We divided each side of an equilateral triangle into 5 equal parts, then drew lines parallel to the sides of the triangle through the division points. The lines and the sides of the triangle have 21 intersection points inside and on the perimeter of the triangle. How many equilateral triangles are there whose all three vertices are one of these intersection points?
We consider regular triangles normal if their vertices are lattice points resulting from the division of the grid, and they are in the same orientation as the original triangle (i.e., they can be obtained by a positive scale reduction of the original triangle).

![](https://cdn.mathpix.com/cropped/2024_05_02_32dc8eaaf976c24af9cdg-1.jpg?height=385&width=436&top_left_y=306&top_left_x=823)

Figure 1

Let $a_{k}$ denote the number of normal triangles with side length $k$ units (clearly $1 \leqslant k \leqslant 5$; assuming the sides of the original triangle are 5 units long, $a_{5}=1$). Let $P Q R$ be any regular sub-triangle. Suppose, for example, that $P$ is closest to the line $A C$, $Q$ to the line $A B$, and $R$ to the line $B C$. By drawing lines through $P$, $Q$, and $R$ parallel to $A C$, $A B$, and $B C$ respectively, we arrive at a normal triangle $U V W$ whose sides $U V$, $U W$, and $V W$ contain $P$, $Q$, and $R$. The regularity of the triangles implies that $U P=V R=W Q=x$ and $P V=R W=Q U=y$. If the side of the triangle $U V W$ is $k$ units, then $0 \leqslant x \leqslant k-1$, so $k$ smaller lattice triangles can be inscribed in $U V W$ in a similar manner to $P Q R$. Since every lattice triangle can be inscribed in exactly one normal triangle, the number of regular lattice triangles in $A B C$ is: $a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_32dc8eaaf976c24af9cdg-1.jpg?height=377&width=418&top_left_y=1167&top_left_x=840)

Figure 2

Let $X Y Z$ be a normal triangle with side length $k$, and $X Y\|A B, Y Z\| B C, Z X \| C A$. The triangle $X Y Z$ is uniquely determined by the distance of $X Y$ from $A B$, $Y Z$ from $B C$, and $Z X$ from $C A$. If these distances are $a, b$, and $c$, then $0 \leqslant a, b, c$ and $a+b+c=5-k$. Any triple of integers $a, b, c$ that satisfies these two conditions determines a normal triangle with side length $k$; thus, $a_{k}$ is equal to the number of ordered triples of integers $a, b, c$ such that $a+b+c=5-k$. Here, the value of $a$ can vary from 0 to $5-k$; given $a$, $b$ can be any of the numbers $0,1 \ldots, 5-k-a$, which gives $6-k-a$ possibilities (with fixed $a$ and $b$, $c$ is uniquely determined). Therefore,

$$
a_{k}=\sum_{a=0}^{5-k}(6-k-a)=(6-k)^{2}-\frac{(6-k)(5-k)}{2}=\frac{(6-k)(7-k)}{2}
$$

Thus, the number of triangles we are looking for is:

$$
15+2 \cdot 10+3 \cdot 6+4 \cdot 3+5 \cdot 1=70
$$

Remark. If we divide the sides of the triangle $A B C$ into $n$ parts instead of 5, then the number of regular lattice triangles is:

$$
\sum_{k=1}^{n} k \cdot \frac{(n+1-k)(n+2-k)}{2}=\frac{n(n+1)(n+2)(n+3)}{24}=\binom{n+3}{4}
$$