# Circle packing in an equilateral triangle

Circle packing in an equilateral triangle is a packing problem in discrete mathematics where the objective is to pack n unit circles into the smallest possible equilateral triangle. Optimal solutions are known for n < 13 and for any triangular number of circles, and conjectures are available for n < 28.[1][2][3]

A conjecture of Paul Erdős and Norman Oler states that, if n is a triangular number, then the optimal packings of n − 1 and of n circles have the same side length: that is, according to the conjecture, an optimal packing for n − 1 circles can be found by removing any single circle from the optimal hexagonal packing of n circles.[4] This conjecture is now known to be true for n ≤ 15.[5]

Minimum solutions for the side length of the triangle:[1]

Number of circles Is triangular Length Area
1 True ${\displaystyle 2{\sqrt {3}}}$ = 3.464... 5.196...
2 False ${\displaystyle 2+2{\sqrt {3}}}$ = 5.464... 12.928...
3 True ${\displaystyle 2+2{\sqrt {3}}}$ = 5.464... 12.928...
4 False ${\displaystyle 4{\sqrt {3}}}$ = 6.928... 20.784...
5 False ${\displaystyle 4+2{\sqrt {3}}}$ = 7.464... 24.124...
6 True ${\displaystyle 4+2{\sqrt {3}}}$ = 7.464... 24.124...
7 False ${\displaystyle 2+4{\sqrt {3}}}$ = 8.928... 34.516...
8 False ${\displaystyle 2+2{\sqrt {3}}+{\dfrac {2}{3}}{\sqrt {33}}}$ = 9.293... 37.401...
9 False ${\displaystyle 6+2{\sqrt {3}}}$ = 9.464... 38.784...
10 True ${\displaystyle 6+2{\sqrt {3}}}$ = 9.464... 38.784...
11 False ${\displaystyle 4+2{\sqrt {3}}+{\dfrac {4}{3}}{\sqrt {6}}}$ = 10.730... 49.854...
12 False ${\displaystyle 4+4{\sqrt {3}}}$ = 10.928... 51.712...
13 False ${\displaystyle 4+{\dfrac {10}{3}}{\sqrt {3}}+{\dfrac {2}{3}}{\sqrt {6}}}$ = 11.406... 56.338...
14 False ${\displaystyle 8+2{\sqrt {3}}}$ = 11.464... 56.908...
15 True ${\displaystyle 8+2{\sqrt {3}}}$ = 11.464... 56.908...

A closely related problem is to cover the equilateral triangle with a fixed number of equal circles, having as small a radius as possible.[6]