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 a^2 + a^2 - 2 a \cdot a \cos \theta = b^2 where \theta 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 \theta is obtuse \cos \theta is negative hence we get the inequality 2a^2 < b^2 which implies \sqrt 2 a < b . … (iii)

Combining (i), (ii) and (iii) we have \sqrt 2 a < 2008 - 2a < 2a . Solving the inequalities we get 502 < a < 588.25 allowing total 86 values of a. Thus there are 86 such triangles.

Back to RMO 2008

Critical Ideas: Cosine rule for measuring sides of a triangle, Pythagorean Inequality

Some Useful Links:

RMO 2002 Problem 2 – Fermat’s Last Theorem as a guessing tool– Video

Our Math Olympiad Program