INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

October 5, 2013

Integer Sided Obtuse angled triangles with perimeter 8

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

One comment on “Integer Sided Obtuse angled triangles with perimeter 8”

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.