This puzzle was suggested by jwezorek in Three triangles passing through every dot of a 5x5 grid

25 dots are drawn as a 5x5 regular square grid. Can you draw 4 non-right angled triangles that pass through every dot? The triangles cannot have a right angle and their corners must lie on the dots. Bonus: can you find multiple solutions that are not rotations/mirrors of each other?

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    $\begingroup$ In my opinion this is the most interesting puzzle of the 3 variants I posted :) $\endgroup$ Jun 29, 2021 at 2:08
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    $\begingroup$ Agree to that. Not too easy, but not that hard, and still interesting. $\endgroup$
    – justhalf
    Jun 29, 2021 at 9:09
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    $\begingroup$ Is there an OEIS sequence: "minimum number of non-right angled triangles that pass through every dot of a NxN grid"? How do you guess this scales with N? sublinear or superlinear in N? $\endgroup$
    – smci
    Jun 29, 2021 at 23:44
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    $\begingroup$ This sequence doesn't exist, but I can make it :) $\endgroup$ Jun 30, 2021 at 1:37
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    $\begingroup$ Actually T(2) doesn't exist, as all triangles are right-angled... $\endgroup$ Jul 2, 2021 at 6:00

1 Answer 1


Here is one solution. I came up with it pretty quickly so I'm guessing there are multiple solutions out there:


Here is a second solution which is similar to the first, but not an exact mirror/rotation of it:


Here is a third solution not at all like the previous two:


What seems to be a common theme in these solutions is:

Each triangle has one "straight" edge along the rim, one "45 degree diagonal edge" and one "other" edge. The bulk of the dots are covered by the straight and diagonal edges, and the "other" edge is usually just wasted.

A 4th solution, using the same techniques:


  • $\begingroup$ Correct and well done! Looks like there are many solutions. $\endgroup$ Jun 29, 2021 at 9:24
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    $\begingroup$ I wonder whether there is a solution where no dot is covered twice. $\endgroup$
    – Gareth McCaughan
    Jun 29, 2021 at 11:16
  • $\begingroup$ @GarethMcCaughan I think it would be hard with the technique I used, because to cover the middle 9 dots, I normally use 3 diagonals going one way and 1 diagonal going the other way, which naturally crosses at least one dot twice. To not do that, the diagonals would need to all go in the same direction. I'll think about if that is possible. $\endgroup$
    – JS1
    Jun 29, 2021 at 18:01
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    $\begingroup$ @GarethMcCaughan, I have checked with my program and I don't think it is possible. However there is a solution with a single dot that is covered twice and all other dots covered once. $\endgroup$ Jun 30, 2021 at 2:01

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