1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2015$ and $a \neq c$ (the numbers $a, b, c, d$ are not given).
1. Answer: 1007.5.

First solution. Expanding the brackets, we get $a b-(a+b) x=c d-(c+d) x$. Substituting $d=2015-a$ and $b=2015-c$, we obtain

$a(2015-c)-(a+2015-c) x=c(2015-a)-(c+2015-a) x . \quad \begin{aligned} & \text { combining } \text { and } \\ & \text { similar terms, }\end{aligned}$

we get $(a-c) 2015=2(a-c) x$. Considering that $a \neq c$, we get $2 x=2015$, that is, $x=1007.5$.

Second solution. Since $c=2015-b$ and $d=2015-a$, the parabolas $y=(x-a)(x-b)$ and $y=$ $(x-c)(x-d)$ are symmetric with respect to the line $x=2015 / 2$. Therefore, they intersect at

a point lying on this line. If these parabolas had another common point, then the symmetric point to it would also lie on both parabolas, meaning that our equation would have at least three distinct roots. But then the parabolas would coincide. This contradicts the condition, as in this case $a=d=2015 / 2$ and $b=c=2015 / 2=a$.