Okay, how many triangles here? Does the left one “interrupt” the right one?
I didn’t understand the “outside border” part, but if inside/outside is important, it still seems like the green circle is interrupting the orange circle from the inside.
(I don’t usually take the opportunity to trolley people on the Internet, really!)
Pretty sure I see 21!
Your pic’s a little too small for me to see clearly, but it looks to me like #8 and #12 are not triangles, but some sort of irregular arrow shape.
No, good point. Assume that the lines of the circles are placed too close to see that one doesn’t interrupt the others, and that the lines in the drawn triangle are not intended to cross the outer boundary triangle. That gives us a dominant outside shape we cannot discard, three dependent uninterrupted interior triangles, and the 6 trapezoids.
So if the triangle were constructed of wire, you could cut a the smallest possible hole in a board that you could fit the entire triangle through (flat to the board surface, not on edge), and also cut triangle shaped blocks that would fill, with little extra space, the 3 interior triangles.
There ya go, everyone else can go count imaginary triangles.
You made my day. Thank you! 
Yeah, I get 18 too
Edit- Fuck, didnt think of the paper lines too.
Way more.
I count 72.
Edit: Now it’s 74. Found 2 more. Hint: Give A hard look.
I remember having a multiple choice test in 9th grade and one of the possible answers was “I dunno.” That meant there were at least two correct answers to that question. Well chalk up yet another grievance against the AISD…
I see dead triangles.
No. There are fork handles.
My first count gives 18 triangles. Did I miss any? I don’t think so. This doesn’t seem like a very tricky puzzle.
I’ll accept that.
3"(1+3+2)=18?
As someone who teaches undergrads… thanks for making my job harder
I’m going with 5 before I check the answer.
As noted above, there are WAY more than 18 triangles. The initial problem does not restrict use of the lines of the paper or text ; it’s quite poorly designed.
Sort of the same thing - I got 18 then added the infinite exterior triangle for 19.
I think you should probably conclude there are around 10^240 triangles since we are looking at the inside of nearly all triangles formed by considering three random atoms in the universe to be the vertices.
There are 385*. I see far fewer.
*For each level, there are 10 triangles one unit wide, 9 triangles 2 units wide, 8 triangles 3 units wide, etc. 10 + 9 + 8 + … + 1 = (10 + 1) + (9 + 2) + (8 + 3) + … = 11 * 5 = 55. There is no comingling of triangles between the levels, and there are 7 levels in total. 7 * 55 = 385.
Atoms are points, not lines. But it’s an intersting calculation. The combination equation is
C=n!/r!(n-r)!
r=3 and I estimate n= ~2.3e17 (quick and dirty estimate using an average atomic weight.
That means there are about 1.9e51 combinations of 3 atoms (and the same number of triangles).
If we decide to use protons, neutron, and electrons, the number jumps to about 1.5e55 triangles.
If we decide those really are points, we could try to argue the fibers of the paper are lines. Ignoring that they are curvy, we wouldn’t have any triangles because the fibers can’t be co-planar and intersect (they’d be in the same place at the same time). But if we used their projection on a plane parallel to the paper, I bet the number would be based on the number Pi. Randomly placed line segments are a way to estimate Pi.
