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 have been proved for n ≤ 15, and for any triangular number of circles, and conjectures are available for n ≤ 34.^[1][2][3][4]

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.^[5] This conjecture is now known to be true for n ≤ 15.^[6] In a paper by Graham and Lubachevsky concerning solutions for 22 ≤ n ≤ 34 they also conjectured seven infinite families of optimal solutions in addition to the one by Erdős and Oler. These families give conjectured solutions for many more numbers, including n = 37, 40, 42, 43, 46, 49.^[3]

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

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