Ant on a Triangular Grid

On a square grid, an ant that crawls $x$ units east and $y$ units north for relatively prime $x$ and $y$ will pass through a total of $1$ starting square (colored gray below), $x - 1$ squares to the right of each vertical intersection (colored blue below), and $y - 1$ squares above each horizontal intersection (colored green below) for a total of $S = 1 + (x - 1) + (y - 1) = x + y - 1$ squares.

If x and y are not relatively prime, the diagram can be split up into $\gcd(x, y)$ identical $\frac{x}{\gcd(x, y)} \times \frac{y}{\gcd(x, y)}$ rectangles.

Applying the above equation, the number of squares the ant would pass through is $S = \gcd(x, y) \times (\frac{x}{\gcd(x, y)} + \frac{y}{\gcd(x, y)} - 1)$ or

$S = x + y - \gcd(x, y)$

A equilateral triangular grid can be transformed into a square grid, where each square is $2$ triangles, and the number of triangular bases east is reduced to $x - \frac{1}{2}y$ squares east.

Substituting $x - \frac{1}{2}y$ and $y$ into the equation above and multiplying it by $2$ , the number of triangles that ant would pass through is $T = 2((x - \frac{1}{2}y) + y - \gcd(x - \frac{1}{2}y, y))$ or

$T = 2x + y - 2 \cdot \gcd(x - \frac{1}{2}y, y)$

Therefore, for $x = 1000$ and $y = 550$ , the ant would pass through $T = 2 \cdot 1000 + 550 - 2 \cdot \gcd(1000 - \frac{1}{2} \cdot 550, 550) = \boxed{2500}$ triangles.

The ant enters a new triangle every time it crosses a line. Things are trickier if the ant's path hits an intersection in the grid - but we get around that by splitting the path up at these points. The ant first hits an intersection after crawling $40$ triangles east and $22$ triangles north (bonus question: why not $20$ triangles east and $11$ triangles north?). During this journey, it crosses $21$ east-west lines, $39$ NW-SE (ish) lines, and $40$ SW-NE (ish) lines, entering a total of $21+39+40=100$ triangles on its way. It then repeats this path $24$ more times, for a total of $25 \times 100=\boxed{2500}$ triangles along its path.