A regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ and an equilateral triangle $A_{1} A_{3} A_{5}$ are inscribed in a circle of radius $r$. From each vertex of this triangle, we measure a distance of $x$ in both directions along the sides. By connecting the endpoints of these $x$ distances in the same direction, the connecting lines form a hexagon.

$1^{0}$. For which value of $x$ will this hexagon be regular?

$2^{0}$. Determine the ratio of the areas of the two hexagons and the equilateral triangle.

$3^{0}$. Determine the ratio of the radii of the circles circumscribed around the two regular hexagons!
On the sides of $\triangle A_{1} A_{3} A_{5}$, starting from the vertices and measuring a distance $x$ in the positive direction of rotation, we get points $C_{1}$, $C_{2}$, $C_{3}$, and in the negative direction of rotation, analogous points $D_{1}$, $D_{2}$, $D_{3}$. The intersection points of the sides of the regular triangles $C_{1} C_{2} C_{3}$ and $D_{1} D_{2} D_{3}$ are $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$, $B_{5}$, $B_{6}$.

$C_{1}$ and $D_{1}$, $C_{3}$ and $D_{3}$ are symmetric point pairs with respect to the altitude $A_{1} A_{4}$ (axis of symmetry) of $\triangle A_{1} A_{2} A_{3}$. Therefore, the intersection point of $C_{1} C_{3}$ and $D_{1} D_{3}$, $B_{1}$, must lie on the line $A_{1} A_{4}$. Similarly, $B_{3}$ must lie on the line $A_{3} A_{6}$, and $B_{5}$ on the line $A_{5} A_{2}$, and it must be that $O B_{1}=O B_{3}=O B_{5}$, where $O$ is the center of the regular hexagon $A_{1} \ldots A_{6}$ and the circle circumscribed around it. $(O A_{i}=r)$. At the same time,

$$
B_{1} B_{3}=B_{3} B_{5}=B_{5} B_{1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_9cc0b3c19b8d166f12e3g-1.jpg?height=648&width=672&top_left_y=527&top_left_x=716)

If the polygon $B_{1} \ldots B_{6}$ is regular, then the points $B_{2}$, $B_{4}$, $B_{6}$ must also lie on the circle with center $O$ and radius $O B_{1}=O B_{3}=O B_{5}=\varrho$. Therefore, the side of the hexagon $B_{1} B_{2} \ldots B_{6}$ is $=\varrho$. Furthermore, $\alpha=O B_{1} B_{2} \varangle=O B_{1} B_{6} \varangle=$ $60^{\circ}$; these angles have the vertex angle: $\varepsilon=A_{1} B_{1} C_{1} \varangle=A_{1} B_{1} D_{1} \varangle=60^{\circ}$. This implies that the triangles $A_{1} B_{1} C_{1}$ and $A_{1} B_{1} D_{1}$ are right-angled and that

$$
\delta=C_{1} B_{1} B_{2} \varangle=D_{1} B_{1} B_{6} \varangle=60^{\circ} \text { s. i. t. }
$$

Accordingly, the triangles $C_{1} B_{1} B_{2}$, $D_{1} B_{1} B_{6}$, etc., are equilateral; their side length is equal to the side length of the regular hexagon $B_{1} \ldots B_{6}$, i.e., $\varrho$: $B_{1} C_{1}=B_{1} D_{1}=\ldots=\varrho=B_{1} B_{2}=\ldots$

However, it follows from the above that the hypotenuse of the right-angled triangle $A_{1} D_{1} D_{3}$, $A_{1} D_{3}$, is twice the length of the smaller leg, $x$, i.e., $A_{1} D_{3}=2 x$ and $A_{1} A_{3}=3 x$, so $x=\frac{A_{1} A_{3}}{3}=\frac{r \sqrt{3}}{3}$.

$2^{\circ} \cdot 3^{\circ}$. Furthermore: $A_{1} B_{1}=B_{1} D_{3}=2 \varrho$; thus $A_{1} O=r=3 \varrho$, i.e., $\varrho=\frac{r}{3}$, so

$$
\varrho: r=1: 3
$$

The area of the polygon $A_{1} A_{2} \ldots A_{6}$: $T_{6}=6 \cdot \frac{r^{2} \sqrt{3}}{4}=\frac{\sqrt{3} r^{2} 3}{2}$.

The area of the polygon $B_{1} B_{2} \ldots B_{6}$: $t_{6}=6 \frac{\varrho^{2} \sqrt{3}}{4}=6 \cdot \frac{r^{2}}{9} \cdot \frac{\sqrt{3}}{4}=\frac{r^{2} \sqrt{3}}{6}=\frac{T_{6}}{9}$.

The area of the triangle $A_{1} A_{3} A_{5}$: $T_{3}=\frac{T_{6}}{2}=\frac{3 r^{2} \sqrt{3}}{4}$.

$$
t_{6}: T_{3}: T_{6}=\frac{1}{9}: \frac{1}{2}: 1=2: 9: 18
$$

György Taksony (Ág. ev. g. VII. o. Bp.)