At most, what fraction of a triangle can be covered by a square whose every vertex lies on a side of the triangle?
Solution. Out of the four vertices of the square, two fall on the same side of the triangle, and one vertex of the square must be on each of the other two sides of the triangle. We can assume that the vertices of the square $X Y Z V$ are positioned on the sides of the triangle $A B C$ as shown in the figure, possibly with the vertex $X$ coinciding with $A$, or the vertex $Y$ coinciding with $B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_73793bcf1ec20c1d31c0g-1.jpg?height=474&width=585&top_left_y=300&top_left_x=759)

The side $V Z$ of the square is parallel to the line $A B$, so the triangles $A B C$ and $V Z C$ are similar. Let the ratio of similarity be $\lambda$. Then, if the side length of the square is $\lambda c$, and if the height from $C$ in the triangle $A B C$ is $m_{c}$, the height from $C$ in the triangle $V Z C$ is $\lambda m_{c}$. The side $V X$ of the square is parallel to the heights from $C$ in the triangles, so

$$
m_{c}=\lambda m_{c}+V X=\lambda m_{c}+\lambda c, \quad \text { or } \quad \lambda=\frac{m_{c}}{m_{c}+c}
$$

and therefore $0<\lambda<1$.

Let the area of the triangle $A B C$ be $T$. Then $m_{c}=2 T / c$, so

$$
\lambda=\frac{m_{c}}{m_{c}+c}=\frac{2 T}{2 T+c^{2}}
$$

From this, we get

$$
1-\lambda=\frac{c^{2}}{2 T+c^{2}}
$$

Using these, the ratio of the area of the square to the area of the triangle $A B C$ is

$$
\begin{aligned}
\frac{(\lambda c)^{2}}{T} & =2 \lambda^{2} \frac{c^{2}}{2 T}=2 \lambda \frac{2 T}{2 T+c^{2}} \cdot \frac{c^{2}}{2 T}=2 \lambda \frac{c^{2}}{2 T+c^{2}}= \\
& =2 \lambda(1-\lambda) \leq 2\left(\frac{\lambda+(1-\lambda)}{2}\right)^{2}=\frac{1}{2}
\end{aligned}
$$

where in the estimation, we used the inequality between the arithmetic and geometric means (which we can do since both $\lambda$ and $1-\lambda$ are positive). Equality holds if and only if $\lambda=1 / 2$. This is exactly the case when

$$
\frac{2 T}{2 T+c^{2}}=\frac{1}{2}
$$

which is true if $c^{2}=2 T$, or $c=m_{c}$.

Thus, the square can cover at most half of the area of the triangle, and it covers exactly half if and only if one of the heights of the triangle equals the length of the corresponding side, and one side of the square lies on this side of the triangle.