Among the triangles with a unit area, which one can cover the largest area square?
Among the variously shaped unit area triangles, we need to find those into which a square of maximum area can be inscribed. As a first step, we show that the vertices of the maximum area square must necessarily lie on the boundary of the triangle. If at least two vertices of the inscribed square are interior points of the triangle (Figure 1), then the triangle has a vertex from which, by magnifying the view, the image of a vertex falls on the perimeter. Repeating this, we can achieve that three vertices of the square lie on the boundary of the triangle, while the area of the square increases.

In Figure 2, three vertices of the square lie on the sides of the triangle. With the notation of the figure, $X Y \| B C$, and we can assume that $P B \leq P Y$ and $C R \geq R X$. The line $P S$ intersects $B C$ at $D$ and $A C$ at $E$. Due to the assumption $P B \leq P Y$, the area of the triangle $P B D$ is not greater than that of the triangle $P S Y$. Therefore, the area of the triangle $C D E$ is smaller than that of the triangle $A B C$. If we now enlarge the triangle $C D E$ - together with the square - so that its area becomes unit, then the area of the square $P Q R S$ will also increase.

From the above, it follows that the maximum area square is placed in the triangle as shown in Figure 3, with one of its sides lying on the side of the triangle that does not have a vertex with an obtuse angle. With the notation of the figure, from the similarity of the triangles $A S R$ and $A B C$, we have $x: d=(x+d): B C$, from which $B C=\frac{d(x+d)}{x}$. Thus, the area of the triangle is $\frac{1}{2} \cdot \frac{d}{x} \cdot(x+d)^{2}$, which, using the arithmetic and geometric mean inequality, can be estimated as:

$$
\frac{1}{2} \cdot \frac{d}{x}(d+x)^{2} \geq \frac{1}{2} \cdot \frac{d}{x} \cdot(2 \sqrt{d x})^{2}=2 d^{2}
$$

This means that $2 d^{2}$ will be the largest when it equals the area of the triangle, and this occurs when $x=d$. In this case, $d^{2}=\frac{1}{2}$, and $B C=x+d=\sqrt{2}$. Therefore, the largest square can be inscribed in unit area triangles that are not obtuse and have a base and height of $\sqrt{2}$.

Note. In the solution of the problem, the inscribed square appeared in the sense of a covered square. In a stricter sense, the inscribed square is the square shown in Figure 3.
![](https://cdn.mathpix.com/cropped/2024_05_02_af4ead404927cefc9d0ag-1.jpg?height=1516&width=652&top_left_y=1189&top_left_x=714)