## Task 12/77

We are looking for (up to similarity) all triangles in which the tangent values of the interior angles are all integers.
We first prove the auxiliary theorem: If $\alpha+\beta+\gamma=180^{\circ}$, then

$$
\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma
$$

Proof: Since $\gamma=180^{\circ}-(\alpha+\beta)$, we have

$$
\tan \gamma=\tan \left[180^{\circ}-(\alpha+\beta)\right]=-\tan (\alpha+\beta)=-\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}
$$

Thus,

$$
\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha+\tan \beta-\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=
$$

$$
=\tan \alpha \tan \beta \cdot(-1) \cdot \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\tan \alpha \tan \beta \tan \gamma
$$

Now, for simplification, let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$ and solve the Diophantine equation

$$
x+y+z=x y z
$$

in integers (where, due to the angle sum theorem, at most one value can be negative and no value can be zero). First, we show that no value is negative. Suppose (without loss of generality), $z$ is negative: $z<0$. Then

$$
x+y-z=x y z \rightarrow x+y=x y z+z \rightarrow x+y=z(x y+1)
$$

(since $x>0, y>0$), so $x y+1>0$ because $z<0$. Thus, $x y<1$. This inequality is not solvable in integers.

Now let $z=1$. Then we have

$$
x+y+1=x y \rightarrow y+1=x y-x=x(y-1) \rightarrow x=\frac{y+1}{y-1}=1+\frac{2}{y-1}
$$

(assuming $y \neq 1$; it is easy to see that for $z=1, y=1$ no $x$ exists). This equation yields $x=3$ for $y=2$ and $x=2$ for $y=3$, and it has no other integer solutions. Thus, one solution is the triplet $(1; 2; 3)$, where the order does not matter.

It is easy to see that this is the only solution. If one tangent value were greater than 3, then (due to the monotonicity of the tangent function in the considered range and the angle sum theorem) at least one of the other two tangent values would have to be less than 1 or 2. The only possible integer triplet would then be $(1; 1; 3+\Delta)$ (with $\Delta \in \mathbb{N}$).

The equation $1+1+(3+\Delta)=1 \cdot 1 \cdot (3+\Delta)$ leads to the contradiction $2=0$.

Thus, the only solution (up to order) is the triplet $(\tan \alpha; \tan \beta; \tan \gamma)=(1; 2; 3)$ with the angles

$$
(\alpha; \beta; \gamma)=\left(45^{\circ}; 63.43^{\circ}; 71.56^{\circ}\right)
$$

Therefore, there is exactly one triangle (up to similarity) with exclusively integer tangent values of the interior angles.