\section*{Problem 4 - 340924}

Above the side \(AB\) of the equilateral triangle \(ABC\) with given side length \(a\), the square \(ABPQ\) is constructed outward.

Then, imagine this square as movable. It is to "roll" around the triangle \(ABC\) in a mathematically positive direction without slipping.

(At the beginning, only point \(B\) remains fixed, while the other points move until the segment \(BP\) comes into the position of \(BC\); then \(C\) remains fixed, and so on).

a) In this way, the square is rolled until it has a side that coincides with \(AB\) for the first time again (this does not have to be the side that was initially \(AB\)).

How long is the path traveled by
- the point \(A\),
- the center \(M\) of the square \(ABCD\),
- the midpoint \(H\) of the side \(AB\) of the square?

b) Starting from the initial state \(ABPQ\), not only one full "round trip around the triangle \(ABC\)" as described in a) was performed, but this was continued until the square had rolled such that each of the four points \(A, B, P, Q\) had returned to their original positions.
- How many full round trips around the triangle \(ABC\) took place in total from the initial state to the described end?
- How many full rotations of the square, relative to its own center \(M\), were performed in total?
}

(a) The full circumnavigation of triangle \(A B C\) takes place in three partial movements (see figure):

The first is the rotation around \(B\) through the obtuse angle \(\angle P B C\) of size \(360^{\circ}-60^{\circ}-90^{\circ}=210^{\circ}\); here, \(P, Q, A, M, H\) and \(C, R, S, N, J\) are transformed.

The second partial movement is the rotation around \(C\) by \(210^{\circ}\); here, \(R, S, B, N, J\) are transformed into \(A, T, U, O, K\). The third partial movement is the rotation around \(A\) by \(210^{\circ}\); here, \(T, U, C, O, K\) are transformed into \(B, P, Q, M, L\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1154.jpg?height=734&width=717&top_left_y=358&top_left_x=675)

The path of \(A\) therefore consists of the three circular arcs \(k_{A 1}=A P S, k_{A 2}=S T \mathrm{~m} k_{A 3}=T Q B\). They are arcs of central angles of size \(210^{\circ}\) in circles with radii \(a, a \sqrt{2}, a\). Thus, \(A\) travels a path of length

\[
\frac{\pi a \cdot 210^{\circ}}{180^{\circ}} \cdot(1+\sqrt{2}+1)=(2+\sqrt{2}) \cdot \frac{7}{6} \pi a \approx 12.5137 a
\]

The path of \(M\) consists of \(k_{M 1}=M N, k_{M 2}=N O, k_{M 3}=O M\); arcs of central angles of \(210^{\circ}\) in circles of radius \(\frac{a}{2} \sqrt{2}\); thus, \(M\) travels a path of length

\[
\frac{\pi a \cdot 210^{\circ}}{180^{\circ}} \cdot\left(3 \cdot \frac{1}{2} \sqrt{2}\right)=\frac{7}{4} \sqrt{2} \pi a \approx 7.7750 a
\]

For the path of \(H\), consisting of \(k_{H 1}=H L J, k_{H 2}=J K, k_{H 3}=K L\), the length is

\[
(1+2 \cdot \sqrt{5}) \cdot \frac{7}{12} \pi a \approx 10.0282 a
\]

(b) In the first circumnavigation, the points \(A, B, P, Q\) are transformed, as described in (a), into \(B, P, Q, A\), i.e., they have the same position as after a quarter rotation around \(M\). To ensure that they return to their initial position for the first time, such a circumnavigation must be performed 4 times. The segment \(M Q\) is transformed in the first partial movement described in (a) into the segment \(N R\). Relative to the center of the rolling square, this is a rotation of the size of the obtuse angle \(\angle Q B R=210^{\circ}\).

The same applies to the subsequent partial movements, so during the first circumnavigation, a rotation of \(3 \cdot 210^{\circ}\) occurs relative to the center of the square. In the determined 4 circumnavigations, this is a rotation of \(12 \cdot 210^{\circ}=2520^{\circ}=7 \cdot 360^{\circ}\), which is 7 full rotations.

Adapted from \([5]\)

\subsection*{6.36.3 Third Round 1994, Class 9}