You want to connect the points A, B, C and D to each other by drawing the connection with a marker. But you want to use the minimum amount of ink of the marker.

What is the shape that you will draw?

You'll make a cross, thus spending [tex: 4\sqrt{2}\approx 5.7] units of ink. Wasn't that too easy?
The "obvious" solution would have been to connect the points ACBD in order thus spending 6 units of ink, I guess.

True chads use 0 ink by making a torus

P13704
There is a better way.

P13704
That seems correct. Can you prove it?

P13708
Lol. I'm sure OP meant to imply that the problem was in an euclidean plane

This puzzle was more interesting than I initially thought.

P13715
I can't, because it isn't. I have, however, found the correct solution. With a proof as well.

First of all, the shape HAS to look like the first image. E and F may be confounded, but additional intersections would be redundant. To minimize the total length, points shall only be linked by straight lines as well.

E (respectively F) is the Fermat point of the triangle AFD (respectively BCE). As such, E (respectively F) must lie somewhere on the [GF] (respectively [EH]) line. As a result, E and F both lie on the vertical axis.

We can then finalize the construction of the Fermat points to obtain the final position of E and F.

To obtain the total length, we observe that all three angles surrounding the Fermat point are 120°. As a result, the angle ADE is 30°. We can now obtain the length of each segment.
[tex: DE=EA=BF=FC=\frac{1}{cos(30°)}=\frac{2}{\sqrt{3}}] and [tex: EF=2-\frac{2}/{sin(30°)}]
Therefore, [tex: L=2+2\sqrt{3}]

P13897
I messed up when copying my notes.
[tex: EF=2-2 tan(30°)=2-\frac{2}{\sqrt{3}}]
[tex: L=2+2\sqrt{3}\approx5.46]

P13897
P13898
Yep, this is the solution.

In an equilateral triangle the shortest way to link all vertex is to pass by the center. In a square, one should use P13897's construction. How could one extend the construction in order to find the shortest way to link all vertex of an n-sided regular polygon?

P14037
Given the [tex: \geq 120^\circ] angles, I conjecture that for hexagons and above, going around the outside is optimal. I can say at least that it's a local minimum in the sense that it's not improved by changes like pic related (which was not true for the square).

That may be true, but your explanation isn't very convincing because there are topologies other than the one you used.
Picrel is [tex: 3\sqrt3 \approx 5.2>5] though

P14041
Yes, thus a conjecture. I haven't gotten anywhere on a general proof yet, but I think for the specific case of the hexagon, I think I can show the only cases we need to consider are these three. The first two options are P14041 and P14038, and the third is the second pic attached. Will post some more details later on how to calculate the points for each case and show that the points found are unique.

P14214
>second pic attached
ended up being the first pic

P14214
>Will post some more details later on how to calculate the points for each case and show that the points found are unique.

The solution has to be some sort of tree, and as P13897 pointed out, we can represent any tree as a tree in which each point except the terminal ones has three neighbors, with some points confounded. Every point except the terminal ones has to be at the Fermat point of its neighbors, which means that the lines to the three neighbors form 120° angles, except in the case where the three points form an angle greater than 120°, in which case the Fermat point is the vertex of that angle. This exceptional case can only occur when the vertex is one of the terminal points, since otherwise we obtain a smaller total length by changing the connections to the confounded points. For the regular hexagon, we don't have to worry about this case, since none of the points can be outside the convex hull of the points we're connecting, which limits any angle at one of the vertices of the hexagon to 120°.

The three possible ways of connecting six points by such a tree are shown in P14214. Let's consider the third case. We know point G which connects to A and B must lie on the circle centered at H, which by the inscribed angle theorem must be constructed to make m∠AHB = 360° - 2*m∠AGB = 120°. If we take the third line segment GJ and extend it in the opposite direction, it must intercept the circle at a point I located 120° away from A and B, again by the inscribed angle theorem. We can repeat this process with I and C to find a point L that must lie on the extension of JP. The same process for the lower half of the hexagon gives us another point R on the extension of JP. Drawing the line between them and finding their intersections with the circles gives us J and P. We can now connect J with I to find G, and P with O to find M.

F = (0, 0)
E = (1, 0)
A = (-1/2, √3/2)
B = (0, √3)
C = (1, √3)
D = (3/2, √3/2)
H = (-1/2, 5√3/6)
I = (-1, √3)
K = (0, 4√3/3)
L = (0, 2√3)
N = (3/2, √3/6)
O = (2, 0)
Q = (1, -√3/3)
R = (1, -√3)
J = (3/7, 5√3/7)
P = (4/7, 2√3/7)
G = (1/14, 11√3/14)
M = (13/14, 3√3/14)
AG = DM = CJ = FP = JP = 2√7/7
BG = EM = GJ = MP = √7/7
total length = 2√7 ≈ 5.3 > 5

So for the specific case of the regular hexagon, going around the outside is indeed the best.