11.5. Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written next to its row and its column (a "multiplication table"). What is the maximum number of products in this table that could be rational numbers?

(O. Podlipsky)
Answer: 1275 products.

Solution. First, let's show that there are no fewer than 1225 irrational numbers in the table. Suppose that among the rational numbers, there is a zero, and it is written along the top side of the table.

Let $x$ be the number of irrational numbers and $50-x$ the number of rational numbers written along the left side of the table. Then, along the top side, there are $50-x$ irrational and $x$ rational numbers (including zero). Note that the product of a non-zero rational and an irrational number is always irrational. Therefore, there are at least $x(x-1) + (50-x)^2$ irrational numbers in the table. Note that $f(x) = x(x-1) + (50-x)^2 = 2x^2 - 101x + 50^2$. The vertex of the parabola $f(x)$ is at the point $101/4 = 25.25$, so the minimum value of $f(x)$ at an integer point is achieved at $x=25$ and is equal to $25 \cdot 24 + 25^2 = 1225$.

If zero is replaced by a non-zero rational number, the number of irrational numbers can only increase. Therefore, in any case, there are no fewer than 1225 irrational numbers in the table. Thus, there are no more than $2500 - 1225 = 1275$ rational numbers in the table.

Exactly 1275 rational numbers can be in the table, for example, in the following case. Along the left side, the numbers are $1, 2, \ldots, 24, 25, \sqrt{2}, 2\sqrt{2}, \ldots, 25\sqrt{2}$, and along the top side, the numbers are $0, 26, 27, \ldots, 49, 26\sqrt{2}, 27\sqrt{2}, \ldots, 50\sqrt{2}$. Then, the only irrational products will be $25 \cdot 24 + 25^2 = 1225$ products of non-zero rational and irrational numbers.

Comment. Only the correct answer - 0 points.

Proof that there are no more than 1275 rational numbers - 4 points.

If the proof is only conducted in a special case where an equal number of rational and irrational numbers are along the left (or top) side - 1 point is given for this part of the solution instead of 4.

The inequality $x(x-1) + (50-x)^2 \geqslant 1225$ (or an equivalent one) is used without proof - 1 point is deducted.

A correct example showing that there could be exactly 1275 rational numbers - 3 points.

Only an example with non-zero numbers showing that there could be exactly 1250 rational numbers - 1 point.

The statement that the product of a non-zero rational and an irrational number is irrational can be used without proof.

If this statement is significantly used for any (not necessarily non-zero) numbers (i.e., an incorrect statement is used!) - no more than 2 points are given for the problem.