The Dozo Board and the Fifty Stars

To mark the $100$th Fiddler, two counting puzzles. A Dozo board has $28$ holes in a triangular array ($7$ per side). How many distinct equilateral triangles have all three vertices at holes? Congruent triangles in different positions count separately.

The Fiddler, Zach Wissner-Gross, July 4, 2025(original post)

Solution

Upright, inverted, and tilted equilateral triangles must all be counted. A classical identity says the number of equilateral triangles with vertices in a triangular array of $n$ points per side is $\binom{n+2}{4}$, every tilted triangle being inscribed in a unique upright bounding one. For $n=7$, $$\binom{9}{4}=\boxed{126}.$$

The computation

Place the $28$ holes at their true coordinates and test every triple: it is equilateral when its three squared side lengths agree (and are nonzero). The brute-force count matches $\binom{9}{4}$.

import itertools as it
from math import sqrt, comb
pts = [(k - r/2, -r*sqrt(3)/2) for r in range(7) for k in range(r + 1)]
d2 = lambda a, b: (a[0] - b[0])**2 + (a[1] - b[1])**2
n = sum(abs(d2(a, b) - d2(b, c)) < 1e-9 and abs(d2(b, c) - d2(a, c)) < 1e-9 and d2(a, b) > 1e-9
        for a, b, c in it.combinations(pts, 3))
print(n, comb(9, 4))                     # 126  126

Extra Credit

The $50$ stars on the American flag sit in the usual offset grid ($9$ rows alternating $6$ and $5$ stars). How many distinct parallelograms have all four vertices at star centres?

Solution

Two point-pairs form a parallelogram exactly when they share a midpoint (the diagonals bisect each other). Bucketing all $\binom{50}{2}$ pairs by midpoint, summing $\binom{k}{2}$ within each bucket, and discarding the degenerate collinear quadruples gives $$\boxed{5918}\ \text{parallelograms}.$$ (The source’s count is paywalled; this is my own enumeration.)

The computation

Group every pair of stars by the midpoint of the segment joining them; any two pairs sharing a midpoint are the diagonals of a parallelogram. Count those, then subtract the collinear (degenerate) cases.

from collections import defaultdict
from itertools import combinations
stars = [(x, y) for y in range(9) for x in range(11) if (x + y) % 2 == 0]
mid = defaultdict(list)
for i, j in combinations(range(50), 2):
    mid[(stars[i][0] + stars[j][0], stars[i][1] + stars[j][1])].append((i, j))
para = degen = 0
for pairs in mid.values():
    para += len(pairs)*(len(pairs) - 1)//2
    for (i, j), (p, q) in combinations(pairs, 2):
        x0, y0 = stars[i]
        cr = lambda A, B: (A[0] - x0)*(B[1] - y0) - (A[1] - y0)*(B[0] - x0)
        if cr(stars[j], stars[p]) == 0 and cr(stars[j], stars[q]) == 0: degen += 1
print(para - degen)                      # 5918