- Jan 16, 2018
I don't see this as a trial and error problem. Let's take A:you need to find such a map, largely by trial and error coupled with a sense of how such routes can vary.
Let's create the layout that corresponds to this answer.
1. Draw segment X--A
2. Where is B? Since we visit X on the way to B, we can conclude that X is on the line AB. Draw B: B--X--A
3. Where is C? Since we visit B on the way to X, we can conclude that B is on the line CX. Draw C: C--B--X--A
That's the layout: C--B--X--A
Now, is the answer the shortest route? Yes. We are using straight lines to go to A and back, and to C and back.
I don't know whether this approach is formal enough, but it works, as demonstrated here for A and for E earlier in the thread.
And regarding C, same approach: B--X--A has to be a straight line. C is not on that line since we don't visit B on the way back. But this tells us that there is a shorter route, based on the triangle inequality: XACBX. AC is shorter than AB + BC.