On the sides of the square $A_{1} A_{2} A_{3} A_{4}$, we measure a distance of $x$ from each vertex in both directions. By connecting the endpoints of these $x$ distances in the same direction, we obtain the squares $B_{1} B_{2} B_{3} B_{4}$ and $C_{1} C_{2} C_{3} C_{4}$. The sides of these two squares intersect at the vertices of the octagon $D_{1} D_{2} \ldots D_{8}$.

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

$2^{0}$. The vertices of this regular octagon that lie on the diagonals of the original square form another square. Repeating the previous construction on this square, we obtain another regular octagon, and so on. What are the sums of the perimeters and areas of the infinitely many regular octagons thus obtained?
$1^{0}$. Measuring any distance $x$ along the sides of the square in the indicated manner, the distances $B_{1} B_{2}$ and $C_{1} C_{2}$ are symmetric with respect to the symmetry axis $K L$ of the square. $B_{1} B_{4}$ and $C_{2} C_{1}$ are symmetric with respect to the diagonal $A_{1} A_{2}$. This implies that the vertices $D_{1}, D_{3}, D_{5}, D_{7}$ lie on the symmetry axes of the opposite sides of the square, and the vertices $D_{2}, D_{4}, D_{6}, D_{8}$ lie on the diagonals, which are the symmetry axes of the square.

![](https://cdn.mathpix.com/cropped/2024_05_02_7c73d705dd708149a39ag-1.jpg?height=488&width=496&top_left_y=347&top_left_x=804)

Furthermore, $B_{1} B_{4}$ and $C_{1} C_{4}$ are symmetric with respect to $K L$; this implies that $B_{1} D_{1}=C_{1} D_{1}$, $B_{1} D_{2}=C_{1} D_{8}$, and $D_{1} D_{2}=D_{1} D_{8}$, symmetric with respect to the diagonals. Therefore, all sides of $D_{1} D_{2} \ldots D_{8}$ are equal, and their vertices lie on the symmetry axes of the square.

For the octagon to be regular, it is necessary and sufficient that each angle is $135^{\circ}$, i.e., $2 \alpha=135^{\circ} \left(\alpha=67.5^{\circ}\right)$. In other words, the angles in the right triangles formed by the sides of the octagon are $45^{\circ}$ (i.e., $B_{1} D_{1}=B_{1} D_{2}=C_{2} D_{2}=C_{2} D_{3}=\ldots$). Since $C_{2} A_{1} D_{2} \varangle=45^{\circ}$ and $A_{1} D_{2} C_{2} \varangle=67.5^{\circ}$, it must be that $A_{1} D_{2} C_{2} \varangle=67.5^{\circ}=\alpha_{1}$, so $A_{1} D_{2}=A_{1} C_{2}=x$. In the isosceles triangle $B_{1} C_{1} D_{1}$, the base angles are each $45^{\circ}/2 = 22.5^{\circ}$. Connecting point $C_{1}$ to point $O$, the common center of the regular octagon and the square, $O C_{1}$ connects the vertices of two isosceles triangles, $O D_{1} D_{8}$ and $C_{1} D_{1} D_{8}$, and thus bisects the angles at these vertices. This implies that

$$
A_{1} O C_{1} \varangle=45^{\circ}+22.5^{\circ}=A_{1} C_{1} O \varangle
$$

and thus

$$
A_{1} O=A_{1} C_{1}
$$

which means:

$$
\frac{1}{2} a \sqrt{2}=a-x, \quad \text { so } \quad x=\frac{a}{2}(2-\sqrt{2})=A_{1} A_{2}-A_{1} O
$$

This gives us the value of $x$, and thus the construction of the regular $D_{1} \ldots D_{8}$ polygon!

