2. Masha drew a rhombus in her notebook and wrote four numbers at the vertices, the sum of which is 2021. Then she wrote the product of the numbers at the ends of each side of the rhombus and calculated the sum of the numbers written on the sides. Katya increased all the numbers written by Masha at the vertices by 1, and then repeated what Masha did and calculated the new sum of the numbers on the sides. By how much is Katya's sum greater than Masha's?
Answer: 4046.

Solution. Let Masha write the numbers $a, b, c, d$ at the vertices of the rhombus, then $a+b+c+d=2021$, and she obtained the sum of the products

$$
a b+b c+c d+d a=(a+c)(b+d)
$$

Since Katya increased each number by 1, her sum is

$$
(a+c+2)(b+d+2)=(a+c)(b+d)+2(a+b+c+d)+4
$$

greater than Masha's by $2(a+b+c+d)+4=2 \cdot 2021+4=4046$.

Criteria. Full solution - 7 points. Considered a particular case with the correct answer - 1 point. Only answer - 0 points.