Expression value, a little harder

[ins-]

Let $a$, $b$, $c$, $d$ are distinct real numbers, such that:

$a+b=cd$

$c+d=ab$

$a-b=c-d$

Find the value of $ab+cd$.

[scwec]

Здесь ответ $-4$

[mihiv]

$a=b=c=d=2?$

[gris]

Нет, там разны е требуются. Например, $(-4,0,-2,2)$

[ins-]

:) yes, "distinct" is a way to ignore the case $ab-cd=0$. The problem have very short and nice solution.

[mihiv]

А, понятно.

[ins-]

$ac+bd$ also have a fixed value.

[waxtep]

ins- в сообщении #1109187 писал(а):

$ac+bd$ also have a fixed value.

$8$ ! (not to be mixed with $8!$)

[ins-]

Yes, it is 8. It can be evaluated in at least 2-3 ways. One of them I saw is a very short and impressive.

[waxtep]

ins- в сообщении #1109201 писал(а):

Yes, it is 8. It can be evaluated in at least 2-3 ways.

(spoiler)

mine was noticing $a+d=b+c=-2$ (based on nice solution for $ab+cd$), substituting it directly into $ac+bd$, and then using $a+b+ab=c+d+cd$. A cute one! Also, at this step, one just can easily get explicit expressions for any three variables, just to make sure that the magic works! (I have to confess I've done it)

[ins-]

Your is a 4-th one :)