Let's examine the accuracy of the following approximate construction for the side of a regular nonagon inscribed in circle $k$.

The regular hexagon inscribed in $k$ has four consecutive vertices $A, B, C, D$. The intersection point of the circle centered at $B$ with radius $BA$ and the circle centered at $D$ with radius $DB$ inside $k$ is $G$. The intersection point of line $BG$ and line $AC$ is $H$. Then the side of the nonagon is approximated by the length of segment $BH$.
Let the radius of the circle be denoted by $r$, so the base $B G = r$ and the legs $D B = D G = r \sqrt{3}$ of the triangle defined by these lengths give us

$$
\cos \angle D B G = \cos \beta = \frac{1}{2 \sqrt{3}}, \quad \sin \beta = \frac{\sqrt{11}}{2 \sqrt{3}}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_2f94af534a2e728263b4g-1.jpg?height=1106&width=1399&top_left_y=374&top_left_x=360)

In the triangle $B C H$, the angle at $H$ is $120^{\circ} - \beta$, since clearly $\angle B C H = 30^{\circ}$ and $\angle C B H = 30^{\circ} + \beta$. Therefore, using the Law of Sines and the angle addition formula, we have

$$
\begin{aligned}
\frac{B H}{B C} & = \frac{\sin 30^{\circ}}{\sin (120^{\circ} - \beta)} = \frac{1}{\sqrt{3} \cos \beta + \sin \beta} \\
B H & = \frac{2 r}{1 + \sqrt{\frac{11}{3}}} = \frac{2 \sqrt{3} r}{\sqrt{11} + \sqrt{3}} = \frac{r}{4} (\sqrt{33} - 3)
\end{aligned}
$$

Since, according to our table, $5.7445 < \sqrt{33} < 5.7455$, we have

$$
0.6861 r < B H < 0.6864 r
$$

On the other hand, the side length of the regular nonagon is $a_9 = 2 r \sin 20^{\circ}$, and here $0.34195 < \sin 20^{\circ} < 0.34205$, so

$$
0.6839 r < a_9 < 0.6841 r.
$$

Thus, $B H$ approximates $a_9$ from above, and the error is less than $(0.6864 - 0.6839) r = 0.0025 r$, which is less than $4\%$ of the constructed segment.

Remarks. 1. On our diagram, the arcs with radius $B H$ around points $B$, $D$, and the point $C^*$ opposite $C$ also mark the vertices of the approximate nonagon.

2. According to more precise tables, the error is less than $0.0022 r$ and does not exceed $3\%$.

3. For clarity, we also provide the angular error of the approximation: the central angle $2 \alpha$ corresponding to the chord $B H$ lies between $40^{\circ} 7'$ and $40^{\circ} 8'$, while the angle corresponding to $a_9$ is $40^{\circ}$. Thus, the angular error is less than 8 minutes of arc, and the relative error is $1/300$.

4. According to the following calculation, we use the sine table only once.

$\sin 3 \alpha = \sin \alpha (3 - 4 \sin^2 \alpha) = \frac{\sqrt{33} - 3}{8} \cdot \frac{3 + 3 \sqrt{33}}{8} = \frac{45 - 3 \sqrt{33}}{32} = 0.8677$.

$3 \alpha = 60^{\circ} 11.5'$.

5. The construction was communicated by Fr. Fronek, who was then a high school student in Steyr, Austria.