## Task A-2.1.

Let $a, b, c$ and $d$ be distinct real numbers. If $a$ and $b$ are solutions to the equation $x^{2}-10 c x-11 d=0$, and $c$ and $d$ are solutions to the equation $x^{2}-10 a x-11 b=0$, determine the sum $a+b+c+d$.
## Solution.

From Viète's formulas, it follows that $a+b=10c$ and $c+d=10a$. By adding the equations $a+b=10c$ and $c+d=10a$, we get

$$
a+b+c+d=10(a+c)
$$

Since $a$ is a solution to the equation $x^{2}-10cx-11d=0$ and $d=10a-c$, it follows that

$$
0=a^{2}-10ac-11d=a^{2}-10ac-11(10a-c)=a^{2}-110a+11c-10ac
$$

Similarly, we obtain

$$
c^{2}-110c+11a-10ac=0
$$

By subtracting these two equations, we get

$$
(a-c)(a+c-121)=0
$$

Dividing by $a-c \neq 0$, we get $a+c=121$. Therefore, $a+b+c+d=1210$.