coverOn Friday I posted this puzzle….

How many triangles are hidden in this image?


If you have not tried to solve it, have a go now.  For everyone else, the answer is after the break.

48! How did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


  1. For each size of triangle, I counted triangles that are repeated by rotational symmetry and then added any centre triangles.

    Dark Grey:
    1x single 5 unit triangle. (0x3+1)
    3x 4 unit triangles. (1×3+0)
    6x 3 Unit triangles. (2×3+0)
    10x 2 unit triangles. (3×3+1)
    15x single unit triangles. (4×3+1×3+0) – two rows of rotational symmetry

    Inverted Light Grey:
    3x 2 unit triangles (1×3+0)
    10x single unit triangles (3×3+1)

    48 Total.

  2. The number of base triangles was n=5. The formula for an odd number is given in a Google search to be (n*(n+2)*(2n+1) -1)/8. For an even number one removes the -1. The reason it is interesting is to see how the math was done to get the answer.
    Thanks for that nifty puzzle.

  3. Thanks to everyone who supported me during the “attack of the clones”.

    Richard’s answer is technically wrong (perhaps the best sort of wrong there is). The correct answer is the one given on Friday: 47. This is easy to see if you consider the puzzle as a two-dimensional drawing. Then the correct perspective is that of an inhabitant of “Flatland” (Russ Abbott 1884). The Flatlander will see the one large triangle. All the others (the other 47) are hidden.

    This clever puzzle by Richard perhaps inadvertently let those of us with a knowledge of the history of Recreational Mathematics see the deeper answer that even he himself missed.

    Such a learning curve is the Internet.

    1. Fantastic! Or maybe even flatastic! 😀

      Perhaps he had meant that the container triangle was hidden. Such is life in quantum dimensions! 😀

    2. Nice try, but obviously popular TV entertainer Russ Abbot didn’t write Flatland and I wouldn’t make that kind of mistake.

  4. The question was:
    “How many triangles are hidden in this image?”
    I can find 48, so I suggest the answer was zero. If they all can be found, none are “hidden”.

    1. Actually, there is no way to tell how many triangles were hidden, since all we can see is the ones that are not.

  5. What I object to is that Richard does not give any discussion or explanation for the answer. What’s his input – just looking up a puzzle on the internet and posting it?

  6. I worked out the number of triangles for n=1 –> n=7, then was able to discover the formula (it helps there’s a website devoted to cataloguing number series). Further digging revealed the formulas for the numbers of point-up and point-down triangles, as well as expanding the brackets to provide the relevant cubic formulae.

    It’s possible to then derive a formula to state the number of triangles of size x for a large triangle of size n, but gave up after realising you’d need different formulae for n=odd and n=even.

    I have them written down… on a different computer to the one I’m using. If anyone’s sufficiently bored, I’ll post them up later 🙂

  7. I simply counted. Starting with black at the bottom, I counted:
    5+4+3+2+1 =15 small black triangles
    4+3+2+1=10 triangles made up of 3 black and 1 white
    3+2+1=6 triangles made up of 6 black and 3 white
    2+1=3 triangles made up of 6 black and 3 white
    1=1 triangle = the whole thing
    For the white triangles ,
    4+3+2+1 =10white triangles
    2+1=3 triangles made up of 3 white and 1 black


  8. I counted. |Got 45. Looked at comments today to realise I missed the three triangles made up of 3 white 1 black !

  9. I haven’t proven it, but I think I developed an equation to determine the number of triangle.
    The trick is to consider the two orientations of triangles separately.

    The various sized ‘right-side-up’ triangles follow the formula: (n)x1 + (n-1)x2 + (n-2)x3 + (n-3)x4 … as many times as the number n is, where n = number of base unit ‘right-side-up’ triangles.

    Then, the various sized ‘up-side-down’ triangles follow this formula: (n-1)x1 + (n-2)x1 + (n-3)x2 + (n-4)x2 obviously starting with cases where n>=2 and having the multiplier in the pattern 1,1,2,2,3,3,4,4



    right-side-up triangles= 4×1 + 3×2 + 2×3 + 1×4 = 4 + 6 + 6 + 4 = 20
    upside-down triangles= 3×1 + 2×1 + 1×2 = 3 + 2 + 2 = 7
    total = 27


    right-side-up triangles = 5×1 + 4×2 + 3×3 + 2×4 + 1×5 = 5 + 8 + 9 + 8 + 5 = 35
    upside-down triangles = 4×1 + 3×1 + 2×2 + 1×2 = 4 + 3 + 4 + 2 = 13
    total = 48

  10. I saw forty eight triangles, five circles, a dog smoking a pipe and a cloud that looked like WG Grace, do I win the prize?

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