The answer is $128$ , achievable by $A=(10,0), B=(0,-63), C=(-54,1)$ . We now show the bound.
We first do the following optimizations:
-if you have a point goes both left and right, we may obviously delete both of these moves and decrease the number of moves by $2$ .
-if all of $A,B,C$ lie on one side of the plane, for example $y>0$ , we shift them all down, decreasing the number of moves by $3$ , until one of the points is on $y=0$ for the first time.
Now we may assume that $A=(a,d)$ , $B=(b,-e)$ , $C=(-c,f)$ where $a,b,c,d,e,f \geq 0$ . Note we may still shift all $A,B,C$ down by $1$ if $d,f>0$ , decreasing the number of moves by $1$ , until one of $d,f$ is on $y=0$ for the first time. So we may assume one of $(a,b)$ and $(d,f)$ is $0$ , by symmetry. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:
Case 1 (where $a=d=0$ ) if $wx-yz=4042$ , find the minimum possible value of $w+x+y+z$ .
Case 2 (else) $wy+xy+xz=(w+x)(y+z)-wz=4042$ , find the minimum possible value of $w+x+y+z$ .
Note that $(m+n)^2=4mn+(m-n)^2$ so if $m+n$ is fixed then $mn$ is maximized exactly when $|m-n|$ is minimized. In particular, if $m+n \leq 127$ then $mn-op \leq mn \leq 63*64 = 4032 <4042$ as desired.
~Lcz