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Saturday, December 13, 2008

Q) Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter 2008 units.
(From RMO - 2008)

A) Let x be the length of the isosceles side.

Perimeter = 2x + k = 2008

Using cosine rule:

k^2 = x^2 + x^2 - 2x^2. cos@ .......... where @ E (90, 180)

k^2 = 2x^2.[2.sin^2 (@/2)] ......... where (@/2) E (45, 90)

k E ( sqrt(2) x, 2x)

Also notice that 2x > k is satisfied automatically here.

Now 2x + k = 2008

Solving,

x E ( 2008/4,
(2008)/(2+ sqrt2) )

x E ( 502, 588.something)

But only integral values are required, so x lies between 503 and 588, inclusive.

So, number of values
= 588 - 503 + 1
= 86 solutions


Thus, there are 86 such triangles.

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