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Cycle chain drawing

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blowlamp19/09/2018 19:42:00
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A similar solution to that of Alan Vos

tangent.jpg

duncan webster19/09/2018 19:57:08
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Posted by Alan Vos on 19/09/2018 17:57:20:

I believe this works. The key is making GH equal to the radius of the two circles.

tangent construction.jpg

AB is the line between the centres of the two circles
CD is the defined line, parallel to AB
E is the centre of the desired circle

Construct the perpendicular bisector of AB, this intersects CD at G)
Construct H such that GH=AF=BI
Construct the perpendicular bisector of AH
Those two perpendicular bisectors intersect at E
The desired circle has radius EF=EG=EI

As a check, construct the perpendicular bisector of BH, which should also pass through E.

Neat, making 2 equilateral triangles. Wish I'd thought of that

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