I am an engineer in the oil field and I am trying to find a socket that can fit around a square drive pin. I am trying to prove this for fun before I just draw it on CAD and measure it. I believe sockets have six or twelve sides. I have had some success calculating the largest hexagonal wrench that can fit on a square drive, but it came out very large and I remember that it was not this large of a tool when I worked in the field. The problem is that when you add this extra point in a hexagon making it a decagon, it allows the square to rotate. The square drive is d=.750 and I came out with x=(sqrt(5)/(3-sqrt(3))d, where x is the width across two sides of a hexagon. The proof of the hexagon starts with the fact that the largest square inside of a hexagon shares the same center. Its a strange proof that I dont understand, but it can be found here: http://www.drking.org.uk/hexagons/misc/deriv3.html.

This web page gives us the result that (1) d=(3-sqrt(3))a. Then I derived (2) a=x/sqrt(5) using the Pythagorean therom. From there, I substituted (2) into (1). and got the result x~1.764d, x~1.323 when d=.750.

I know this cant be right. I could calculate the diagonals of the square drive and set that equal to the diagonals of the dodecagon. From there I could probably find the distance between two sides of the dodecagon and that would be a starting point, but I'd rather find a relation and proof for largest square inside a dodecagon that cannot rotate. I know aligning the square diagonals with the dodecagon diagonals would not be correct because I do not think this is how you'd want to start this proof.enter image description here

  • $\begingroup$ @john wayland bales I overlooked the difference that adding another 6 sides would make. It looks correct. $\endgroup$ – Mike K Apr 30 '19 at 2:02

Wouldn't it be easier with a regular dodecagon, since 4 divides evenly into 12?

The largest square which will not rotate will fit snugly into four corners of the dodecagon.

Square inscribed in dodecagon

Note that in $\triangle AOM$, side $OM$ must be the largest side. Since $\angle MOA<30^\circ$ then $\angle MAO>75^\circ$.

Proof in pictures


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