1 How many different sets of three points in this equilateral triangular grid are the vertices of an equilateral triangle? Justify your answer.
Solution:
Call the distance between two adjacent dots on the same row 1 unit. Call an equilateral triangle size $x$ if each of its sides are $x$ units long. So the smallest triangle that can be drawn on the grid is size 1 , and the largest triangle has size 4 .
It is easy to see that in this grid there are only a few possible distances between pairs of points: $1,2,3,4, \sqrt{3}, \sqrt{7}$, $2 \sqrt{3}$, and $\sqrt{13}$, so every equilateral triangle that fits in the grid must have its sides equal to one of these lengths. First, consider triangles with one side horizontal. They can point up or down, and the figure on the left above illustrates a size 1 triangle pointing down and a size 2 triangle pointing up. Each such triangle is completely determined by its uppermost or lowermost point. For upward-pointing triangles there are $4+3+2+1=10$ size 1 triangles, $3+2+1=6$ size 2 triangles, $2+1=3$ size 3 triangles, and 1 size 4 triangle, for a total of 20 upwardpointing triangles. Next, consider triangles with one side horizontal pointing down. There are $3+2+1=6$ size 1 triangles and 1 size 2 triangle of this type, for a total of 7 such triangles.
Next, consider tilted triangles that do not have any horizontal sides. There are two possible sizes illustrated by the middle and right-most figures above. The smaller type with side $\sqrt{3}$ have a vertical edge and point either right or left. There are 3 right-pointing triangles a three left-pointing triangles for a total of 6 . There are only two of the larger triangles with no horizontal line having side $\sqrt{7}$.
Th only other distances between pairs of points in the grid are $2 \sqrt{3}$ and $\sqrt{13}$, and it is easy to see that triangles with these side lengths will not fit.
Thus there are $20+7+6+2=35$ total triangles.