Reflect the sides of the acute-angled $A B C$ triangle over the altitudes that are not perpendicular to them. Of the six resulting reflections, three form the triangle $A_{1} B_{1} C_{1}$ inside the $A B C$ triangle, while the other three form the triangle $A_{2} B_{2} C_{2}$. The latter triangle has a $20^{\circ}$ and a $70^{\circ}$ angle. Calculate the angles of the $A_{1} B_{1} C_{1}$ triangles.
From the formulation of our task, it follows that no two sides of the $ABC$ triangle are equal. If there were two equal sides, then they would be mirror images of each other with respect to the altitude of the third side. In this case, the $A_1B_1C_1$ triangle could not be entirely inside the $ABC$ triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_3c2fc57bda87e5cfe78bg-1.jpg?height=510&width=531&top_left_y=279&top_left_x=797)

Let's denote the angles of the $ABC$ triangle in the usual way as $\alpha, \beta, \gamma$. We can assume that $\alpha > \gamma > \beta$ holds. Let $A', B', C'$ denote the points $BC \cap B_1C_1, CA \cap C_1A_1$, and $AB \cap A_1B_1$ respectively (1st figure). Since the lines $AA', BB'$, and $CC'$ are the reflections of the sides of the $ABC$ triangle over the altitudes of the triangle, considering the order of the angles, we get that $AB'B \varangle = B'AB \varangle = \alpha$, $AC'C \varangle = C'AC \varangle = \alpha$, and $A'AC \varangle = A'CA \varangle = \gamma$. In the quadrilateral $B'AC'A_1$, three angles are $\alpha$, so the angle at $A_1$ is $360^\circ - 3\alpha$.

Thus, $B_1A_1C_1 \varangle = 180^\circ - (360^\circ - 3\alpha) = 3\alpha - 180^\circ$. The exterior angle at $A'$ in the triangle $A'AB$ is $\gamma$, and the angle at $B$ is $\beta$, so the interior angle at $A$ is $A'AB \varangle = \gamma - \beta$. Therefore,

$$
A_1B_1C_1 \varangle = AB_1C' \varangle = 180^\circ - (B_1AC' \varangle + AC'B_1 \varangle) = (\alpha + \beta + \gamma) - ((\gamma - \beta) + \alpha) = 2\beta.
$$

The angle $A_1C_1B_1$ is $180^\circ - ((3\alpha - 180^\circ) + 2\beta) = 2\gamma - \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3c2fc57bda87e5cfe78bg-1.jpg?height=790&width=1120&top_left_y=1272&top_left_x=489)

# 2nd figure

Similarly, we can calculate the angles of the $A_2B_2C_2$ triangle (2nd figure). $AA''B \varangle = \beta$, $CC''B \varangle = \beta$, and $BB''C \varangle = \gamma$ (where $A'', B''$, and $C''$ are the points of intersection of the reflections and the sides of the original triangle), so

$$
\begin{gathered}
A_2B_2C_2 \varangle = B_2AC \varangle + B_2CA \varangle = (\gamma - \beta) + (\alpha - \beta) = 180^\circ - 3\beta; \\
B_2C_2A_2 \varangle = 180^\circ - A''C_2B \varangle = 180^\circ + \beta - 2\gamma; \\
C_2A_2B_2 \varangle = \beta + \gamma - \alpha.
\end{gathered}
$$

Since $\alpha > \gamma > \beta$, the smallest of these three angles, $\beta + \gamma - \alpha$, cannot be greater than $60^\circ$, so it must be the $20^\circ$ angle of the triangle. We cannot determine which of the other two angles is $70^\circ$, so we need to consider two cases.
i)

$$
A_2B_2C_2 \varangle = 180^\circ - 3\beta = 90^\circ, \quad B_2C_2A_2 \varangle = 180^\circ + \beta - 2\gamma = 70^\circ, \quad C_2A_2B_2 \varangle = \beta + \gamma - \alpha = 20^\circ.
$$

Solving the system of equations, we get $\alpha = 80^\circ$, $\beta = 30^\circ$, and $\gamma = 70^\circ$. From these, we obtain

$$
A_1B_1C_1 \varangle = 2\beta = 60^\circ, \quad B_1A_1C_1 \varangle = 3\alpha - 180^\circ = 60^\circ, \quad \text{and} \quad A_1C_1B_1 \varangle = 2\gamma - \alpha = 60^\circ.
$$

ii)

$$
A_2B_2C_2 \varangle = 180^\circ - 3\beta = 70^\circ, \quad B_2C_2A_2 \varangle = 180^\circ + \beta - 2\gamma = 90^\circ, \quad C_2A_2B_2 \varangle = \beta + \gamma - \alpha = 20^\circ.
$$

From this system of equations, we get $\alpha = 80^\circ$, $\beta = \left(\frac{110}{3}\right)^\circ$, and $\gamma = \left(\frac{190}{3}\right)^\circ$, so

$$
A_1B_1C_1 \varangle = \left(\frac{220}{3}\right)^\circ, \quad B_1A_1C_1 \varangle = 60^\circ, \quad \text{and} \quad A_1C_1B_1 \varangle = \left(\frac{140}{3}\right)^\circ.
$$

It can be easily verified that both cases can occur if we start from an appropriate $ABC$ triangle. Therefore, the angles of the $A_1B_1C_1$ triangle are either all $60^\circ$, or they are $\left(\frac{140}{3}\right)^\circ$, $60^\circ$, and $\left(\frac{220}{3}\right)^\circ$.

Based on the work of Béla András Rácz (Fazekas M. Fóv. Gyak. Gimn., 9th grade)