In the rectangle $A B C D$, the midpoint of side $B C$ is $F$, and the point $H$ is the closer to $C$ of the two trisection points on side $C D$. How large can the angle $H A F$ be?
I. solution. In similar rectangles, the angle $H A F$ is equal. Fix the side $B C$ and change the side $A B$, so we can find a similar rectangle for every rectangle in the plane among these rectangles. Let the length of the segment $B C$ be 2, and the length of the segment $A B$ be $3 x$. Then the tangent of the angle $\beta=F A B$ is: $\operatorname{tg} \beta=\frac{1}{3 x}$, and the tangent of the angle $\alpha+\beta=H A B$ is:

$$
\operatorname{tg}(\alpha+\beta)=\frac{2}{2 x}=\frac{1}{x}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e7154378161c8fc713cbg-1.jpg?height=254&width=420&top_left_y=445&top_left_x=842)

Using the addition formula for $\operatorname{tg} \alpha=\operatorname{tg}((\alpha+\beta)-\beta)$:

$$
\operatorname{tg}((\alpha+\beta)-\beta)=\frac{\operatorname{tg}(\alpha+\beta)-\operatorname{tg} \beta}{1+\operatorname{tg}(\alpha+\beta) \operatorname{tg} \beta}=\frac{\frac{1}{x}-\frac{1}{3 x}}{1+\frac{1}{x} \cdot \frac{1}{3 x}}=\frac{2 x}{3 x^{2}+1}=\frac{2}{3 x+\frac{1}{x}}
$$

In the denominator of the expression, there is almost the sum of a number and its reciprocal. Dividing the numerator and the denominator by $\sqrt{3}$, the following is obtained:

$$
\frac{\frac{2}{\sqrt{3}}}{\sqrt{3} x+\frac{1}{\sqrt{3} x}}
$$

The denominator of the fraction is now the sum of a positive number and its reciprocal, so its value set is the interval $[2, \infty)$. Thus, the value of the fraction, and therefore the tangent of the angle $H A F$, can be any element of the interval $\left(0 ; \frac{1}{\sqrt{3}}\right]$. Since the tangent function is monotonically increasing for acute angles, the angle $H A F$ can take any value between 0 and $\frac{\pi}{6}$. Its maximum value is $\frac{\pi}{6}$.

II. solution. Fix the points $D$ and $C$, and let $D^{\prime}$ and $H^{\prime}$ be the reflections of $D$ and $H$ over $C$. Since $D^{\prime}$ lies on the line $A F$, the angle $H A D^{\prime}$ is equal to the angle $H A F$. Imagine that the size $\gamma$ of the angle $H A F$ is given, and examine the position of the point $A$. The point $A$ is located on the perpendicular to $D C$ at $D$, and on a viewing circle with an angle $\gamma$ whose endpoints are $H$ and $D^{\prime}$, and whose center lies on the perpendicular to $C D$ at $H^{\prime}$. The radius of the circle is therefore at least $H D^{\prime}=D H^{\prime}$; if it is exactly this, then the central angle corresponding to the segment $H D^{\prime}$ is $\frac{\pi}{3}$, i.e., $\gamma=\frac{\pi}{6}$. If we increase the radius of the circle, the value of $\gamma$ decreases, and in every case the point $A$ can be constructed. Therefore, the size of the angle $H A F$ can be any positive value not greater than $\frac{\pi}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e7154378161c8fc713cbg-1.jpg?height=363&width=486&top_left_y=1808&top_left_x=814)