I've solved this problem using this logic.

1. Definitions.

integer GetLineNumber(x) - returns a line number of the 'x' element in the triangle.

boolean IsTop(x, linex) - returns TRUE, if the 'x' element, which is situated on 'linex' line, has a common edge with a neighbour from 'linex - 1' line. [define IsTop(1, 1) = FALSE]

integer RightMove(x)
begin
  if (IsTop(x, GetLineNumber(x)) x = x + 1;
  return (x + 2 * GetLineNumber(x));
end;

integer LeftMove(x)
begin
  if (IsTop(x, GetLineNumber(x)) x = x - 1;
  return (x + 2 * GetLineNumber(x));
end;

2. Statement.
a) 'm' and 'n' ('n' > 'm') are situated on the lines 'linem' and 'linen'.
b) rightp = RightMove(RightMove(...RightMove(m)...)) [('linen' - 'linem') times repeated].
leftp = LeftMove(LeftMove(...LeftMove(m)...)) [('linen' - 'linem') times repeated].
c) If 'leftp' <= 'n' <= 'rightp' than the length of the path from 'm' to 'n' is equal to the length of the path from 'm' to any 'p' such, that 'leftp' <= 'p' <= 'rightp'.

All you need is to add some code which will count the edges while using RightMove and LeftMove functions and understand, what to do, if the condition 'leftp' <= 'n' <= 'rightp' is FALSE.