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in Olympiad by (65.1k points)

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

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Explanation:

Let the sides be x, x, y, where x, y are positive integers. Since we are looking for obtuse-angled triangle, y > x. Moreover ,2x + y = 2008 shows that y is even. But y < x + x, by triangle inequality. Thus y < 1004.Thus the possible triples (y, x, x) = (1002, 503, 503), (1000, 504, 504), (998, 505, 505), and so on. The general form is (y, x, x) = (1004 - 2k, 502 + k, 502 + k) , where k = 1, 2, 3,…., 501.

But the condition that the triangle is obtuse leads to

(1004 - 2k)2 > 2(502 + k)2 

This simplifies to

5022 + k2 - 6(502)k > 0. 

Solving this quadratic inequality for k, we see that 

k <502 (3 - 22), or k > 502 (3 + 22),

Since k  501, we can rule out the second possibility. Thus k < 502(3 - 2√2), which is approximately 86.1432. We conclude that k ≤ 86. Thus we get 86 triangles

(y, x, x) = (1004 - 2k, 502 + k, 502 + k), k = 1, 2, 3, ………, 86.

The last obtuse triangle in this list is : (832, 588, 588). (It is easy to check that 8322 - 5882 - 5882 = 736 > 0, whereas 8302 - 5892 - 5892 = -4982 < 0).

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