Counting the Triangles in this image individually would take far too long. Can anyone come up with an algorithm to figure out how many triangles are in this shape? Answers should include the number of triangles you find and the algorithm you used to get that number.

That is far too many triangles.

  • $\begingroup$ Are you looking for a human method or a computer method? The folks over at PCG could help with the computer part. $\endgroup$ – Engineer Toast May 5 '15 at 19:55
  • $\begingroup$ Human method preferrably $\endgroup$ – Chase Sandmann May 5 '15 at 20:09
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    $\begingroup$ I don't get the "Counting them individually would take far too long" because the accepted answer is simply doing that $\endgroup$ – Ivo Beckers May 6 '15 at 7:51
  • $\begingroup$ Accepted answer doesn't seem very easy to do. Finding symmetries and using properties of the triangles can get you to a provable solution much quicker. $\endgroup$ – Trenin Feb 18 '16 at 16:43


Count all the singles, then all the triangles made of 4 triangles, and so on, up to the huge ones. It's a methodical and easy way to do it.

size 1: 48
size 4 : 30
size 9 : 20
size 16 : 12
size 25 : 6
size 36 : 2

Total: 118

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  • $\begingroup$ Can you also state how many triangles you find with that algorithm? I clarified my question to include that requirement. $\endgroup$ – Chase Sandmann May 4 '15 at 22:40
  • $\begingroup$ ok. hold on... let me count them. $\endgroup$ – JLee May 4 '15 at 22:41
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    $\begingroup$ An easy way to count how many size 4 triangles there are is to count all the single triangles bordered on all three sides by another triangle (as opposed to the edge) $\endgroup$ – Engineer Toast May 6 '15 at 14:40
  • $\begingroup$ But how do you make sure you counted correctly? $\endgroup$ – Trenin Feb 18 '16 at 16:43

There are two types of triangles; those with the point up, and those with the point down. Since everything is symmetric, so all we need to do is count triangles pointing up (or down) and double it. We will count the ones pointing up.

Next, see that the six pointed star is just the superposition of one large triangle pointing up over one large triangle pointing down.

Now, we can start counting the up triangles.

Individual single-unit triangles pointing up are found in the large triangle pointing up, plus $3$ left over around the edge. The large triangle pointing up has a base of $6$. So, the number of individual triangles is $T_1=6+5+4+3+2+1+3=21+3=24$.

The next size of triangles are ones with a base of 2. These triangles contain 4 triangles, with an upside down triangle at the centre. Looking at the figure, we therefore only need to count the interior upside down triangles. This time, there are no triangles around the edge, and less of them can fit in the large triangle. We get $T_2=5+4+3+2+1=15$.

Repeat with the rest, and you get $T_3=4+3+2+1=10$, $T_4=3+2+1=6$, $T_5=2+1=3$ and $T_6=1$.

The sum of these is $T=24+15+10+6+3+1=59$. Double this to get the total and you have $118$.

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