Where did all these dowels come from?

[Prof. Templeton]

This may have been asked before since it was taken straight from bn’s playbook.

You have an unlimited supply of dowels all of the same length and they can only be connected at their ends. A triangle made by joining three dowels will be a rigid form, but a square made by connecting four dowels can be easily distorted into other shapes without breaking or bending a dowel or disconnecting the ends. As if the end connections were hinged. In three dimensional space the simplest way to brace the square would be to add eight more dowels the form an octahedron.

Let’s assume, however that we are limited to only two dimensions. Without bending, breaking or overlapping dowels, only connecting them at the ends and remaining on a two dimensional plane we can still brace the square by attaching triangles to it. Here is an example with 35 total dowels.

The question is: What is the fewest number of dowels required to hold a square firm?

[plainglazed]

My question about the 33 solution is that I imagine the top and bottom squares collapsing to the left and right respectively, while still remaining pinched between the crab claws. So I think the squares are not actually firmed up.

Yeah, had similar concerns. Finally just went with it assuming the two squares shared a dowel and acted as if hinged as per the op. so if they collapsed, they would need to remain congruent parallelagrams and the claws would not permit. Even if it were to qualify, strongly doubt it's optimal.

[curr3nt]

I don't think the triangles would prevent the squares from moving right or left. The triangles only keep the distance between them from exceeding one unit but collapsed the distance between those two lines would still be one unit. It would stop collapsing when the top line is inline with the triangles point.

[Prof. Templeton]

Yes, it looks like that 33 solution is unstable. Although not optimal, I will give a hint to a better solution.

Three tessellated hexagons. Remove some of the dowels and squeeze two of them closer together until you can put a square in the middle of them in a way that it is stable. There is another, better solution, but this one is almost as good.