3. Given an isosceles right triangle with legs equal to $\sqrt{2}$. Find the straight-line cut of the smallest length that divides the area of the triangle in half.

In the answer, specify the length of the cut, rounded to 2 decimal places after the decimal point. Answer: $0.91$.
Solution: Let in triangle $ABC$ angle $B$ be a right angle. The cut intersects two sides of the triangle at points $P$ and $Q$.

First, assume that these points are located on the legs. Let $x = BP, y = BQ$.

Then $\frac{1}{2} xy = \frac{1}{2} S(ABC) = \frac{1}{2}$, and $PQ^2 = x^2 + y^2 = x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$. Clearly, the minimum length of $PQ = \sqrt{2} \approx 1.41$.

Now, let points $P, Q$ be located on the leg $AB$ and the hypotenuse $AC$, respectively.

Let $AP = x, AQ = y$. Then $\frac{\sqrt{2}}{4} xy = \frac{1}{2}$, i.e., $xy = \sqrt{2}$. By the cosine rule, $PQ^2 = x^2 + y^2 - 2 \cdot \frac{\sqrt{2}}{2} xy = x^2 + \frac{2}{x^2} - 2 = \left(x - \frac{\sqrt{2}}{x}\right) + 2\sqrt{2} - 2$.

The minimum value of $PQ = \sqrt{2\sqrt{2} - 2} \approx 0.91$.