redrogue wrote:Spoiler:31, assuming 'overlap' can mean 'share a single point,' and that circles 2 and up must overlap the one prior
The largest chord of a circle is its diameter. So, 15 circles of radius 1 = 30, plus the radius of circle 0.
Pelli wrote:I think I have a solution:Spoiler:By rotational symmetry of circles, the optimal placement is to have all centres on a line.
If circle #n has radius rn and distance dn to the next circle (between centres), then we have the relationship rn2 = dn2 + rn+12 . Hence we have 1 = r02 = d02 + d12 + ... + d152 , where d15 = r15 is the distance from the centre of the last circle to its farthest point. We want to maximise d0 + d1 + ... + d15 . By AM-QM the maximum is attained for all d equal, i.e. d0 = d1 = ... = d15 = 1/4. The answer is 16*1/4 = 4.
Edit: Inserted sub and sup tags as suggested
skeptical scientist wrote:Pelli has the right solution, and a nice exposition as well (although you should really use [sub] and [sup] tags!)
Moose Hole wrote:Pro-tip: don't click spoilers if you don't want to know the answer.Avin wrote:I'm still disappointed that I was just given the answer without continuing to have the chance to work it out myself.
Avin wrote:I'm not talking about this forum. I already knew the answer when someone incorrectly marked my answer correct on TheConfoundry.
Lenoxus wrote:Anyway, it turns out the real answer involves basic geometry. I'm almost picturing it, but I'd love it if someone had a picture somewhere.
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