9.1. For non-negative numbers $a, b, c, d$, the following equalities are satisfied:

$\sqrt{a+b}+\sqrt{c+d}=\sqrt{a+c}+\sqrt{b+d}=\sqrt{a+d}+\sqrt{b+c}$. What is the maximum number of distinct values that can be among the numbers $a, b, c, d$?
Answer. Two.

Solution. Square the first equality, combine like terms, cancel by 2, square again, combine like terms once more, ultimately obtaining the equality $a c+b d=a b+c d$, equivalent to $(a-d)(c-b)=0$, from which either $a=d$ or $b=c$. Performing similar manipulations with the second equality, we get $a=b$ or $d=c$. Any equality from the first pair shares a common letter with any equality from the second pair, so three out of the four numbers must be equal to each other, and among the numbers $a, b, c, d$ there cannot be more than two distinct numbers. On the other hand, any set of four non-negative numbers, where three are the same and the fourth is arbitrary, clearly satisfies the condition of the problem, so among the numbers $a, b, c, d$ there can be exactly two distinct numbers.

Grading criteria. Proof that among the numbers $a, b, c, d$ there cannot be more than two distinct numbers: 5 points. Example where there are exactly two distinct numbers: 2 points.