \perp B U^{*}$, so

$$
\begin{aligned}
& T S^{*}=U^{*} M=U^{*} B-M B=(a+r)-a \sin \left(30^{\circ}+\varphi\right) \quad \text { and } \\
& T U^{*}=S^{*} M=S^{*} A-M A=(a+r)-a \cos \left(30^{\circ}+\varphi\right)
\end{aligned}
$$

The minimum value of $T S^{*}$ occurs at $\varphi=30^{\circ}$, when the subtracted $M B$ is the largest, because $30^{\circ} \leq 30^{\circ}+\varphi \leq 60^{\circ}$ and $\sin \left(30^{\circ}+\varphi\right)$ is increasing for the values of $\varphi$ in question, and $T S_{\min }^{*}=r+a(1-\sqrt{3} / 2)=p$; similarly, $T U_{\min }^{*}$ is also $p$ at $\varphi=0^{\circ}$, because the cosine function is decreasing in the interval under consideration.

II. For $30^{\circ} \leq \varphi \leq 60^{\circ}$, $S T \perp C S^{*}$, and still $T U \perp B U^{*}$, so

$$
T S^{*}=U^{*} N=(a+r)-a \cos \varphi
$$

$\varphi=30^{\circ}$ gives $T S_{\mathrm{min}}^{*}=p$, and

$$
T U^{*}=S^{*} C+C N=r+a \sin \varphi
$$

and $T U_{\min }=r+a / 2=q$, also at $\varphi=30^{\circ}$, because the added term is the smallest then.

The $\varphi=30^{\circ}$ case is the common boundary point of the I. and II. intervals; according to this, in the vicinity of $\varphi=30^{\circ}$, $S^{*}$ moves from the midpoint $S_{0}$ of the side $S T$ towards $T$, then turns back towards $S_{0}$. Meanwhile, $q=(a+2 r) / 2$, so $U^{*}$ falls on the midpoint $U_{0}$ of the side $T U$ (this will not turn out to be the smallest value for all possible values), and $U^{*}$ moves from $T$ towards $U$.

III. Finally, for $60^{\circ} \leq \varphi \leq 90^{\circ}$, $S T \perp C S^{*}$ remains unchanged, but now $T U \perp A U^{*}$, so

$$
T S^{*}=U^{*} P=U^{*} A+A P=r+a \sin \left(\varphi-30^{\circ}\right)
$$

and $T S_{\min }^{*}=q$, because $30^{\circ} \leq \varphi-30^{\circ} \leq 60^{\circ}$ implies that $A P$ is the smallest when $\varphi-30^{\circ}=30^{\circ}, \varphi=60^{\circ}$. ($S^{*}$ moves from $S_{0}$ towards $S$.) On the other hand,

$$
T U^{*}=S^{*} P=S^{*} C+C P=r+a \cos \left(\varphi-30^{\circ}\right)
$$

and $T U_{\min }^{*}=q$, because $C P_{\min }$ occurs at $\varphi=90^{\circ}$ ($U^{*} \equiv U_{0}$).

The smallest values of $T S^{*}$ and $T U^{*}$, which appeared in sequence as $p, p, q$ and $p, q, q$, respectively, are smaller for $p$ because $1-\sqrt{3} / 2<1 / 2$. $T U_{\min }^{*}$ also gives the smallest value of $S S^{*}$, so the length of the segment of the side $S T$ (and any side) that has been rubbed is $S T-2 p=(a+2 r)-2 r-a(2-\sqrt{3})=a(\sqrt{3}-1)$, independent of $r$.

Gábor Dávid (Székesfehérvár, József A. g. III. o. t.)

Remarks. We could have also found the maximum values of $T S^{*}$ and $T U^{*}$. - Many solutions did not justify why we consider three cases with the rotational process: when both $S T$ and $T U$ touch the large arc of $\mathbf{R}$ with radius $a+r$, or both "small" - or one large and one small - although several types are indeed not possible.