2016/081) find the number of integer-sided isosceles obtuse angled triangles with perimeter 2008.

let the sides by x,x,y.
so 2x + y = 2008
now 2x >y so y < 1004.
y has to be even.
x cannot be the longest side becuase if x is longest we have isosceles triangle with 2 large angles so cannot be obtuse
so the 3 sides are (y= 1004-2k, x= 502+k, x= 502 + k) and $k < 502$ and k is positive
because it is obtuse
$x^2+x^2 < y^2$ or $2(502+k)^2 < (1004-2k)^2$
solving these we get $k < 502(3-2\sqrt{2})$ or  $k > 502(3+2\sqrt{2})$
because k < 502 so we heve  $k < 502(3-2\sqrt{2})$ giving $k < 86.1432$
so there are 86 triangles