A triangle has angles $\alpha, \beta$ and $\gamma$. What are the angles of the triangle formed by the tangents drawn to the circumcircle at the vertices?
Solution. Let's denote the vertices of the triangle as $A, B, C$ in the usual way. If the triangle is right-angled, for example, $\gamma=90^{\circ}$, then $AB$ is the diameter of the circumcircle, so the tangents to the circumcircle at $A$ and $B$ are parallel (Figure 1), hence in this case the tangents do not form a triangle. In the following, we assume that the triangle $ABC$ is not right-angled. Let the vertices of the triangle formed by the tangents to the circumcircle at the vertices be $A', B'$, and $C'$, and the center of the circumcircle be $K$. Based on the relationship between the inscribed and central angles, the central angle corresponding to the arc $BC$ not containing $A$ is $2\alpha$, the central angle corresponding to the arc $CA$ not containing $B$ is $2\beta$, and the central angle corresponding to the arc $AB$ not containing $C$ is $2\gamma$. The tangent to the circle is perpendicular to the radius at the point of tangency, so $KA \perp B'C'$, $KB \perp C'A'$, and $KC \perp A'B'$. In the following, we distinguish between the cases of an acute and an obtuse triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd2fb1ec30fcc10631e0g-1.jpg?height=371&width=374&top_left_y=541&top_left_x=865)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_dd2fb1ec30fcc10631e0g-1.jpg?height=504&width=599&top_left_y=1033&top_left_x=752)

If $ABC$ is acute (Figure 2), then the circumcircle of $ABC$ is the incircle of the triangle $A'B'C'$. The quadrilaterals $KAC'B$, $KBA'C$, and $KCB'A$ are cyclic, because two of their opposite angles are right angles. The angles we are looking for are precisely the angles opposite to $K$ in these cyclic quadrilaterals. In any cyclic quadrilateral, the sum of the opposite angles is $180^{\circ}$, so in this case the angles of the triangle $A'B'C'$ are $180^{\circ}-2\alpha$, $180^{\circ}-2\beta$, and $180^{\circ}-2\gamma$.

If $ABC$ is obtuse, we can assume that $\alpha>90^{\circ}$ (Figure 3). In this case, the circumcircle of $ABC$ is the excircle of the triangle $A'B'C'$ opposite to the side $B'C'$. In this case, the quadrilaterals $KAC'B$, $KAB'C$, and $KCA'B$ are cyclic, because two of their opposite angles are right angles. The angles we are looking for are now the angles opposite to $K$ in the cyclic quadrilateral $KCA'B$, and the exterior angles at the vertices opposite to $K$ in the cyclic quadrilaterals $KAB'C$ and $KAC'B$. In any cyclic quadrilateral, the exterior angle at a vertex is equal to the interior angle at the opposite vertex, so in this case the angles of the triangle $A'B'C'$ are $180^{\circ}-(360^{\circ}-2\alpha)=2\alpha-180^{\circ}$, $2\gamma$, and $2\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd2fb1ec30fcc10631e0g-1.jpg?height=623&width=393&top_left_y=2033&top_left_x=861)

Figure 3

Note. Many solvers (more than 60%) forgot to examine the non-acute triangles. They should have suspected something when they wrote that the angle sought, for example, is $180^{\circ}-2\alpha$, which is negative for $\alpha>90^{\circ}$, i.e., in the case of an obtuse triangle.