Let's determine the angles of the "most irregular" acute triangle. We call a triangle the most irregular if the smallest difference between its angles and $90^{\circ}$ is as large as possible.
Let the angles of the sought triangle be $\alpha, \beta$, and $\gamma$, and let us choose the notation so that there is no smaller angle than $\alpha$, and no larger angle than $\gamma$, that is,

$$
\alpha \leq \beta \leq \gamma \leq 90^{\circ}
$$

The smallest difference $\delta$ in question can only be one of the differences between adjacent pairs in (1)

$$
\delta_{1}=\beta-\alpha, \quad \delta_{2}=\gamma-\beta, \quad \delta_{3}=90^{\circ}-\gamma
$$

or it can be more than one if they are equal, because the difference between non-adjacent pairs cannot be smaller, so

$$
\delta \leq \delta_{1}, \quad \delta \leq \delta_{2}, \quad \delta \leq \delta_{3}
$$

The sum of the differences in (2) is

$$
\delta_{1}+\delta_{2}+\delta_{3}=90^{\circ}-\alpha
$$

On the other hand,

$$
\begin{gathered}
\beta=\alpha+\delta_{1}, \quad \gamma=\beta+\delta_{2}=\alpha+\delta_{1}=\alpha+\delta_{2}, \text { so } \\
\alpha+\beta+\gamma=3 \alpha+2 \delta_{1}+\delta_{2}=180^{\circ}
\end{gathered}
$$

Expressing $\alpha$ from (4) and substituting it into (5)

$$
\delta_{1}+2 \delta_{2}+3 \delta_{3}=90^{\circ}
$$

Substituting the not larger $\delta$ for $\delta_{1}, \delta_{2}, \delta_{3}$ in (3), the value on the left side will be smaller or remain unchanged

$$
6 \delta \leq 90^{\circ}
$$

Accordingly, $\delta$ cannot be greater than $15^{\circ}$. However, it can take this value if all three differences are equal, that is, if $\delta_{1}=\delta_{2}=\delta_{3}=15^{\circ}$. In this case,

$$
\gamma=90^{\circ}-\delta_{3}=75^{\circ}, \quad \beta=\gamma-\delta_{2}=60^{\circ}, \quad \alpha=\beta-\delta_{1}=45^{\circ}
$$

these are the angles of the most irregular acute-angled triangle.

István Karsai (Szeged, Radnóti M. g. I. o. t.)