What is the minimum number of sides of a regular polygon that approximates the area of its circumscribed circle with an error less than one thousandth?
Let the radius of the circle circumscribed around the polygon be $r$, and the number of sides of the polygon be $n$. Then the polygon can be divided into $n$ isosceles triangles, each with a vertex angle of $\frac{2 \pi}{n}$ and legs of length $r$ (Fig. 1). Therefore, the area of the polygon is $T_{n}=n \cdot \frac{r^{2}}{2} \sin \frac{2 \pi}{n}$. We know that the area of the circle is $T_{k}=r^{2} \pi$. The polygon approximates the area of its circumscribed circle with an error less than one thousandth if the following holds:

$$
\frac{T_{k}-T_{n}}{T_{k}} < \frac{1}{1000}
$$

This can be rewritten as:

$$
1 - \frac{T_{n}}{T_{k}} < \frac{1}{1000}
$$

or

$$
\frac{T_{n}}{T_{k}} > \frac{999}{1000}
$$

Substituting the expressions for $T_{n}$ and $T_{k}$, we get:

$$
\frac{n \cdot \frac{r^{2}}{2} \sin \frac{2 \pi}{n}}{r^{2} \pi} > \frac{999}{1000}
$$

Simplifying, we have:

$$
\frac{n}{2 \pi} \sin \frac{2 \pi}{n} > \frac{999}{1000}
$$

or

$$
n \cdot \sin \frac{2 \pi}{n} > \frac{999}{500} \pi \approx 6.276717
$$

If $n=81$, then (1) cannot hold because $81 \cdot \sin \frac{2 \pi}{81}$ is an integer.

Let $x=\frac{2 \pi}{n}$. It suffices to show that $\frac{\sin x}{x}$ is monotonically increasing as $x$ decreases (Fig. 2). This is because $\sin x < x$, and the graph of $\sin x$ is concave from below. Therefore, the line $O B_{2}$ intersects the segment $A_{1} B_{1}$ at an internal point $P$. Thus, if $x_{1} < x_{2}$, then:

$$
\frac{A_{1} P}{O A_{1}} = \frac{A_{2} B_{2}}{O A_{2}} = \frac{\sin x_{2}}{x_{2}}
$$

which is what we wanted to prove.

Therefore, regular polygons with at least 82 sides approximate the area of their circumscribed circle with an error less than one thousandth.

Based on the work of Kiss-Tóth Christian (Fazekas M. Főv. Gyak. Gimn., 8th grade)

Note. It is not entirely obvious how to find the value $n=81$. With a calculator and some patience, we can reach this value relatively quickly, but with some background knowledge, we can find the threshold more quickly.

It is known that if $x$ is a real number, then the sine of the angle whose radian measure is $x$ is the sum of the following infinite series:

$$
x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots + \ldots
$$

This is not really necessary here, but it helps to obtain upper and lower bounds (which become increasingly accurate) by including alternating terms.

For all positive $x$, the following inequality holds:

$$
\sin x > x - \frac{x^{3}}{3!}
$$

Dividing by $x > 0$, we get:

$$
\frac{\sin x}{x} > 1 - \frac{x^{2}}{6}
$$

If we rewrite the inequality $n \cdot \sin \frac{2 \pi}{n} > \frac{999}{500} \pi$ as:

$$
\frac{\sin \frac{2 \pi}{n}}{\frac{2 \pi}{n}} > 0.999
$$

then it certainly holds if:

$$
1 - \frac{\left(\frac{2 \pi}{n}\right)^{2}}{6} > 0.999
$$

Rearranging, we get:

$$
n > \frac{2 \pi}{\sqrt{0.006}} \approx 81.12
$$

Therefore, (1) certainly holds if $n \geq 82$, and it can be quickly determined that (1) does not hold for $n=81$, either by using a calculator or by considering the error in the inequality (*), which is smaller than the absolute value of the next term, $\frac{x^{5}}{5!}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_61aa8ebde2ec7a9dd7d9g-2.jpg?height=1290&width=1110&top_left_y=270&top_left_x=496)