Problem 1333

1. Identify the problem and the constraints:
We need to find the least possible area of a convex set in the plane that intersects both branches of the hyperbolas $xy = 1$ and $xy = -1$. A convex set is defined such that for any two points in the set, the line segment connecting them is also contained in the set.

2. Determine the points of intersection:
The convex set must contain at least one point on each of the four branches of the hyperbolas. Let these points be:
$$(x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)$$
where $x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4$ are positive numbers and satisfy the hyperbola equations:
$$x_1 y_1 = 1, \quad x_2 y_2 = 1, \quad x_3 y_3 = 1, \quad x_4 y_4 = 1$$

3. Calculate the area of the convex hull:
The smallest convex set containing these points is the convex hull, which in this case is a quadrilateral. The area $A$ of a quadrilateral with vertices $(x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)$ can be calculated using the shoelace formula:
$$A = \frac{1}{2} \left| x_1 y_2 + (-x_2) y_3 + (-x_3) (-y_4) + x_4 y_1 - (y_1 (-x_2) + y_2 (-x_3) + y_3 x_4 + y_4 x_1) \right|$$

4. Simplify the area expression:
Substitute $y_i = \frac{1}{x_i}$ for $i = 1, 2, 3, 4$:
$$A = \frac{1}{2} \left| x_1 \frac{1}{x_2} + (-x_2) \frac{1}{x_3} + (-x_3) \frac{1}{x_4} + x_4 \frac{1}{x_1} - \left( \frac{1}{x_1} (-x_2) + \frac{1}{x_2} (-x_3) + \frac{1}{x_3} x_4 + \frac{1}{x_4} x_1 \right) \right|$$
$$= \frac{1}{2} \left| \frac{x_1}{x_2} - \frac{x_2}{x_3} - \frac{x_3}{x_4} + \frac{x_4}{x_1} - \left( -\frac{x_2}{x_1} - \frac{x_3}{x_2} + \frac{x_4}{x_3} + \frac{x_1}{x_4} \right) \right|$$
$$= \frac{1}{2} \left| \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \right|$$

5. Apply the AM-GM inequality:
For any positive $u$, we have $u + \frac{1}{u} \geq 2$. Applying this to each term:
$$\frac{x_1}{x_2} + \frac{x_2}{x_1} \geq 2, \quad \frac{x_2}{x_3} + \frac{x_3}{x_2} \geq 2, \quad \frac{x_3}{x_4} + \frac{x_4}{x_3} \geq 2, \quad \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 2$$
Summing these inequalities:
$$\frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 8$$

6. Determine the minimum area:
Therefore, the minimum area is:
$$A \geq \frac{1}{2} \times 8 = 4$$
Equality holds when $x_1 = x_2 = x_3 = x_4$, meaning the quadrilateral is a rectangle with sides parallel to the axes.

The final answer is $\boxed{4}$.