Among other things, LinuxLibrarian wrote:

> Jellby wrote:
>> 
>> - Circles with radius R.
>> - Centers of circles form a equilateral triangle ABC of side L
>> - All 3 circles are tangent to a "circumscribed" triangle abc of side l
>> - Circle centered in A touches abc in P and Q.
>> - aPA is a triangle with angles of 30, 90 and 60 degrees, one leg (PA) is
>> R, the other leg (aP) should be easy to calculate: sin(30)/R=sin(60)/aP
>> - The outer's triangle side (ab) is: aP+AB+Sb = 2*aP+L
>> - The relationship between R and L should be easy too: sin(60)=L/(2*R)
>> - I may be mistaken.
>> 
> 
> Looks great, and thanks, but... What's the "S" in Sb? :-}

It would be one the point where the circle centered on B touches the ab 
side. Have you drawn it?

-- 
light_source{9+9*x,1}camera{orthographic look_at(1-y)/4angle 30location
9/4-z*4}light_source{-9*z,1}union{box{.9-z.1+x clipped_by{plane{2+y-4*x
0}}}box{z-y-.1.1+z}box{-.1.1+x}box{.1z-.1}pigment{rgb<.8.2,1>}}//Jellby

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