For a triangle with angles $\alpha, \beta, \gamma$, $\operatorname{ctg} \frac{\alpha}{2}, \operatorname{ctg} \frac{\beta}{2}, \operatorname{ctg} \frac{\gamma}{2}$ are consecutive integers. What is the largest angle of the triangle?
I. solution. Let $\operatorname{ctg} \frac{\alpha}{2}=n, \operatorname{ctg} \frac{\beta}{2}=n-1, \operatorname{ctg} \frac{\gamma}{2}=n-2$, where $n$ is an integer. We will show that (up to similarity) at most one triangle can satisfy the conditions of the problem. Suppose that a triangle with angles $\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime}$ also satisfies the conditions, that is, for some integer $k$, $\operatorname{ctg} \frac{\alpha^{\prime}}{2}=k, \operatorname{ctg} \frac{\beta^{\prime}}{2}=k-1, \operatorname{ctg} \frac{\gamma^{\prime}}{2}=k-2$. The half-angles of the triangles belong to the interval $\left(0, \frac{\pi}{2}\right)$, and in this interval, the $\operatorname{ctg} x$ function is strictly decreasing. If we assume that $k=n$, then $\operatorname{ctg} \frac{\alpha^{\prime}}{2}=\operatorname{ctg} \frac{\alpha}{2}$ implies $\alpha^{\prime}=\alpha$, and similarly, $\beta^{\prime}=\beta$ and $\gamma^{\prime}=\gamma$. If this is not the case, then we can assume that $k>n$. However, then $\operatorname{ctg} \frac{\alpha^{\prime}}{2}>\operatorname{ctg} \frac{\alpha}{2}$ implies $\alpha^{\prime}<\alpha$, and $\beta^{\prime}<\beta$ and $\gamma^{\\prime}<\gamma$, which is impossible.

Therefore, it is sufficient to find a single triangle that satisfies the conditions. It is easily readable from the diagram that such a triangle is the right-angled triangle with sides $3,4,5$. The largest angle of this triangle is $\gamma=\frac{\pi}{2}$.

II. solution. We use the notation from the previous solution. According to the addition formula for the tangent function,

$$
\operatorname{ctg} \frac{\beta}{2}=\operatorname{ctg}\left(\frac{\pi}{2}-\frac{\alpha+\gamma}{2}\right)=\operatorname{tg}\left(\frac{\alpha}{2}+\frac{\gamma}{2}\right)=\frac{\operatorname{tg} \frac{\alpha}{2}+\operatorname{tg} \frac{\gamma}{2}}{1-\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\gamma}{2}}
$$

that is,

$$
\begin{aligned}
n-1=\frac{\frac{1}{n}+\frac{1}{n-2}}{1-\frac{1}{n(n-2)}} & =\frac{2(n-1)}{n^{2}-2 n-1} \\
n^{2}-2 n-1 & =2 \\
(n-3)(n+1) & =0
\end{aligned}
$$

Since the half-angle of any angle of a triangle lies between 0 and $\frac{\pi}{2}$ and the cotangent of such angles is positive, the only solution is $n=3$, so for the largest angle, $\operatorname{ctg} \frac{\gamma}{2}=1$, that is, $\gamma$ is a right angle.

Szemjon Viktória (Carpathia, Jánosi Jánosi Secondary School, 11th grade)