Let's determine the angles of a right-angled triangle where the centroid lies on the inscribed circle. Construct such a right-angled triangle given the hypotenuse.
I. Solution: Place the right-angled triangle, which we assume to be constructed, in a coordinate system such that the coordinate axes coincide with the legs; and take the length of the hypotenuse as the unit of length. The lengths of the legs can be expressed using the sine and cosine of one of the acute angles, $\sin \alpha$ and $\cos \alpha$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_5a2611f83c9746a5610bg-1.jpg?height=559&width=686&top_left_y=306&top_left_x=709)

If we denote the radius of the inscribed circle by $\varrho$, both coordinates of its center will be $\varrho$, so the equation of the circle is:

$$
(x-\varrho)^{2}+(y-\varrho)^{2}=\varrho^{2}
$$

or

$$
x^{2}+y^{2}-2 \varrho(x+y)+\varrho^{2}=0
$$

The tangent segments from the acute vertices to the inscribed circle of radius $\varrho$ are equal in pairs, so the sum of the tangent segments on the legs equals the hypotenuse:

$$
(\sin \alpha-\varrho)+(\cos \alpha-\varrho)=1
$$

from which

$$
\varrho=\frac{\sin \alpha+\cos \alpha-1}{2} .
$$

The coordinates of the centroid can be calculated from the coordinates of the vertices of the triangle as $\left(\frac{\cos \alpha}{3}, \frac{\sin \alpha}{3}\right)$. According to the problem, these coordinates satisfy the equation of the inscribed circle. Substituting the value of $\varrho$ from (2) and the coordinates of the centroid into (1), and using the identity $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we get the following equation after rearrangement:

$$
\frac{1}{9}-\frac{1}{3}(\sin \alpha+\cos \alpha-1)(\sin \alpha+\cos \alpha)+\left(\frac{\sin \alpha+\cos \alpha-1}{2}\right)^{2}=0
$$

Let $(\sin \alpha+\cos \alpha)$ be denoted by $z$, and multiply the equation by 36:

$$
4-12(z-1) z+9\left(z^{2}-2 z+1\right)=0
$$

From this, we get the following quadratic equation for $z$:

$$
3 z^{2}+6 z-13=0
$$

We only take the positive root for $z$ (since both $\sin \alpha$ and $\cos \alpha$ are positive):

$$
z=\frac{-6+\sqrt{36+12 \cdot 13}}{6}=-1+\frac{\sqrt{48}}{3}=\frac{4 \sqrt{3}}{3}-1
$$

Thus,

$$
\sin \alpha+\cos \alpha=\frac{4 \sqrt{3}}{3}-1
$$

Squaring both sides,

$$
\sin ^{2} \alpha+2 \sin \alpha \cos \alpha+\cos ^{2} \alpha=\frac{16}{3}-\frac{8 \sqrt{3}}{3}+1
$$

Using known trigonometric identities,

$$
\sin 2 \alpha=\frac{16-8 \sqrt{3}}{3} \sim \frac{2.1436}{3}=0.7145
$$

Reversing the approximate value,

$$
2 \alpha=45^{\circ} 36' \quad \underline{\underline{\alpha=22^{\circ}} 18'}
$$

The other angle is

$$
\underline{\underline{\beta=67^{\circ} 12'}}
$$

With this, we have determined the angles of the right-angled triangle. It remains to construct the triangle. We can substitute the calculated value of $\sin \alpha+\cos \alpha$ into the relationship for the radius of the inscribed circle obtained at the beginning of the problem:

$$
\varrho=\frac{2 \sqrt{3}}{3}-1
$$

From this, we can construct $\varrho$. The radius of the inscribed circle and the hypotenuse uniquely determine the right-angled triangle. The center $O$ of the inscribed circle is on a line parallel to the hypotenuse at a distance of $\varrho$, and also on the $135^{\circ}$-angle circle determined by the hypotenuse (Figure 2). (The center of the angle circle is the point on the perpendicular bisector of the hypotenuse at a distance of $\frac{c}{2}$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_5a2611f83c9746a5610bg-2.jpg?height=594&width=794&top_left_y=1015&top_left_x=649)

Figure 2

After drawing the inscribed circle, we draw tangents from the endpoints of the hypotenuse: these will be the legs of the right-angled triangle. The problem is now completely solved.

II. Solution: Let the center of the inscribed circle in the right-angled triangle $ABC$ to be constructed be $O$, the centroid $S$, the midpoint of the hypotenuse $F$, and the radius of the inscribed circle $\varrho$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_5a2611f83c9746a5610bg-2.jpg?height=345&width=704&top_left_y=1912&top_left_x=689)

Figure 3

Since $F$ is the center of the circumcircle (Thales circle) of the triangle, the triangle $ACF$ is isosceles, with the angles at $A$ and $C$ both being $\alpha$. The point $O$ is on the angle bisector of the right angle, so $\angle OCS = 45^{\circ} - \alpha$. The length of the segment $CF$ is half the hypotenuse, $\frac{c}{2}$, and the length of the segment $CS$ is $\frac{2}{3}$ of this, i.e., $\frac{c}{3}$. The side $OS$ of the triangle $COS$ is $\varrho$, and the third side, $OC$, is the hypotenuse of an isosceles right triangle with a leg of $\varrho$, so its length is $\varrho \sqrt{2}$.

Applying the cosine rule to the triangle $COS$,

$$
\varrho^{2}=\frac{c^{2}}{9}+2 \varrho^{2}-\frac{2 c \varrho \sqrt{2}}{3} \cos \left(45^{\circ}-\alpha\right)
$$

Solving for $\cos \left(45^{\circ}-\alpha\right)$, and immediately using the formula for the cosine of the difference of two angles,

$$
\cos \left(45^{\circ}-\alpha\right)=\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha=\frac{c^{2}+9 \varrho^{2}}{6 c \varrho \sqrt{2}}
$$

Since $\sin \alpha=\frac{a}{c}$, $\cos \alpha=\frac{b}{c}$, and the equality of the tangent segments to the inscribed circle, $a+b=c+2 \varrho$, we can simplify using $\frac{1}{c}$:

$$
c+2 \varrho=\frac{c^{2}+9 \varrho^{2}}{6 \varrho}
$$

From this, we get the quadratic equation for $\varrho$ (we only need the positive root):

$$
3 \varrho^{2}+6 c \varrho-c^{2}=0
$$

Solving this quadratic equation,

$$
\varrho=\frac{-6 c+\sqrt{36 c^{2}+12 c^{2}}}{6}=c\left(\frac{2 \sqrt{3}}{3}-1\right)
$$

Substituting the value of $\varrho$ into the expression for $\cos \left(45^{\circ}-\alpha\right)$,

$$
\cos \left(45^{\circ}-\alpha\right)=\frac{c}{6 \varrho \sqrt{2}}+\frac{3 \varrho}{2 c \sqrt{2}}=\frac{c}{\sqrt{2} c(4 \sqrt{3}-6)}+\frac{c(2 \sqrt{3}-3)}{2 c \sqrt{2}} \approx 0.9259
$$

From this,

$$
45^{\circ}-\alpha=22^{\circ} 12' \\
\under