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 $latex a^2 + a^2 - 2 a \cdot a \cos \theta = b^2 $ where $latex \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 $latex \theta $ is obtuse $latex \cos \theta $ is negative hence we get the inequality $latex 2a^2 < b^2 $ which implies $latex \sqrt 2 a < b $. ... (iii)
Combining (i), (ii) and (iii) we have $latex \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.
Critical Ideas: Cosine rule for measuring sides of a triangle, Pythagorean Inequality
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