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5 circles in a hexagon. (Posted on 2018-11-25) Difficulty: 5 of 5
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:


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

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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. 


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. 


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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