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This puzzle is part of the Monthly Topic Challenge #11: Now in 3D.

You are given a collection of sticks which are straight lines of length 1. Two such sticks can be attached to each other at their end points either to form a straight line of length two or at a 90 degree angle. Your task is to build a closed path with an odd length (that is using an odd number of sticks).

Part a) Show that this is impossible if all sticks remain in the 2-dimensional plane.

Part b) What is the shortest solution (using the smallest amount of sticks) in 3 dimensions?

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    $\begingroup$ In that closed path, does the last stick have to be connected to the first stick at a right angle too? Or do they not have to be connected, and just touch at any angle? $\endgroup$
    – Jaap Scherphuis
    Jun 5 at 8:36
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    $\begingroup$ @JaapScherphuis Yes. Once the loop is closed there shouldn't be a distinguishable first or last stick anymore. $\endgroup$
    – quarague
    Jun 5 at 8:40

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So...

If all sticks are in a flat plane, they all must be either — or | (nothing else!).

Now, to build a closed path, the sticks must return to where they came from. Let's call this the origin point $(0,0)$. Furthermore, we can think of — as the X-axis and | as the Y-axis.

Turns out, each — changes our X-axis by $1$ and | changes our Y-axis by $1$! Since we need both our coordinates to be at $0$, the number of — and |s each have to be even.

When we are given access to the third dimension...

We get WAY more options. Instead of choosing left or right when rotating 90 degrees, we can also choose to go up, down, 45-degree up, 88-degree up, and everything in-between. Wow!

What's the shortest solution, then?

Probably 7 sticks! Here's a visualization in GeoGebra.

Why's that the shortest solution?

Well... With 2 sticks, the best we can do is form a 90-degree angle with distance $\sqrt2$ from the origin. So, 3 sticks isn't enough.

Also, I previously did this for a 5-stick solution, but the first and last sticks don't form a 90-degree angle. All I did was add 2 more sticks, and that's it!

Though...

I can't (yet) prove how we need at least 7 sticks and not 5. Maybe you can prove/disprove it 🙂

Edit: Seems like @Herbert Kociemba has proven it!
Edit 2: @Jaap Scherphuis did it manually.

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    $\begingroup$ I found the same solution, but then realised that where the loop closes, the sticks do not meet at a right angle. $\endgroup$
    – fljx
    Jun 5 at 8:35
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    $\begingroup$ Breaking the length-2 segment into 4 length-1 segments creates a 9-gonal solution with only right angles. Unclear to me whether there is a 7-gonal solution with only right angles. $\endgroup$
    – Edward H
    Jun 5 at 10:16
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    $\begingroup$ There does not exist a solution with 5 sticks in 3D. Put the endpoints p0..p4 of the sticks s01, s12, s23, s34 and s40 into a coordinate system. Without loss of generality you can assume p0=(0,0,0), p1=(1,0,0) and p4= (0,0,1). p2= (1,y2,z2) and p3=(x3,y3,1). With these 4 variables 5 equations have to be true simultanuously: (p2-p1)*(p2-p1)=1, (p3-p2)*(p3-p2)=1, (p4-p3)*(p4-p3)=1, (p3-p2)*(p2-p1)=0 and (p4-p3)*(p3-p2)=0 where * denotes the scalar product. With Mathematica for example you can see that you can solve 4 but not all 5 equations simultanuously. $\endgroup$
    – Herbert Kociemba
    Jun 5 at 14:17
  • $\begingroup$ @HerbertKociemba ah, that's right! Is some programming needed to prove that, though? Can we do it manually? $\endgroup$
    – ThePuzzler- or rather APuzzler
    Jun 5 at 14:39
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    $\begingroup$ You can prove it manually, but it is a bit tricky. I'd set $p0=(-1/2,0,0)$, $p1=(1/2,0,0)$, and assume without loss of generality that $p2_z=p4_z$, and $p2_y=-p4_y$, i.e. hold it so that these three sticks are rotationally symmetric about the z axis. If the two sticks meeting at $p3$ are straight, then the distance $p2-p4$ is $2$, which fixes them but gives them wrong angles. If the two sticks at $p3$ are at right angles, then distance $p2-p4$ is $\sqrt2$. Again those points $p2$,$p4$ can be solved, but to make their angles equal symmetry forces $p3_x=0$, but that also gives wrong angles. $\endgroup$
    – Jaap Scherphuis
    Jun 5 at 15:06

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