Isn't it a<589 ?
And since a > 502, the number of triangles = 86 ?
Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter 2008.
same as
2a+b=2008
such that b>√2a
a(2+√2)<2008
no of integers a such that
a<2008/(2+√2) = 588.**
a=1,.....588
for each a, we get a unique b..
yes.. it is the right method...
basically there is a way to get it to one inequaltiy.. but ur solution is perfect :)