11.1. Let $a^{2}+b^{2}=c^{2}+d^{2}=1$ and $a c+b d=0$ for some real numbers $a, b, c, d$. Find all possible values of the expression $a b+c d$.
Answer: 0.

Solution: Let's first assume $b \neq 0$. From the second equation, express $d=\frac{-a c}{b}$ and substitute it into the equation $c^{2}+d^{2}=1$. Eliminating the denominator, we get $c^{2}\left(a^{2}+b^{2}\right)=b^{2}$, from which, given $a^{2}+b^{2}=1$, we obtain $b= \pm c$. Substituting this into the equation $a^{2}+b^{2}=c^{2}+d^{2}$, we get $a= \pm d$. Given the equation ${ }^{a c}+b d=0$, we have either $b=c, a=-d$, or $b=-c, a=d$. In both cases, $a b+c d=0$.

If $b=0$, then from $a c+b d=0$ it follows that ${ }^{a=0}$ or ${ }^{c=0}$. The first is impossible due to $a^{2}+b^{2}=1$, so $c=0$, from which we again get $a b+c d=0$.

Grading criteria. Only the answer even with the corresponding $a, b, c, d$: 0 points. Failure to consider the case ${ }^{b=0}$ (if necessary in the solution): minus 1 point.