10. In the answer, indicate $n_{1}+n_{k-4}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Answer: 36.

II. Among all triangles inscribed in a circle of fixed radius, with a known sum of the squares of all angles $\left(\alpha^{2}+\beta^{2}+\gamma^{2}=\pi^{2} / 2\right)$, find all triangles with the maximum possible area. For each such triangle, find the smallest value of all pairwise products of angles. In the answer, write the smallest of these values, rounding to two decimal places if necessary. All angles are expressed in radians.

Solution. We solve the problem of finding the maximum area $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, with the constraints $\alpha+\beta+\gamma=\pi, \alpha^{2}+\beta^{2}+\gamma^{2}=\pi^{2} / 2, \alpha, \beta, \gamma>0$. Order the angles: $\alpha \leqslant \beta \leqslant \gamma$.

We will show that in this case, the triangle is either obtuse or right, from which it follows that $\alpha+\beta \leqslant \pi / 2$. Indeed,

$$
\frac{\pi^{2}}{2}=\alpha^{2}+\beta^{2}+\gamma^{2} \leqslant(\alpha+\beta+\gamma) \cdot \gamma=\pi \cdot \gamma \Longrightarrow \gamma \geqslant \frac{\pi}{2}
$$

Excluding $\gamma$ from the condition $\alpha+\beta+\gamma=\pi$ and substituting into $\alpha^{2}+\beta^{2}+\gamma^{2}=\pi^{2} / 2$, we get $\alpha^{2}+\beta^{2}+(\pi-\alpha-\beta)^{2}=$ $\pi^{2} / 2$ or

$$
3(\alpha+\beta)^{2}-4 \pi(\alpha+\beta)+(\alpha-\beta)^{2}+\pi^{2}=0
$$

Introduce the notation $x=\alpha+\beta$ and $y=\alpha-\beta$. Then $x \in\left(0 ; \frac{\pi}{2}\right], y \leqslant 0$ and the following holds:

$$
3\left(x-\frac{2 \pi}{3}\right)^{2}+y^{2}=\frac{\pi^{2}}{3}, \quad x \in\left[\frac{\pi}{3} ; \frac{\pi}{2}\right], \quad y \in\left[-\frac{\pi}{\sqrt{3}} ; 0\right] .
$$

Here, the lower bounds on $x$ and $y$ are taken from equation (1) since, in particular, from it follow the inequalities $\left|x-\frac{2 \pi}{3}\right| \leqslant \frac{\pi}{3}$ and $|y| \leqslant \frac{\pi}{\sqrt{3}}$. The inequality $x=\alpha+\beta \leqslant \pi / 2$ is justified above. Note that from (1) we can find the explicit form $x(y)=\frac{2 \pi}{3}-\sqrt{\frac{\pi^{2}}{9}-\frac{y^{2}}{3}}$.

From equation (1), find the derivative of the function $x(y)$

$$
x^{\prime}(y)=\frac{y}{2 \pi-3 x}, \quad y \in\left[-\frac{\pi}{\sqrt{3}} ; 0\right]
$$

Note that $x^{\prime}(y) \leqslant 0$ for the considered $y$. Return to the extremal problem

$$
\begin{array}{r}
S=2 R^{2} \sin \alpha \sin \beta \sin \gamma=R^{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta)) \sin (\pi-\alpha-\beta)= \\
=R^{2}(\cos (y)-\cos (x)) \sin (x) \mapsto \max
\end{array}
$$

Denote $f(y)=(\cos (y)-\cos (x(y))) \sin (x(y))$ and show that the maximum of this function on the interval $y \in\left[-\frac{\pi}{\sqrt{3}} ; 0\right]$ is achieved at the point 0. This will mean that the extremal is an isosceles triangle. For this, it is sufficient to prove that $f^{\prime}(y) \geqslant 0$ for the considered arguments $y$. The following relations hold (in the future, for brevity, we will write $x$ instead of $x(y)$):

$$
\begin{aligned}
f^{\prime}(y)=-\sin y \sin (x)+ & x^{\prime}(y)(\cos x \cos (y)-\cos (2 x)) \geqslant \frac{2 x}{\pi} \cdot \frac{-2 y}{\pi}+x^{\prime}(y)(\cos x-\cos (2 x))= \\
& =\frac{4 x \cdot(-y)}{\pi^{2}}+\frac{y(\cos x-\cos (2 x))}{2 \pi-3 x}=\frac{(-y)}{2 \pi-3 x}\left(\frac{4 x(2 \pi-3 x)}{\pi^{2}}+\cos (2 x)-\cos (x)\right)
\end{aligned}
$$

Here, in obtaining the inequality, we used the inequality $\sin x \geqslant \frac{2 x}{\pi}$, which follows from the convexity of the function $\sin x$ on the interval $x \in[0 ; \pi / 2]$ and the analogous inequality $\sin (-y) \geqslant \frac{2(-y)}{\pi}$ on the interval $y \in[-\pi / 2 ; 0]$, as well as the inequality $\cos y \leqslant 1$.

Since in our case $\frac{(-y)}{2 \pi-3 x} \geqslant 0$ for $x \in\left[\frac{\pi}{3} ; \frac{\pi}{2}\right], y \in\left[-\frac{\pi}{\sqrt{3}} ; 0\right]$, to complete the proof that $f^{\prime}(y) \geqslant 0$ for the considered $y$, it is sufficient to prove that

$$
g(x)=\frac{4 x(2 \pi-3 x)}{\pi^{2}}+\cos (2 x)-\cos (x) \geqslant 0, \quad \forall x \in\left[\frac{\pi}{3} ; \frac{\pi}{2}\right]
$$

Assume that this inequality is proven (the proof will be given below). Then it follows that the function $f(y)$ achieves its maximum value at zero, and therefore $\alpha=\beta$, i.e., the triangle is isosceles.

Thus, returning to the original variables, we arrive at the equation:

$$
\alpha^{2}+\alpha^{2}+(\pi-2 \alpha)^{2}=\frac{\pi^{2}}{2}
$$

Solving the equation, we find: $\alpha=\frac{\pi}{6}$ and $\alpha=\frac{\pi}{2}$. Only the first option is possible since $\alpha$ is the smallest angle in the triangle. Therefore, $\alpha=\beta=\frac{\pi}{6}, \gamma=\frac{2 \pi}{3}$. Consequently, the smallest value of all possible pairwise products of angles is $\pi^{2} / 36 \approx 0.27$.

Answer: 0.27.

Proof of inequality (2). Further, all inequalities will be considered only on the interval $x \in A=\left[\frac{\pi}{3} ; \frac{\pi}{2}\right]$. We will show that $g^{\prime}(x)=\frac{4(2 \pi-6 x)}{\