[edited] [deleted] answer is 1

It's a very good reason for which I took $ab + bc + cd + da + ac + bd = 0$. During such competition, that choice can save the time and the energy of the participants.

Anyone else?

Not if anything to say about it I have. Should I call you emperor of inequalities? I am very charmed by the problem. Search for minimum AM-GM $$1=\frac{a^2+b^2+c^2+d^2}{4}\ge\sqrt[4]{a^2b^2c^2d^2}=\sqrt{|abcd}|\implies 1\ge|abcd|\ge -abcd\implies abcd\ge -1$$Indeed achievable for $(a,b,c,d)$ being a permutation of $(1,1,1,-1)$. Search for maximum There exist numbers $a,b,c,d$ such that $abcd>0$, e.g. $\left(\frac{\sqrt3-1}{2},\frac{\sqrt3-1}{2},\frac{\sqrt3+1}{2},\frac{\sqrt3+1}{2}\right)$. Thus we assume here $abcd>0$. Observe that if $a,b,c,d\in R_+$ or $a,b,c,d\in R_-$ then $ab + bc + cd + da + ac + bd >0$. So $abcd>0$ gives that exactly two of $a,b,c,d$ are positive and two are negative numbers. All given expressions are symmetric so let's take $a,b>0\wedge c,d<0$ and there won't be any harm for the proof. AM-GM $$8=2(a^2+b^2)+2(c^2+d^2)\ge (a+b)^2+(c+d)^2=(a+b)^2+(2-a-b)^2$$$$0\ge \left(a+b-1+\sqrt{3}\right)\left(a+b-1-\sqrt{3}\right)$$$$ a+b\le1+\sqrt3$$$$|c+d|=-(c+d)=(a+b)-2\le \sqrt3-1$$Again AM-GM $$ab\cdot cd\le \frac{(a+b)^2}{4}\cdot\frac{|c+d|^2}{4}\le \frac{\left(\sqrt3+1\right)^2\cdot \left(\sqrt3-1\right)^2}{16}=\frac{1}{4}$$Maximum achievable for $(a,b,c,d)$ being a permutation of $\left(\frac{\sqrt3-1}{2},\frac{\sqrt3-1}{2},\frac{\sqrt3+1}{2},\frac{\sqrt3+1}{2}\right)$.

Bravo! Thank you. No, I'm simply an amateur. The authors solution. Let $a,b,c$ and $d$ be real numbers such that $a+b+c+d=2$ and $ab+bc+cd+da+ac+bd=0$. Find the minimum value and the maximum value of the product $abcd$. Let's find the minimum first. \[a^2+b^2+c^2+d^2=(a+b+c+d)^2-2(ab+bc+cd+da+ac+bd)=4 \]By AM-GM, $4=a^2+b^2+c^2+d^2\geq 4\sqrt{|abcd|}\Rightarrow 1\geq |abcd|\Rightarrow abcd\geq -1$. Note that if $a=b=c$ and $d=-1$, then $abcd=-1$. We'll find the maximum. We search for $abcd>0$. Obviously, the numbers $a,b,c$ and $d$ can not be all positive or all negative. WLOG $a,b>0$ and $c,d<0$. Denote $-c=x, -d=y$. We have $a,b,x,y>0, a+b-x-y=2$ and $a^2+b^2+x^2+y^2=4$. We need to find $\max (abxy)$. We get: $x+y=a+b-2$ and $x^2+y^2=4-(a^2+b^2)$. Since $(x+y)^2\leq 2(x^2+y^2)$, then $2(a^2+b^2)+(a+b-2)^2\leq 8$; on the other hand, $(a+b)^2\leq 2(a^2+b^2)\Rightarrow (a+b)^2+(a+b-2)^2\leq 8$. Let $a+b=2s\Rightarrow 2s^2-2s-1\leq 0\Rightarrow s\leq \frac{\sqrt{3}+1}{2}=k$. But $ab\leq s^2\Rightarrow ab\leq k^2$. Now $a+b=x+y+2$ and $a^2+b^2=4-(x^2+y^2)$. Since $(a+b)^2\leq 2(a^2+b^2)$, then $2(x^2+y^2)+(x+y+2)^2\leq 8$; on the other hand, $(x+y)^2\leq 2(x^2+y^2)\Rightarrow$ $\Rightarrow (x+y)^2+(x+y+2)^2\leq 8$. Let $x+y=2q\Rightarrow 2q^2+2q-1\leq 0\Rightarrow q\leq \frac{\sqrt{3}-1}{2}=\frac{1}{2k}$. But $xy\leq q^2\Rightarrow xy\leq \frac{1}{4k^2}$. In conclusion, $abxy\leq k^2\cdot \frac{1}{4k^2}=\frac{1}{4}\Rightarrow abcd\leq \frac{1}{4}$. Note that if $a=b=k$ and $c=d=-\frac{1}{2k}$, then $abcd=\frac{1}{4}$ In conclusion, $\min (abcd)=-1$ and $\max (abcd)=\frac{1}{4}$.

