Under what conditions can a regular triangle be inscribed in the rectangle $A B C D$, with one vertex at $A$, and the other two vertices lying on the sides $B C$ and $C D$ respectively?
I. solution. Let's assume that a regular triangle can be inscribed in the rectangle in the prescribed manner.

![](https://cdn.mathpix.com/cropped/2024_05_02_65319d40d3ebaacadab1g-1.jpg?height=474&width=483&top_left_y=224&top_left_x=802)

Using the notation of Figure 1, $b = d \cdot \cos \alpha$ and $a = d \cdot \cos (30^\circ - \alpha)$, where $0 \leq \alpha \leq 30^\circ$. From these two equations,

$$
\frac{a}{b} = \frac{\cos (30^\circ - \alpha)}{\cos \alpha}
$$

Transforming the right-hand side:

$$
\frac{a}{b} = \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \tan \alpha
$$

Given the conditions on $\alpha$, $0 \leq \tan \alpha \leq \frac{1}{\sqrt{3}}$, and thus

$$
\frac{\sqrt{3}}{2} \leq \frac{a}{b} \leq \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}
$$

Therefore, a regular triangle can be inscribed in the rectangle $ABCD$ in the desired manner if and only if the ratio of the sides of the rectangle satisfies condition (3).

Conversely: Since $\tan \alpha$ is continuous, for every case where (3) holds, there exists an $\alpha$ such that (2) and (1) hold, and thus there exists a positive real number $d$ such that $a = d \cdot \cos (30^\circ - \alpha)$ and $b = d \cdot \cos \alpha$.

In this case, $d$ is the length of the side of the regular triangle inscribed in the rectangle $ABCD$ in the desired manner.

Károly Podoski (Budapest, Árpád Gymnasium, 4th grade)

II. solution. First, assume that $a \geq b$. If we rotate the vertex $P$ of the regular triangle $APQ$ in Figure 1 by $60^\circ$ counterclockwise around $A$, $P$ will move to $Q$. Therefore, the regular triangle can be inscribed in the rectangle in the desired manner if and only if the $60^\circ$ counterclockwise rotation of segment $BC$ around $A$ has a common point with segment $DC$. Based on Figure 2, we can see that this happens if and only if $AR \geq a$.

![](https://cdn.mathpix.com/cropped/2024_05_02_65319d40d3ebaacadab1g-1.jpg?height=623&width=523&top_left_y=1854&top_left_x=779)

From the triangle $AB'R$ in the figure,

$$
AR = \frac{b}{\cos 30^\circ} = \frac{2b}{\sqrt{3}}
$$

Thus, the required condition is:

$$
a \leq \frac{2b}{\sqrt{3}}, \quad \text{which implies} \quad \frac{a}{b} \leq \frac{2}{\sqrt{3}}
$$

Since the sides of the rectangle can have different lengths and their roles are symmetric, in the case $b > a$, similarly, it must hold that

$$
\frac{b}{a} \leq \frac{2}{\sqrt{3}}
$$

From (1) and (2),

$$
\frac{\sqrt{3}}{2} \leq \frac{a}{b} \leq \frac{2}{\sqrt{3}}
$$

Gábor Kún (Budapest, Berzsenyi Dániel Gymnasium, 3rd grade)

Note. The method of the second solution also works for inscribing an isosceles triangle, where the common vertex of the legs is $A$, the angle between the legs is $\varphi$, and $\varphi \leq 90^\circ$. In this case, the required condition is:

$$
\sin \varphi \leq \frac{a}{b} \leq \frac{1}{\sin \varphi}
$$