Happy 100th Fiddler!

Fiddler

From Samuel Klein comes a triangular tribulation:

Dozo is a strategy game with a rather distinctive board:

The board features 28 holes in which players place markers, with the goal of making an equilateral triangle of any size with one color.

How many distinct equilateral triangles can you find whose vertices are the centers of holes on the board? (If two triangles are congruent but have different vertices, they should still be counted as distinct.)

Solution

I determined the following groups and subgroups of equilateral triangles :

1. Pointing down or up. (78 Triangles)

   1. $s = 1$.
      1. There are 21 pointing down.
      2. There are 15 pointing up.
   1. $s = 2$.
      1. There are 15 pointing down.
      2. There are 6 pointing up.
   1. $s = 3$.
      1. There are 10 pointing down.
      2. There is 1 pointing up.
   1. $s = 4$.
      1. There are 6 pointing down.
   1. $s = 5$.
      1. There are 3 pointing down.
   1. $s = 6$.
      1. There is 1 pointing down.

2. Pointing up and left, or up and right. (20 Triangles)

   1. $s = \dfrac{\sqrt{29}}{2}$.
      1. There are 6 pointing up and right.
      2. There are 6 pointing up and left.
   1. $s = \dfrac{\sqrt{53}}{2}$.
      1. There are 3 pointing up and right.
      2. There are 3 pointing up and left.
   1. $s = \dfrac{\sqrt{85}}{2}$.
      1. There is 1 pointing up and right.
      2. There is 1 pointing up and left.

3. Pointing right or left. (22 Triangles)

   1. $s = \sqrt{3}$.
      1. There are 10 pointing right.
      2. There are 10 pointing left.
   1. $s = 2\sqrt{3}$.
      1. There is 1 pointing right.
      2. There is 1 pointing left.

Answer

There are $$\boxed{120}$$

Take a gander!

Extra Credit

Happy Fourth of July! In celebration of America’s birthday, let’s count more shapes—not in a board game, but in the American flag:

In particular, consider the centers of the 50 stars depicted on the flag. How many distinct parallelograms can you find whose vertices are all centers of stars? (If two parallelograms are congruent but have different vertices, they should still be counted as distinct.)

Solution

Solution

I wrote code that did the following :

1. Create the array of stars, numbered $1-50$, from top left to bottom right.
2. The first loop limits the first vertex of the parallelogram to be any point from Star 1 to Star 43.
3. The second loop limits the next vertex of the parallelogram to be any point from the first vertext to Star 44.
4. The third loop limits the next vertex of the parallelogram to be any point from the second vertex to Star 49.
5. The fourth loop limits the last vertex of the parallelogram to be any point from the third vertex to Star 50.
6. The length of each pair of opposite sides of the quadrilateral are compared.
7. The slopes of each pair of opposite sides of the quadrilateral are compared.
8. The slopes of the first two sides are compared.
9. If the criteria satisfy a paralellogram, the star indices are saved.
10. The first parallelogram is formed from Stars 1, 2, 8, 7.
11. The last parallelogram is formed from Stars 43, 44, 50, 49.
12. There are several ways to speed this code, but it runs fast enough as is.

Answer

There are $$\boxed{5918}$$

Take a gander!

Rohan Lewis

2025.07.07

Code can be found here.