Q: An equilateral triangle, partitioned into n2 congruent equilateral triangles by n-1 equidistant parallels to each of its three sides. Two chess-like bishops placed at any two vertices of the small triangles are said to menace one another if they lie on a same parallel. The problem is to determine the largest number of bishops that can be placed so that none menaces another.

A: A bishop may be assigned three coordinates a; b; c, namely the numbers of sides of small triangles they are off each of the sides of the big triangle.

In other words the condition above is that if i != j; then ai != aj ; bi != bj and ci != cj .
If any pair of these axes coordinates were same, then the bishops would be on the same triangle. If they are on the same triangle, they share the same parallel that made the triangle and hence would menace each other.

In fact the only solution that divides the number of congruent triangles is when the smaller triangle formed by the bishops is itself an equilateral triangle.
The first condition gives us the inequality that pairwise parallel-coordinates are distinct so we have their sum as N(N-1)/2
summing all the three co-ordiantes we have 3N(N-1)/2 <= sum of all( ai+bi+ci)
Since all the triangles are congruent,N = 4
therefore a + b + c = n

By solving this problem, we have the obtained the first condition as in the problem earlier. And again showed that the limiting case of equality is also included as a solution.

#codingexercise
Determine the type of a triangle given its edge lengths
Void printTriangle( int a, int b, int c)
{
If (a == b && b==c)  printf ("equilateral");
If (a ==b || b == c || c == a) printf ("isosceles");
printf ("scalene");
}