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)

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 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

## Some Useful Links:

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

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