The triangle 5,12,13 has an area A=30 and a perimeter P=30, so A/P is 1.
The triangle 9,75,78 has an area A=324 and a perimeter P=162, so A/P is 2.
Find the smallest and largest integer-sided triangles where A/P is 10.
As Larry noted, the triangles become thinner and thinner as they get larger. So the limiting case here is when the triangle inequality is satisfied by the smallest possible margin - the largest side is only 1 unit shorter than the sum of the other two sides: a+b=c+1.
So I tried substituting this into the problem but eventually hit a wall where I had an even quantity set equal to an odd integer.
Then the next case would be a+b=c+2. This does get an answer.
Substituting c=a+b-2 gets these stats:
Perimeter = 2a+2b-2
Area = sqrt[(a+b-1)*(a-1)*(b-1)]
Then A=10*P becomes sqrt[(a+b-1)*(a-1)*(b-1)] = 20*(a+b-1).
Squaring each side; canceling a common factor of a+b-1; and grinding through some algebra yields (a-401)*(b-401) = 160400.
Now to max-out the perimeter I want to split these factors as far out as possible, which is done with the trivial factorization 160400*1.
Then a-401=160400 and b-401=1, which makes a=160801, b=402, and c=161201.
Checking:
Perimeter = 160801+402+161201 = 322404
Area = sqrt[161202*160800*401] = sqrt[10394433921600] = 3224040.
I also noticed that if I take the less extreme factorization 160400=32080*5 then I get Charlie's largest answer (a=32885 b=406 c=32481) found in his initial search .
Possible maximum