If only more people on this forum were so creative and modest at once... I look forward to your next olympic proposal.

Of course I'll let you know.

Since expressions' derivatives are easy enough, we can just calculus bash, Let $a, b,c$ and $d$ be real numbers such that $a + b + c + d = 2$ and $ab + bc + cd + da + ac + bd = 0$. Find the minimum value and the maximum value of the product $abcd$. Let $f,g,h: \mathbb{R} \rightarrow \mathbb{R}$ $g(a,b,c,d)=a+b+c+d-2$,$h(a,b,c,d)=ab+bc=cd+da+ac+bd=0$, $f(a,b,c,d)=abcd$ We have: $$ \nabla g = \langle 1,1,1,1 \rangle$$$$\nabla h= \langle b+d+c,a+c+d,b+d+a,c+a+b \rangle$$$$ \nabla f= \langle bcd,acd,abd,abc \rangle$$By Lagrange multiplers, there are some real numbers $\lambda, \mu$ satisfying: $$\nabla f= \lambda \nabla g+ \mu \nabla h\leftrightarrow \nabla f= \lambda +\mu \nabla h$$ We have to solve system of equations: $$bcd=\lambda+ \mu(b+d+c)$$$$acd=\lambda+ \mu (a+c+d)$$$$abd= \lambda + \mu (b+d+a)$$$$abc= \lambda + \mu (c+a+b)$$Subtracting first and second rows we get $cd(b-a)=\mu(b-a)$. Now assume $b\neq c$.It gives $cd= \mu$. Subtracting other rows, we can get $cd=bd=bc=ad=ab$ one of the trivial solution is $a=b=c=d$. but since these variables can be $0$ or there can be other special solutions, we have to check cases (checking cases where some variables are $0$). Case 1. $a=0$ means $cd=bd=bc=0$. We have subcases. Subcase 1.1 $b=0$. Then we have either $c=0$ or $d=0$, giving solutions $(0,0,2,0),(0,0,0,2)$ Subcase 1.2 and 1.3 $d=0$,$c=0$. Gives similar result as Subcase 1.1 gives a new solution $(0,2,0,0)$ Case 2.$b=0$. means $cd=ad=0$ Subcase 2.1 $d=0$ gives solution $(1,0,1,0)$ Subcase 2.2 Nothing new, same conclusion as case 1. Now let $b=c$. subtracting second and third rows we get: We have 2 cases. Either $b=c=d$ or $d=a$. The first case gives solution $(-1,1,1,1)$. ( basically putting to the equations). The second case, we have to solve $a+b=1,a^2+b^2+3ab=(a+b)^2+2ab=0$. Putting $a=1-b$ to the second equation, we get: $1+2(1-b)b=0$ which is just quadratic formula and we get solutions $(\frac{1+\sqrt{3}}{2},\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2},\frac{1-\sqrt{3}}{2})$ and its repetations. All these solutions we found are critical points, we have to put them on $abcd$, and choose the minimum and maximum values. The minimum value occurs at solution $(1,1,1,-1)$ and repetations, gives -1. The maximum value occurs at solution $(\frac{1+\sqrt{3}}{2},\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2},\frac{1-\sqrt{3}}{2})$ ( not suprised) and gives $\frac{1}{4}$. Wasted time on $0$ cases which is obviously, not minimum/maximum, duh

Similar, but harder https://artofproblemsolving.com/community/q1h2850319p25265454

The authors (one of which is me) wish to thank to all colleagues involved in this topic. The solutions are beautiful