Let n be a natural number.
Determine the minimal number of equilateral triangles of side 1 to cover the surface of an equilateral triangle of n + 1/2n
A simple lower bound is given by the ceiling of the squared area of the larger triangle.
(n+1/(2n))^2 = n^2 + 1 + 1/(4n^2)
The last term is positive and smaller than 1, so the ceiling is n^2+2.
This gives a lot of room to spare and it achievable for any n. Basically you can start at a corner and tile the first (n-1)^2. Then put one in each of the other corners. The remaining equiangular hexagon can be covered by the 2n-1 triangles with minimal overlap.
It's slightly different for even/odd n. I'll make a picture if I get a chance.
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Posted by Jer
on 2021-01-28 08:30:10 |