OFFSET
3,1
COMMENTS
Without the minimal area stipulation, the result differs for some n. (See n = 12 in the examples.) - Peter Munn, Nov 17 2022
LINKS
Hugo Pfoertner, Illustrations of optimal polygons for n <= 26 (2018).
EXAMPLE
For n = 5, the polygon with minimal area A070911(5) = 5 and enclosing circle of least diameter is
2 D
| + +
| + +
| + +
1 E C
| + +
| + +
| + +
0 A + + + B
0 ----- 1 ----- 2 ---
.
The enclosing circle passes through points A (0,0), C (2,1) and D (1,2). Its diameter is sqrt(50/9). Therefore a(5) = 50 and A322029(5) = 9.
For n = 11, a strictly convex polygon ABCDEFGHIJKA with minimal area and enclosing circle of least diameter is
0 ----- 1 ----- 2 ------ 3 ------ 4 ------ 5 ------ 6
5 J ++++++ I
| + +
| + . +
| + +
4 K . H
| + +
| + . +
| + +
3 A . +
| + . +
| + . . +
| + . +
2 B O G
| + . +
| + . +
| + . +
1 C F
| + +
| + +
| + +
0 D ++++++ E
0 ----- 1 ----- 2 ------ 3 ------ 4 ------ 5 ------ 6
.
The diameter d of the enclosing circle is determined by points A and F, with I also lying on this circle. d^2 = 6^2 + 2^2 = 40. Therefore a(11) = 40 and A322029(11) = 1.
n = 12 is a case where the minimal area stipulation is significant. If we take the upper 6 edges in the n = 11 illustration above and rotate them about the enclosing circle's center to generate another 6 edges, we get a 12-gon with relevant squared diameter a(11) = 40 that meets all criteria except minimal area. This 12-gon's area is 26, and to meet the minimal area A070911(12)/2 = 24, the least squared diameter achievable is 52 (see illustration in the Pfoertner link). So a(12) = 52 and A322029(12) = 1. - Peter Munn, Nov 17 2022
CROSSREFS
KEYWORD
nonn,frac,hard
AUTHOR
Hugo Pfoertner, Nov 21 2018
EXTENSIONS
a(21)-a(26) from Hugo Pfoertner, Dec 03 2018
STATUS
approved