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September 15, 2018

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, $\sqrt2 x<y<2x$

Now add 2x to each,

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

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 $86$.

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