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?
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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?
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