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 5 circles in a hexagon. (Posted on 2018-11-25)
Pack five unit circles in the smallest regular hexagon possible.

The exact solution for the smallest possible side length of the hexagon is given by the largest real root of a fourth degree polynomial:

P(x)=ax4+b√(3)x3+cx2+d√(3)x+e

Where a,b,c,d,e are integers. Find them.

 No Solution Yet Submitted by Jer No Rating

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 soln | Comment 5 of 9 |

While the centers of the circles do not make a regular pentagon, they do make an interesting pentagon. It is one with bilateral symmetry across the horizontal bisector of the leftmost circle, with the remaining centers forming an isosceles trapezoid.

I believe a bona fide mathematical proof here would have to show that this is the optimal arrangement. I simply adopted it.

I have put an annotated version of the figure here for reference and wrote my work out longhand here.  (The algebra got a bit gory, likely due to the method and coordinate system I chose.)

The bottom line is:

-63 x^4 + 219 sqrt(3) x^3 - 729 x^2 + 286 sqrt(3) x - 64 = 0 has roots:

x = (0.16660, 1.3586, 1.4964, 2.9994) with the last root being our man.

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Another interesting property of this pentagon is that the line "ce", extended downward, is tangent to the circle below. This is seen from the fact that the angle downward at c is 30 degrees, making a 1, 2, sqrt(3) triangle, using circle radii as sides.

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A personal opinion regarding this site: I believe problem authors should take the responsibility to check whether their problems (new and old) have been solved, and to then: post solutions, or mark solved by pointing to one or more solutions in the comments, or, if not solved, perhaps give hints.
Having a clear indication of which problems remain unsolved will spur visitor interest and serve to revitalize this site.

Edited on December 1, 2018, 12:52 am
 Posted by Steven Lord on 2018-11-29 11:55:22

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