Let's discuss a problem based on Integer Sided Obtuse angled triangles with Perimeter.
Find the number of integer-sided isosceles obtuse-angled triangles with perimeter 2008. (Indian RMO 2008)
Let the three sides be a, a and b. Hence 2a + b = 2008 ... (i)
Using the triangular inequality we have 2a > b ...(ii)
Also using the cosine rule for finding sides of a triangle we note that where is the angle between the two equal sides and obtuse (none of the equal angles can be obtuse as a triangle cannot have more than one obtuse angle). Since is obtuse is negative hence we get the inequality which implies . ... (iii)
Combining (i), (ii) and (iii) we have . Solving the inequalities we get 502 < a < 588.25 allowing total 86 values of a. Thus there are 86 such triangles.
Critical Ideas: Cosine rule for measuring sides of a triangle, Pythagorean Inequality