Emelyanov L.A.

The median $A A_{0}$ of triangle $A B C$ is laid off from point $A_{0}$ perpendicular to side $B C$ to the exterior of the triangle. Denote the second end of the constructed segment by $A_{1}$. Similarly, points $B_{1}$ and $C_{1}$ are constructed. Find the angles of triangle $A_{1} B_{1} C_{1}$, if the angles of triangle $A B C$ are $30^{\circ}, 30^{\circ}$, and $120^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-07.jpg?height=312&width=392&top_left_y=1273&top_left_x=839)

Since triangle $ABC$ is isosceles, $BB_0$ is the perpendicular bisector of the base $AC$. Therefore, $B_1$ lies on this perpendicular and $CB_0 \perp BB_1$. Thus, $CB_0$ is the height and median of triangle $BCB_1$, from which it follows that $BC = B_1C$. Since $\angle B_1BC = \frac{1}{2} \angle B = 60^\circ$, triangle $B_1BC$ is equilateral.

$A_0$, being the midpoint of side $BC$, is the base of the height of this triangle. Therefore, points $A_1$ and $B_1$ lie on the perpendicular bisector of segment $BC$. Similarly, $C_1$ and $B_1$ lie on the perpendicular bisector of segment $AB$. Thus, $\angle C_1B_1A_1 = 180^\circ - B = 60^\circ$, $B_1A_1 = B_1A_0 + A_0A_1 = B_1A_0 + A_0A = B_1C_0 + C_0C = B_1C_0 + C_0C_1 = B_1C_1$, which means triangle $A_1B_1C_1$ is isosceles with the angle at the vertex being $60^\circ$.

## Answer

All angles are $60^\circ$.