Find the number of integer-sided isosceles obtuse-angled triangles with perimeter 2008. (Indian RMO 2008)

Discussion:

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

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