Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.
Find the minimum value and the maximum value of the product $a b c d$.
Let's find the minimum first.

$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$

By AM-GM, \(4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1\).
Note that if \(a=b=c=1\) and \(d=-1\), then \(a b c d=-1\).
We'll find the maximum. We search for \(a b c d>0\).
Obviously, the numbers \(a, b, c\) and \(d\) can not be all positive or all negative.
WLOG \(a, b>0\) and \(c, d<0\).
Let \(x=-c\) and \(y=-d\). Then \(x, y>0, a+b-x-y=2\) and \(a^{2}+b^{2}+x^{2}+y^{2}=4\). We need to find \(\max (a b x y)\). We get: \(x+y=a+b-2\) and \(x^{2}+y^{2}=4-\left(a^{2}+b^{2}\right)\). Since \((x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)\), then \(2\left(a^{2}+b^{2}\right)+(a+b-2)^{2} \leq 8\); on the other hand, \((a+b)^{2} \leq 2\left(a^{2}+b^{2}\right) \Rightarrow (a+b)^{2}+(a+b-2)^{2} \leq 8\).
Let \(a+b=2 s \Rightarrow 2 s^{2}-2 s-1 \leq 0 \Rightarrow s \leq \frac{\sqrt{3}+1}{2}=k\).
But \(a b \leq s^{2} \Rightarrow a b \leq k^{2}\).
Now \(a+b=x+y+2\) and \(a^{2}+b^{2}=4-\left(x^{2}+y^{2}\right)\). Since \((a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)\), then \(2\left(x^{2}+y^{2}\right)+(x+y+2)^{2} \leq 8\); on the other hand, \((x+y)^{2} \leq 2\left(x^{2}+y^{2}\right) \Rightarrow\)
\(\Rightarrow(x+y)^{2}+(x+y+2)^{2} \leq 8\). Let \(x+y=2 q \Rightarrow 2 q^{2}+2 q-1 \leq 0 \Rightarrow q \leq \frac{\sqrt{3}-1}{2}=\frac{1}{2 k}\).
But \(x y \leq q^{2} \Rightarrow x y \leq \frac{1}{4 k^{2}}\).
In conclusion, \(a b x y \leq k^{2} \cdot \frac{1}{4 k^{2}}=\frac{1}{4} \Rightarrow a b c d \leq \frac{1}{4}\).
Note that if \(a=b=k\) and \(c=d=-\frac{1}{2 k}\), then \(a b c d=\frac{1}{4}\)
In conclusion, \(\min (a b c d)=-1\) and \(\max (a b c d)=\frac{1}{4}\).