$\mathrm{Az} A B C$ triangle has no right angle. At each vertex, we draw the tangent to the circumcircle of the triangle. Express the perimeter of the triangle determined by these tangents in terms of the sides of the $A B C$ triangle.
Let the intersection of the tangents at $B$ and $C$ be $D$, and let the tangent at $A$ intersect the previous two at $F$ and $E$ respectively. The exclusion of right angles ensures that these intersection points always occur, since no side of triangle $ABC$ is a diameter of the circumcircle $k$, and thus the tangents are not parallel. The tangent segments are equal: $DB = DC$, $EC = EA$, $FA = FB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_908031d5c86f4784146ag-1.jpg?height=1003&width=767&top_left_y=355&top_left_x=663)

The sides of triangle $DEF$ are obtained by adding the tangent segments exclusively if triangle $ABC$ is acute (Figure 1), or by a combination of addition and subtraction if triangle $ABC$ has an obtuse angle (as shown at vertex $A$ in Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_908031d5c86f4784146ag-1.jpg?height=610&width=982&top_left_y=1560&top_left_x=558)

2. Figure

In other words: this distinction is given by the fact that $k$ is the incircle of triangle $DEF$ in Figure 1, while in the setup of Figure 2, it is the excircle opposite to side $EF$. In both cases, $EF = EA + AF$, and in Figure 1, similarly, $FD = FB + BD$ and $DE = DC + CE$, whereas in Figure 2, $FD = BD - BF$ and $DE = CD - CE$. Thus, the perimeter $p$ of triangle $DEF$ in the cases of Figure 1 and Figure 2 is respectively

$$
p_{1}=2(BD + CE + AF), \quad p_{2}=2 \cdot BD
$$

The length of the tangent segment $DB$ is calculated from the right triangle $DBD'$, where $D'$ is the projection of $D$ onto $BC$, i.e., the midpoint of $BC$. The angle $\angle D'BD$ is a peripheral angle in $k$, specifically the arc $BC$ is between its sides, which does not contain, respectively contains, vertex $A$. Thus, in Figure 1, $\angle D'BD = \alpha$, and in Figure 2, $\angle D'BD = 180^\circ - \alpha$. The angle $\angle B'AB$ is the same, where $B'$ is the projection of vertex $B$ onto line $AC$ (on segment $AC$, or on its extension beyond $A$), so from triangles $BDD'$ and $ABB'$ with the usual short notations, in both cases

$$
BD = \frac{BD'}{AB'} \cdot AB = \frac{ac}{2 \cdot AB'}
$$

The length of the projection $AB'$ is obtained by applying the Pythagorean theorem to the common leg of triangles $BB'C$ and $BB'A$. According to the two situations, $CB' = CA \mp AB' = b \mp AB'$, thus

$$
\begin{aligned}
B'B^2 = BC^2 - B'C^2 & = BA^2 - B'A^2 \\
a^2 - (b \mp AB')^2 & = c^2 - B'A^2 \\
AB' & = \frac{b^2 + c^2 - a^2}{\pm 2b}
\end{aligned}
$$

Thus, for the acute and obtuse triangles respectively,

$$
BD = \frac{abc}{c^2 + b^2 - a^2}, \quad \text{or} \quad BD = \frac{abc}{a^2 - b^2 - c^2}
$$

By cyclically permuting the letters $A, B, C$, $a, b, c$, and $D, E, F$,

$$
CE = \frac{bca}{a^2 + c^2 - b^2}, \quad AF = \frac{cab}{b^2 + a^2 - c^2}
$$

and thus the expressions for the perimeter are:

$$
p_{1} = 2abc \left( \frac{1}{c^2 + b^2 - a^2} + \frac{1}{a^2 + c^2 - b^2} + \frac{1}{b^2 + a^2 - c^2} \right), \quad p_{2} = \frac{2abc}{a^2 - b^2 - c^2}
$$

All denominators are positive, for the vanishing of any one of them would imply, by the converse of the Pythagorean theorem, that one of the angles of triangle $ABC$ is a right angle.

## Zelhofer Walter (Mosonmagyaróvár, Kossuth L. Gymnasium)

Note. The expression for $AB'$ could have been made more mechanical by applying the cosine rule. But this would not have made the separate examination of the case in Figure 2 unnecessary.