The coordinates of the vertices of a triangle are: $A(p, q), B(q, r), C(r, p)$, where $p<q<r$. What are the bounds within which the angles at each vertex lie?
The triangle under investigation is within the square defined by the vertices $C(r, p), D(p, p), E(p, r), F(r, r)$. Specifically, $A$ is located on the side $DE$, and $B$ is on the side $EF$. We also denote the points with coordinates $(q, p)$ and $(r, q)$ as $A'$ and $B'$, which are the reflections of $A$ and $B$ across the diagonal $DF$ of the square. In the rectangle $AB'CD$, $\angle B'AC = \angle ACD = \angle A'BD$, and since this rectangle is congruent to the rectangle $A'BE D$, these angles are also equal to:

$$
\angle B'AC = \angle ACD = \angle A'BD.
$$

Similarly, we obtain that

$$
\angle B'AF = \angle BCF = \angle A'BC.
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_774f70bce0c7bd0f1201g-1.jpg?height=600&width=1108&top_left_y=534&top_left_x=498)

According to these, the angle $\gamma$ at $C$ in the triangle $ABC$ and the angle at $B$ in the triangle $BCD$ complement each other to $90^\circ$. The latter is certainly greater than $45^\circ$, since $B$ lies within the circle circumscribed around the square, so the angle subtended by the side $CD$ from $B$ is greater than the angle subtended from the vertices $E$ and $F$. The viewing angle from $B$ is maximized when $B$ coincides with the midpoint $G$ of the side $EF$, which occurs when $q = \frac{1}{2}(p + r)$. Otherwise, $B$ lies outside the circle circumscribed around the triangle $CDG$, since this circle is tangent to the line $EF$ at $G$. Since $\cos \angle CGD = 0.6$, $\angle CGD = 53.13^\circ$, we have:

$$
36.87^\circ \leq \gamma < 45^\circ
$$

and $\beta + \gamma > 90^\circ$, so the angle $\alpha$ at $A$ in the triangle $ABC$ is less than $90^\circ$. Since $\alpha$ is greater than the angle at $A$ in the triangle $ACF$, $\alpha > 45^\circ$ and $\alpha + \gamma > 90^\circ$, so $\beta < 90^\circ$. Therefore,

$$
45^\circ < \alpha, \beta < 90^\circ.
$$

**Second Solution.** Let the angles $\angle ACD$ and $\angle BCF$ be denoted by $\delta$ and $\varphi$, respectively, and their tangents by $\lambda$ and $\mu$:

$$
\lambda = \tan \delta = \frac{q - p}{r - p}, \quad \mu = \tan \varphi = \frac{r - q}{r - p}
$$

so $\lambda$ and $\mu$ are positive and $\lambda + \mu = 1$. According to the addition formula for the tangent function,

$$
\cot \gamma = \tan(90^\circ - \gamma) = \tan(\delta + \varphi) = \frac{\lambda + \mu}{1 - \lambda \mu}
$$

thus $\tan \gamma = 1 - \lambda \mu$. Therefore, $\tan \gamma$ is always less than 1, and its value is minimal when $\lambda = \mu = \frac{1}{2}$, in which case $\tan \gamma = \frac{3}{4}$, and we again obtain the bounds for $\gamma$ as in (1). Similarly, by the addition formula,

$$
\tan \alpha = \frac{\tan \angle CAB' + \tan \angle B'AB}{1 - \tan \angle CAB' \tan \angle B'AB} = \frac{\lambda + \frac{\mu}{\lambda}}{1 - \mu} = 1 + \frac{\mu}{\lambda^2},
$$

and similarly, we obtain that $\tan \beta = 1 + \frac{\lambda}{\mu^2}$. Therefore, $\tan \alpha$ and $\tan \beta$ are greater than 1, which gives the bounds for $\alpha$ and $\beta$ as in (2). If $\lambda$ approaches 0, $\tan \alpha$, $\tan \beta$, and $\tan \gamma$ approach $\infty$, 1, and 1, respectively, so $\alpha$ approaches $90^\circ$, and $\beta$ and $\gamma$ approach $45^\circ$. If $\mu$ approaches 0, then $\beta$ approaches $90^\circ$, and $\alpha$ and $\gamma$ approach $45^\circ$. Therefore, the bounds in (1) and (2) cannot be improved.