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# RMO 2008 Problem 6 Solution - Pythagoras Extended!

In this post, we are discussing an RMO 2008 Problem 6. This Problem is based on Pythagoras Theorem. Try it yourself and utilize the hints, if required.

## The Problem

Find the number of integer-sided isosceles obtuse-angled triangles with perimeter 2008.

Solution of RMO 2008 Problem 1

• Cosine Rule: If $\mathrm{ABC}$ is any triangle, $\angle B A C=\theta$ then $A B^{2}+A C^{2}-2 \times A B \times A C \times \cos \theta=B C^{2}$
• Pythagoras Theorem: If $\mathrm{ABC}$ is a right angled triangle with $\angle B A C=90^{\circ}$ then $A B^{2}+A C^{2}=B C^{2}$
• Triangular Inequality: Sum of two sides of a triangle is greater than the third side.

By cosine rule,

$AB^2+AC^2-2.AB.BC\cos\theta=BC^2$

$x^2+x^2-2.x^2.\cos\theta=y^2$

Since $\theta$ is greater than $90^{\circ}$ so $\cos\theta$ is negative.

$x^2+x^2$ + positive number = $y^2$

$x^2+x^2<y^2$ $\Rightarrow 2x^2<y^2$

$\Rightarrow \sqrt2 .x<y$

By the triangle inequality,

$y<2x$

So, $\sqrt2 x<y<2x$

$\sqrt2 x+2x<y+2x<2x+2x$

Since the perimeter is $2x+y$ so $2x+y=2008$

Then, $4x>2008$ $\Rightarrow x>502$

$x<\frac{2008(2-\sqrt2)}{2}$

$x<1004(2-\sqrt2)$

$x<588$ so, $502<x<588$

Thus x can take $86$ values.

So the answer is $86$.