![](https://cdn.mathpix.com/cropped/2024_05_02_7c73d705dd708149a39ag-1.jpg?height=588&width=585&top_left_y=1728&top_left_x=759)

$2^{0}$. As we have seen, $A_{1} D_{2}=x$. Thus, $O D_{2}=A_{1} O-x=\frac{a \sqrt{2}}{2}-\frac{a}{2}(2-\sqrt{2})$, so $O D_{2}=a(\sqrt{2}-1)$ and $a_{8}=D_{1} D_{2}=2 O D_{2} \sin 22.5^{\circ}=2 a(\sqrt{2}-1) \sin 22.5^{\circ}$.

The apothem of the regular octagon is $h=O D_{2} \cos 22.5^{\circ}=a(\sqrt{2}-1) \cos 22.5^{\circ}$. Therefore, the perimeter of the $D_{1} D_{2} \ldots D_{8}$ polygon is

$$
k_{1}=16 a(\sqrt{2}-1) \sin 22.5^{\circ}=16 a(\sqrt{2}-1) \frac{\sqrt{2-\sqrt{2}}}{2}=8 a \sqrt{10-7 \sqrt{2}}
$$

and the area is

$$
\begin{gathered}
t_{1}=8 \cdot \frac{D_{1} D_{2}}{2} h=4 \cdot 2 a(\sqrt{2}-1) \sin 22.5^{\circ} \cdot a(\sqrt{2}-1) \cos 22.5^{\circ}=4 a^{2}(3-2 \sqrt{2}) \sin 45^{\circ}= \\
=2 a^{2}(3-2 \sqrt{2}) \sqrt{2}=a^{2}(6 \sqrt{2}-8)
\end{gathered}
$$

The vertices of the octagon on the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$ form a square, $D_{2} D_{4} D_{6} D_{8}$. The half-diagonal of this square is $O D_{2}=a(\sqrt{2}-1)$, and thus the side of this square, for example,

$$
D_{2} D_{4}=a^{\prime}=a(\sqrt{2}-1) \sqrt{2}=a(2-\sqrt{2})
$$

This implies that the side of the regular octagon inscribed in this square is $(2-\sqrt{2})$ times the side of the previous octagon, and so on. Therefore, the perimeters of the regular octagons formed in this manner form a geometric progression with a common ratio $q=2-\sqrt{2}$, and the areas form a geometric progression with a common ratio $q^{\prime}=(2-\sqrt{2})^{2}=6-4 \sqrt{2}$. It is clear that $0<q<1$ and $0<q^{\prime}<1$; thus, both infinite geometric series are convergent.

Therefore, the sum of the perimeters of the infinite number of octagons is:

$$
K=\frac{k_{1}}{1-q}=\frac{8 a \sqrt{10-7 \sqrt{2}}}{1-2+\sqrt{2}}=\frac{8 a \sqrt{10-7 \sqrt{2}}}{\sqrt{2}-1}=\frac{8 a(\sqrt{2}-1) \sqrt{2-\sqrt{2}}}{\sqrt{2}-1}=8 a \sqrt{2-\sqrt{2}}
$$

The sum of the areas is:

$$
T=\frac{t_{1}}{1-q^{\prime}}=\frac{a^{2}(6 \sqrt{2}-8)}{1-6+4 \sqrt{2}}=\frac{a^{2}(6 \sqrt{2}-8)}{4 \sqrt{2}-5}=\frac{a^{2}(6 \sqrt{2}-8)(4 \sqrt{2}+5)}{32-25}=\frac{a^{2}(8-2 \sqrt{2})}{7}
$$

Nagy Elemér (Ciszterci Szent Imre g. VII., Bp. XI.)

${ }^{2} \sin 22.5^{\circ}=\sqrt{\frac{1-\cos 45^{\circ}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}$ and $(\sqrt{2}-1)^{2}(2-\sqrt{2})=(3-2 \sqrt{2})(2-\sqrt{2})=10-7 \sqrt{2}$. Therefore, $(\sqrt{2}-1) \sqrt{2-\sqrt{2}}=\sqrt{10-7 \sqrt{2}}$.


[^0]:    ${ }^{1}$ The angles, however, are only equal alternately. The sum of two adjacent angles is $270^{\circ}$.