A right-angled triangle's inscribed circle passes through the triangle's centroid. What are the acute angles of the triangle?
Let the midpoint of the hypotenuse be $F$, and use the additional notations from the diagram. The circle drawn around vertex $C$ with radius $CE = CH$ contains the intersection point $N$ of the median $CF$ and the incircle closer to $C$, thus

$$
C N < C E = \frac{a+b-c}{2} \leqslant \sqrt{\frac{a^{2}+b^{2}}{2}} - \frac{c}{2} = c\left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) < \frac{c}{3} = \frac{2}{3} C F
$$

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$$

Since the triangle is right-angled and $M$ is its centroid, $C M = \frac{2}{3} C F = \frac{c}{3}$.

First, let's write the power of point $C$ and then the power of point $F$ with respect to the incircle using the secant theorem:

$$
C N \cdot \frac{c}{3} = \left(\frac{a+b-c}{2}\right)^{2}
$$

$$
\frac{c}{6}\left(\frac{c}{2} - C N\right) = F G^{2} = \left(\frac{b-a}{2}\right)^{2}
$$

Adding the (1) equation to twice the (2) equation:

$$
\frac{c^{2}}{6} = \frac{(b-a)^{2}}{2} + \left(\frac{a+b-c}{2}\right)^{2}
$$

Multiplying both sides by $\frac{12}{c^{2}}$:

$$
\begin{gathered}
2 = 6(\sin \beta - \cos \beta)^{2} + 3(\cos \beta + \sin \beta - 1)^{2} \\
6(\cos \beta + \sin \beta) = -3 \sin 2 \beta + 10
\end{gathered}
$$

Here, both sides are positive for any value of the acute angle $\beta$, so squaring both sides gives an equivalent equation:

$$
36(1 + \sin 2 \beta) = 9 \sin^{2} 2 \beta - 60 \sin 2 \beta + 100
$$

The solution to the quadratic equation in $\sin 2 \beta$ that lies between 0 and 1 is $\sin 2 \beta = \frac{16 - 8 \sqrt{3}}{3}$, thus the sought acute angles are: $22.8^{\circ}$ and $67.2^{\circ}$.