Let's select one vertex of a regular 400-sided polygon. Among the diagonals originating from this vertex, find those for which, when subtracting twice their distance from the center, the absolute value of the difference deviates the least from the radius of the circle circumscribed around the polygon.
Let the consecutive vertices of the given polygon be $A_{0}, A_{1}, \ldots, A_{399}$, and let one of the diagonals starting from the selected vertex $A_{0}$ and satisfying the condition be $A_{0} A_{n}$. Due to axial symmetry, it is sufficient to consider indices $0 < n \leq 200$. The distance from the center $O$ to this diagonal is twice the length of $A_{200} A_{n}$, since the angle $A_{0} A_{n} A_{200}$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_05_02_33d6477469b75c61b012g-1.jpg?height=488&width=856&top_left_y=344&top_left_x=621)

As $n$ decreases from 200 to 1, $A_{0} A_{n}$ monotonically decreases, $A_{200} A_{n}$ monotonically increases, and thus the difference $A_{0} A_{n} - A_{200} A_{n}$ monotonically decreases. Therefore, $n$ can be determined by the condition that for the deviations

$$
d_{k} = \left|A_{0} A_{k} - A_{200} A_{k}\right| - A_{0} O
$$

the following must hold:

$$
\left|d_{n+1}\right| \geq \left|d_{n}\right| \quad \text{and} \quad \left|d_{n-1}\right| \geq \left|d_{n}\right|.
$$

From the symmetry of the expression, we can see that if these conditions are satisfied for a certain $n$, they will also be satisfied for $200 - n = n'$.

As a preparatory step, let's find a point $B$ on the circumscribed circle $k$ such that the absolute difference between the length of the chord $A_{0} B$ and twice the distance from $A_{200} B$ is exactly equal to $A_{0} O$; then $B$ will lie on the arc cut by one of the polygon's sides, and due to monotonicity, $A_{n}$ will be one of the endpoints of the side, or $B$ will coincide with a vertex, which will serve as $A_{n}$. Our requirement is:

$$
\begin{gathered}
\left|A_{0} B - A_{200} B\right| = \frac{A_{0} A_{200}}{2} \\
4 A_{0} B^{2} - 8 A_{0} B \cdot A_{200} B + 4 A_{200} B^{2} = A_{0} B^{2} + A_{200} B^{2}
\end{gathered}
$$

Thus, for the value

$$
\frac{A_{0} B}{A_{200} B} = \tan A_{0} A_{200} B = \tan \beta
$$

the following holds:

$$
\tan^{2} \beta - \frac{8}{3} \tan \beta + 1 = 0, \quad \tan \beta = \frac{4 \pm \sqrt{7}}{3} = \left\{\begin{array}{l}
0.4514 \\
2.215
\end{array}\right.
$$

(The product of the two roots is 1, and the angles corresponding to them are complementary, which is essentially the same as what was said about $n$ and $n'$.) Thus, the values of $\beta$ are: $65^{\circ} 42'$ (rounded) and $24^{\circ} 18'$ (rounded up), and the values of $A_{0} O B \angle = 2 \beta$ are $131^{\circ} 24'$ and $48^{\circ} 36'$.

On the other hand, $A_{0} O A_{1} \angle = \omega = 360^{\circ} / 400 = 0^{\circ} 54'$ and

$$
\frac{131^{\circ} 24'}{\omega} = 146 \text{ (rounded)}, \quad \frac{48^{\circ} 36'}{\omega} = 54 \text{ (rounded up)},
$$

so the appropriate points $B$ are on the arcs $A_{146} A_{147}$ and $A_{53} A_{54}$, much closer to $A_{54}$ than to $A_{53}$.

Let $A_{0} A_{200} = 1$, then

$$
A_{0} A_{k} = \sin k \cdot (0^{\circ} 27'), \quad A_{200} A_{k} = \cos k \cdot (0^{\circ} 27'), \text{ and indeed }
$$

$$
\left|d_{54}\right| = \left|d_{146}\right| = 0.0001
$$

but

$$
\left|d_{53}\right| = \left|d_{147}\right| = 0.0102, \quad \left|d_{55}\right| = \left|d_{145}\right| = 0.0105
$$

(We calculated the latter only because we could determine the position of $B$ only by considering the direction of rounding.) Therefore, the sought diagonals are: $A_{0} A_{54}, A_{0} A_{146}, A_{0} A_{254}$, and $A_{0} A_{346}$.