We set the radius of the circle to 1, and we determine the chord in question by the central angle $2 \alpha (<\pi)$ associated with it. The diameter of the circle inscribed in the sector - which touches both the chord and the arc - cannot be larger than the segment of the perpendicular erected at the point of tangency on the chord that falls within the sector. This segment is the largest when it lies on the axis of symmetry of the sector, and in this position, the circle inscribed in the sector is the circle whose diameter is the segment itself; this is what we need to consider.

![](https://cdn.mathpix.com/cropped/2024_05_02_a8fd33143cec1c62db1fg-1.jpg?height=708&width=854&top_left_y=394&top_left_x=625)

The area of the sector is $\pi - (\alpha - \sin \alpha \cos \alpha)$, and the area of the inscribed circle is $\pi (1 + \cos \alpha)^2 / 4$. From these, the expression for the area in question can be transformed as follows:

$$
t = \frac{3 \pi}{4} - \alpha - \frac{\pi}{2} \cos \alpha - \frac{\pi}{4} \cos^2 \alpha + \sin \alpha \cos \alpha
$$

We seek the desired maximum based on the vanishing of the derivative.

$$
\begin{aligned}
t' = & -1 + \frac{\pi}{2} \sin \alpha + \frac{\pi}{2} \sin \alpha \cos \alpha + \cos^2 \alpha - \sin^2 \alpha = \\
& = \frac{\pi}{2} \sin \alpha \left(1 + \cos \alpha - \frac{4}{\pi} \sin \alpha \right) = \\
& = \frac{\pi}{2} \sin \alpha \left(2 \cos^2 \frac{\alpha}{2} - \frac{8}{\pi} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \right) = \\
& = \pi \sin \alpha \cos \frac{\alpha}{2} \left(\cos \frac{\alpha}{2} - \frac{4}{\pi} \sin \frac{\alpha}{2} \right)
\end{aligned}
$$

The zeros of the first two (variable) factors are of no interest to us: in the case of $\sin \alpha = 0$, $\alpha = 0$, the length of the chord is 0, the sector is identical to the given circle, the inscribed circle completely fills it, and the area under consideration is $0$; the root of $\cos \alpha / 2 = 0$ does not fall within the considered domain. The factor in parentheses vanishes if and only if

$$
\tan \frac{\alpha}{2} = \frac{\pi}{4} \approx 0.5708
$$

At this point, the derivative transitions from decreasing to increasing, because $\cos \alpha / 2$ is decreasing, and the second term is also decreasing, since $\sin \alpha / 2$ is increasing, and the factors in front are positive. Therefore, the area under consideration has a maximum at the angle defined by (1).

At the maximum point, the length of the chord is

$$
2 \sin \alpha = \frac{4 \tan \alpha / 2}{1 + \tan^2 \alpha / 2} = \frac{16 \pi}{16 + \pi^2} \approx 1.943
$$

$\left(\alpha = 76^\circ 17.5'\right)$.
In a given circle, we draw a chord and inscribe the largest possible circle in the larger of the resulting circular segments. For what choice of the chord's length will the area of the segment outside the inscribed circle be the largest?