# Making triangles out of a triangular lattice?

First off: Yes, I am well aware that this question has been posted before. However, the answer in that one was incorrect, so I decided to make another thread to gain more input, as well as provide what I have figured out on the problem so far.

The problem asks: How many equilateral triangles in the plane have AT LEAST TWO of the points in the triangular lattice below as vertices?

Using the formula found here: Link, I found out that the number of triangles with THREE vertices in the triangular lattice is 48.

However, I am having trouble finding the number of triangles with TWO vertices in the triangular lattice. Would the two vertices in the triangular lattice come from the points bordering the triangular lattice? If so, I came up with 24.

I am also worried about overcounting...and I also do not think I counted ALL of the possibilities. Can someone guide me in the right direction and provide me with some hints?

• If the third point can be anywhere (not necessarily on a lattice point), couldn't every pair of points make two equilateral triangles in the plane? One in each direction. So you would have $\frac{15\times 14}{2}\times 2$ Dec 11, 2014 at 0:31
• @turkeyhundt The problem is you're double-counting. Taking the top point and the leftmost point to make a triangle would be the same as taking the top point and the rightmost point. Dec 11, 2014 at 0:32
• good point. So each triangle that is made entirely of these 15 lattice points will be counted 3 times, so take my value and subtract 2 times the number of triangles entirely made of lattice points maybe. Dec 11, 2014 at 0:37
• @turkeyhundt If I am understanding correctly...that would give us (15*14)-48 = 162? Dec 11, 2014 at 1:17

First things first: let's make sure that we're counting all the grid triangles correctly. To do this, we'll bucket them by their shapes:

There are clearly $1+3+5+7=16$ triangles like this in the grid.

There are six triangles like this with their point 'up' in the grid (one with its apex at each of the top six points), plus exactly one 'upside-down' triangle with its nadir in the middle of the bottom edge, for a total of $7$ triangles of this shape in the grid.

There's no room for an upside-down version of this triangle, and exactly $3$ regular versions (one for each of the top three points).

Clearly there's only $1$ of this triangle in the grid!

That account for the $16+7+3+1=27$ 'classical' grid-aligned triangles here, but there are also going to be 'skew' triangles whose edges don't align with the grid edges. By trying to look at what non-grid-aligned edges can make triangles in the grid, we can see two different varieties:

You should be able to convince yourself that any triangle of this shape must have one of its edges vertical, so we can count these by their top point: there are two copies of this triangle with their top points on the second row of the grid (this one and its mirror-image in the grid) and four more with their top points on the third row of the grid (one each with their top points at the left and right of that row, and two with their top point in the middle, one 'pointing' each way), for a total of $6$.

Finally, there's this sneaky skew triangle and its mirror image, for $2$ more. This gives a total of $27+6+2=35$ triangles total in the grid.

From here, the math goes much like it does in the other answers: to each of the ${15\choose 2}=105$ pairs of points in the grid there should be two triangles (one with its third vertex on either side of the line between the two points in the pair), for a total of $210$ - but each of our $35$ grid triangles is being triple-counted here (once for each of its three edges), so we need to subtract away two of those copies, giving a final tally of $210-2\cdot35=140$ triangles total.

• You're right. There are more triangles with all points on the lattice than the 27 I counted. Dec 15, 2014 at 9:55

First of all, the answer to three points is not 48 nor 27. Stop thinking an equation is the be all end all of mathematics. If you go into math thinking you can apply some third party equation to every situation without further analysis, you will be wrong. This is NOT a triangle within a triangle question; this is forming a triangle with 3 DOTS, not 3 lines. I am not trying to sound harsh, I just think that people should learn math differently.

There are 27 "normal" triangles you can find and 6 special ones that only exist in this example. These 6 triangles have b = 2 * sin(60 degrees), see if you can find them. There's 2 more with b = ??? (too lazy to calculate). So there's 35 total triangle if you use all 3 dots.

As for your original question, (15*14)/2 * 2 is the absolute maximum counting triples. How many triples do you have? That's where you need the first answer. 27 + 8 = 35. 35 triples means -70. Thus the full answer is ((15*14)/2 * 2) - 70 = 140.

Edit: Bad counting and missed two triangles, 140 should work.

• You do realize you used a few equations in here, right? Dec 14, 2014 at 19:39
• Yes, but they are not copied pasted or inherit for math geniuses from time ago. Math is universal, if some guy 100 years early can come up with an equation, you are expected to be able to do the some... or else you will never excel in math. When I was competing in the ARML (American Regional Mathematics League), the kind of problems we have do not come with pre-calculated equations. You are supposed to "figure things out". Dec 14, 2014 at 19:41
• Why is the equation used in the other answer wrong, and why is yours right? Dec 14, 2014 at 19:42
• Why don't you read it... I didn't use an equation, I just pointed out that this situation is not the same as what the original equations is designed for. There are 6 triangles unaccounted for in the original equation that got 27. Is it not obvious if the answer is 33, then the equation that obtained 27 is incorrect? Dec 14, 2014 at 19:44
• You wrote that $b=2 \sin (60)$, which is an equation - but I suppose that arguing that point wouldn't be constructive. So is the answer 33 or 144? Dec 14, 2014 at 19:45

Well, first of all, the no. of triangles with all vertices in the lattice is 27, not 48. You had to take the length of the base (which is 4), not the no. of points.

Second, each triangle that has 3 vertices is being counted THRICE, not twice. So the answer should be $\frac{15\times14}{2}\times2-27\times2=156$

• I tested it out and it said that it was incorrect. Dec 12, 2014 at 21:22
• @MathyPerson Why? Dec 13, 2014 at 7:23
• Not sure. Once I input it it just says that it's incorrect. If it's correct, the solution will be uploaded. Can you further explain your answer? Where did the 15*14 come from? Dec 13, 2014 at 21:34
• @MathyPerson There are 15 points. Each point can be connected to 14 other points. So this would lead us to assume that there are $15\times14$ points possible. However, this would mean that a point A connecting to B, and point B connecting to A as 2 different combinations, so we divide by 2 as every line has been counted twice. We further multiply by this by 2, as every line segment can be used to make 2 eq. triangles, 1 on either side. Dec 14, 2014 at 12:41
• @MathyPerson I've given the answer for triangles with ATLEAST 2 points in the lattice. If you meant EXACTLY 2 points, subtract another 27 from the answer, to get 129. Dec 14, 2014 at 12:44

I counted triangles in three stages: as pointing up, inverted, and oblique. For equilateral triangular grids, with $$n$$ lattice points on a side:

For straight vertical triangles, I’m getting $$T_{n-1}$$, the $$(n-1)$$-th tetrahedral number.
For inverted triangles: $$\left\lfloor\frac{\left(n^2-1\right)(2n-3)}{24}\right\rfloor$$, by adapting OEIS A002623 Ref: Radu Grigore For oblique triangles: $$\left\lfloor\frac{\left((n-1)^2-1\right)\left((n-1)^2-3\right)}{24}\right\rfloor$$, by using OEIS: A001752 first formula (no source given)

Adding the inverted triangles and obliques I find: $$\left\lfloor\frac{n^4-2n^3-n^2+2n}{24}\right\rfloor$$, which numerically seems to be the $$(n-2)$$-th pentatopic number, call it $$H_{n-2}$$.

Combined with the vertical triangles, this gives $$H_{n-1}$$, since this equals $$T_{n-1} + H_{n-2}$$.

Disclaimer: In hindsight I realize this doesn’t quite answer your question because I assumed all three vertices are on the grid.