Let the circumcenter of triangle $ABC = H$ be denoted by $T$ with respect to the reflection over the line $AB$. Calculate the angles of $H$ in the following three cases:
I. $T$ coincides with the incenter of $H$;

II. $T$ coincides with the excenter of $H$ opposite to the side $AB$;

III. $T$ coincides with the centroid of $H$.

(The excenter is the center of the circle that is externally tangent to the side $AB$ and the extensions of the other two sides. - For first-year students, a complete answer to the third question will also be accepted if it is based on construction and measurement.)
In all three cases, denoting the center of the circumscribed circle of $H$ by $O$, we have $O A=O B$, and based on the reflection, the quadrilateral $A O B T$ is a rhombus. In none of these cases does it degenerate, because in cases I and III, $T$ is definitely an interior point of $H$, and in case II, it is definitely an exterior point. Accordingly, $O$ is an exterior point in cases I and III, and an interior point in case II.

Furthermore, it always holds that $O A B \varangle=T A B \varangle=T B A \varangle=O B A \varangle$, which we denote by $\varepsilon$.

In case I, according to the properties of the incircle center, $T A$ bisects the angle $B A C$, and $T B$ bisects the angle $A B C$, so the angles at $H$ at $A$ and $B$ are $2 \varepsilon$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_532e4b6bdc0122410b9dg-1.jpg?height=645&width=379&top_left_y=434&top_left_x=862)

Figure 1

From $C B A \varangle=2 \varepsilon$, by the theorem of central and inscribed angles, we have $C O A \varangle=4 \varepsilon$. Furthermore, $C A O \varangle=3 \varepsilon$, and since the triangle $A C O$ is isosceles, the sum of its angles gives $3 \varepsilon+3 \varepsilon+4 \varepsilon=180^{\circ}, \varepsilon=18^{\circ}$. Thus, the angles of $H$ are: $36^{\circ}, 36^{\circ}$, $108^{\circ}$.

In case II, according to the properties of the excircle center, $T A$ and $T B$ bisect the external angles of $B A C$ and $A B C$, respectively, and these two angles are $2 \varepsilon$, so the angles at $H$ at $A$ and $B$ are equal and have a size of $180^{\circ}-2 \varepsilon$, and the angle at $C$ is $180^{\circ}-2(180^{\circ}-2 \varepsilon)=4 \varepsilon-180^{\circ}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_532e4b6bdc0122410b9dg-1.jpg?height=651&width=376&top_left_y=1469&top_left_x=861)

Figure 2

Based on these, in the triangle $A C O$, the angles at $C$ and $A$ are

$$
2 \varepsilon-90^{\circ}, \quad \text{and} \left(180^{\circ}-2 \varepsilon\right)-\varepsilon
$$

respectively (in the latter, we used the above finding that $O$ is in the interior of the angle $B A C$), and from their equality, $\varepsilon=54^{\circ}$, and thus the angles of $H$ are: $72^{\circ}, 72^{\circ}, 36^{\circ}$.

Finally, in case III, according to the properties of the centroid, $C$ lies on the line connecting the midpoint $F$ of the side $A B$ with $T$ and $F C=3 \cdot F T$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_532e4b6bdc0122410b9dg-2.jpg?height=385&width=659&top_left_y=122&top_left_x=711)

Figure 3

From $A O=B O$, the perpendicular line $T O$ to $A B$ also passes through $F$, and thus through $C$, making $H$ isosceles, and $F T=F O$, and

$$
C O=C F+F O=4 F O
$$

From this,

$$
\sin \varepsilon=\frac{F O}{A O}=\frac{F O}{C O}=\frac{1}{4}, \quad \varepsilon=14^{\circ} 29^{\prime}
$$

(rounded), and subsequently $A O C \varangle=90^{\circ}-\varepsilon=75^{\circ} 31^{\prime}$,

$$
\begin{gathered}
A C B \varangle=2 \cdot A C O \varangle=2\left(90^{\circ}-\frac{A O C \varangle}{2}\right)=104^{\circ} 29^{\prime}, \\
B A C \varangle=A B C \varangle=37^{\circ} 46^{\prime} .
\end{gathered}
$$

Without trigonometric knowledge, based on the above analysis, the radius $O C$ of the circumscribed circle is intersected by the perpendicular line at the quarter point of $F$, and the positions of $A$ and $B$ are determined, and the angles of the resulting $H$ are measured.