Given a regular octagonal frame and an equilateral triangular plate with a side length equal to the side of the octagon. The plate is attached to the frame with one of its sides on the inner side. The plate is rolled around the inner side of the frame. What area does the plate scrape as it reaches the side of the frame opposite to its starting position, and what area does it scrape until it returns to its original position?
Let the vertices of the frame be $A_{1}, A_{2}, A_{3}, \ldots, A_{s}$, the side lengths be $c$, the initial position of the disk: $A_{1} A_{2} B_{1}$, and the phases of rolling when it rests on each side: $A_{2} A_{3} B_{2}, A_{3} A_{4} B_{3}, A_{4} A_{5} B_{4}$, and at the first stop: $A_{5} A_{6} B_{5}$. During the rotation around $A_{2}$, the farthest points from $A_{2}$, the vertices, trace the $c$-radius arcs $A_{1} B_{2}$ and $B_{1} A_{3}$, and the entire disk rubs the circular segment corresponding to the shorter arc $A_{1} A_{3}$ of the $c$-radius circle centered at $A_{2}$. During the subsequent rotations, the rubbed areas are bounded by the arcs $A_{2} A_{4}, A_{3} A_{5}, A_{4} A_{6}$ centered at $A_{3}, A_{4}, A_{5}$, respectively, intersecting the previous boundary lines at $B_{2}, B_{3}, B_{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_55596655f58f368a2628g-1.jpg?height=494&width=501&top_left_y=436&top_left_x=801)

The angle of the octagon is $135^{\circ}$, so the angle between the adjacent sides of the triangular disk when it rests on two adjacent frame sides is $135^{\circ} - 2 \cdot 60^{\circ} = 15^{\circ}$. The area $t_{1}$ rubbed until the $A_{5} A_{6}$ side is reached can be decomposed into four $15^{\circ}$-angled circular segments, the $60^{\circ}$-angled circular segments corresponding to the initial and final positions $A_{1} A_{2} B_{1}$ and $A_{6} A_{5} B_{5}$, and three equilateral triangles, thus

$$
t_{1}=\frac{c^{2} \pi}{360^{\circ}}\left(4 \cdot 15^{\circ}+2 \cdot 60^{\circ}\right)+3 \frac{c^{2} \sqrt{3}}{4}=\frac{c^{2}}{4}(2 \pi+3 \sqrt{3}) \approx c^{2} \cdot 2.870
$$

In the case of rolling back to the initial position, the initial $A_{1} A_{2} B_{1}$ circular segment lacks the circular segment outside the chord $A_{1} B_{1}$, as the disk rubs this during the final rotation around $A_{1}$. The total rubbed area consists of 8 triangles and 8 small circular segments:

$$
t_{2}=\frac{8 c^{2} \sqrt{3}}{4}+8 \frac{c^{2} \pi}{360^{\circ}} \cdot 15^{\circ}=\frac{c^{2}}{3}(\pi+6 \sqrt{3}) \approx c^{2} \cdot 4.511
$$

Nagy Péter Tibor (Kiskunhalas, Szilády Á. g. II. o. t.)