Let the angles of a triangle be $\alpha, \beta, \gamma$. Find the smallest and largest values of $\alpha$, rounded to the nearest tenth of a degree, for which $\beta-\gamma=30^{\circ}$, and

$$
10 \sin \alpha \sin \beta \sin \gamma-2 \sin 60^{\circ}\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)>0
$$

given that the values of $\alpha$ between $23.0^{\circ}$ and $100.0^{\circ}$ satisfy the inequality. (It is advisable to first transform the left side in such a way that during the trial and error process, there is no need to multiply table values and approximate values of irrational numbers.)
We need to find the pairs of $\alpha$ values rounded to the tenth of a degree, differing by $0.1^{\circ}$, between $23^{\circ}$ and $100^{\circ}$, such that one value still satisfies the given inequality (i.e., the left side is positive), while the other does not (i.e., the left side is negative). To simplify the calculations, we can write the left side of the inequality as follows:

1)

$$
(10 \sin \alpha-2 \sqrt{3})[\cos \alpha+\cos (\beta-\gamma)]-2 \sqrt{3}\left\{\sin ^{2} \alpha+(1-\cos \alpha)[1-\cos (\beta-\gamma)]\right\}
$$

By substituting $2 \cos (\beta-\gamma)=\sqrt{3}$ and $-2 \sin ^{2} \alpha=\cos 2 \alpha-1$, and then expanding and rearranging:

$$
\begin{gathered}
(10 \sin \alpha \cos \alpha+\sqrt{3} \cos 2 \alpha)+(5 \sqrt{3} \sin \alpha-3 \cos \alpha)-3 \sqrt{3}= \\
=(4 \sin 2 \alpha+\sin 2 \alpha+\sqrt{3} \cos 2 \alpha)+(5 \sqrt{3} \sin \alpha-5 \cos \alpha+2 \cos \alpha)-3 \sqrt{3}
\end{gathered}
$$

Finally, by recognizing the sum of sines with $\sqrt{3}=2 \sin 60^{\circ}=2 \cos 30^{\circ}$ and $1=2 \cos 60^{\circ}=2 \sin 30^{\circ}$:

$$
\begin{gathered}
4 \sin 2 \alpha+\left(2 \cos 60^{\circ} \sin 2 \alpha+2 \sin 60^{\circ} \cos 2 \alpha\right)+\left(10 \cos 30^{\circ} \sin \alpha-10 \sin 30^{\circ} \cos \alpha\right)+2 \cos \alpha-3 \sqrt{3}= \\
=4 \sin 2 \alpha+2 \sin \left(2 \alpha+60^{\circ}\right)+10 \sin \left(\alpha-30^{\circ}\right)+2 \cos \alpha-3 \sqrt{3}
\end{gathered}
$$

Thus, the values of $\alpha$ can be found by examining the sign of the expression:

$$
f(a) \equiv 2 \sin 2 \alpha+\sin \left(2 \alpha+60^{\circ}\right)+5 \sin \left(\alpha-30^{\circ}\right)+\cos \alpha-3 \sqrt{3} / 2
$$

Here, we can easily take values from the table, multiplying by integers as needed.

Already now, $f\left(30^{\circ}\right)=1$ (an easily calculable value) and at the given point: $f\left(23^{\circ}\right)=0.1128$, so for a $30^{\circ}-23^{\circ}=7^{\circ}$ decrease, the value of $f$ decreased by $1-0.1128=0.8872$. To make $f(\alpha)$ negative, a further decrease of more than 0.1128 is needed, which is roughly one-eighth of the previous decrease. Therefore, let's decrease $\alpha$ by a round $1^{\circ}$. Thus, $f\left(22^{\circ}\right)=-0.00720$, so the smallest sought value is $\alpha_{\min }=22.1^{\circ}$.

Similarly, $f\left(90^{\circ}\right)=\sqrt{3} / 2=0.8660$ (an easily determinable value), $f\left(100^{\circ}\right)=0.2580$, so for a $10^{\circ}$ increase, $f$ decreased by about 0.6. The further decrease needed is less than half of this, so let's increase $\alpha$ by roughly half of $10^{\circ}$. Thus, $f\left(105^{\circ}\right)=-0.0273$, so for a $5^{\circ}$ increase, a decrease of 0.2853 occurred. To reach $f\left(105^{\circ}\right)$, we need to increase by about one-tenth of this, so let's decrease $\alpha$ by $5^{\circ} / 10=0.5^{\circ}$. Then, $f\left(104.5^{\circ}\right)=+0.0002$, still positive, but $f\left(104.6^{\circ}\right)<0$, so the largest sought value of $\alpha$ is $\alpha_{\max }=104.5^{\circ}$.

Ferenc Kolonits (Budapest, Piarist g. IV. o. t.)

Note. If, for example, the angles of the triangle $A_{1} B_{1} C_{1}$ are $\alpha=22.8^{\circ}, \beta=93.6^{\circ}, \gamma=63.6^{\circ}$, or $\alpha=104.4^{\circ}, \beta=52.8^{\circ}$, $\gamma=22.8^{\circ}$, then the triangle $A_{2} B_{2} C_{2}$ constructed according to the article will have the same orientation.

${ }^{1}$ See Pál Lásincz: Remarks on a competition problem, expressions (2) and (5) in the KML. XVII. vol. 131-132. p. (December 1958).