What is the minimum area of a triangle that can contain a unit square?
A minimal area triangle must have at least one point of the square on each of its sides, otherwise, the corresponding side could be moved parallel to the direction of the square, reducing the area of the triangle. It is clear that if one side of the triangle contains an interior point of one side of the square, then it must contain the entire side (since the triangle contains the square), so we can assume that each side of the minimal area triangle must contain at least one vertex of the square. If two sides of the triangle contain the same vertex of the square, then the triangle and the square share a vertex. Around this common vertex, the sides of the triangle can be rotated to coincide with the sides of the square, without increasing the area of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_c53e1b75a78a527dcf55g-1.jpg?height=783&width=897&top_left_y=506&top_left_x=603)

Overall, we have found that the diagram is generally valid, with the addition that the quadrilateral $C P F Q$ may degenerate into a line segment or a point (and the triangles $A T_{1} G$ and $B T_{2} E$ may also naturally degenerate), but these do not affect our reasoning. Let us use the notation in the diagram, and let $D T_{1} = a, A T_{1} = x; D T_{2} = b$ and $B T_{2} = y$. Then $G T_{1} = 1 - a$ and $E T_{2} = 1 - b$. Due to the equality of their angles, $A T_{1} G \triangle \sim G F P \triangle$ and $B T_{2} E \triangle \sim E F Q \triangle$ hold, from which $P F = (1 - a) / x$ and $F Q = (1 - b) / y$ follow. Furthermore,

$$
A D T_{1} \triangle \sim D B T_{2} \triangle
$$

implies $a b = x y$. Without loss of generality, we can assume that $x / a \leq 1$. (Otherwise, $y / b \leq 1$, and the roles are symmetric.)

Let us estimate the area $T$ of the triangle $A B C$ from below:

$$
\begin{aligned}
T & \geq T - T_{F Q C P} = T_{D E F G} + T_{A D G} + T_{B E D} + T_{F P G} + T_{Q F E} = \\
& = 1 + \frac{x}{2} + \frac{y}{2} + \frac{\frac{1 - a}{x}}{2} + \frac{\frac{1 - b}{y}}{2} = 1 + \frac{x + y + \frac{1 - a}{x} + \frac{1 - b}{y}}{2}
\end{aligned}
$$

Notice that

$$
\frac{1 - a}{x} + \frac{1 - b}{y} - \frac{1 - x y}{y} = \frac{1}{x} - \frac{a}{x} + \frac{1}{y} - \frac{1}{y} - \frac{b}{y} + x = x + \frac{1}{x} - \left(\frac{a}{x} + \frac{x}{a}\right) \geq 0
$$

since by assumption $a \leq 1$, and thus $x \leq \frac{x}{a} \leq 1$, and the function $f(t) = t + 1 / t$ is monotonically decreasing on the interval $(0, 1]$. From this, continuing (1), we can further estimate $T$ from below:

$$
T \geq 1 + \frac{x + y + \frac{1 - x y}{y}}{2} = 1 + \frac{y + \frac{1}{y}}{2} \geq 1 + \frac{2}{2} = 2
$$

Equality holds only if $T_{F Q C P} = 0$, that is, one side of the square coincides with a side of the triangle (the remaining two vertices of the square are on the other two sides of the triangle). Simple calculations show that in this case, indeed, $T = 2$.

Therefore, the area of a triangle that contains a unit square is at least 2 square units.