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Integer Triangle Trial (Posted on 2015-04-14) Difficulty: 3 of 5
Determine the total count of isosceles obtuse-angled triangles with perimeter 2015 having the proviso that each of the three sides has integer lengths.

No Solution Yet Submitted by K Sengupta    
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Solution solution Comment 1 of 1
The key is to make the apex angle larger than 90°. So at what size base does the triangle become 90°?

2*((2015-b)/2)^2 = b^2
(2015-b)^2 = 2*b^2
2015^2 - 4030*b + b^2 = 2*b^2

b^2 + 4030*b - 2015^2 = 0

b = (-4030 + sqrt(4030^2 + 4*2015^2)) / 2 ~= 834.64

So the minimum base length is 835.

The base can't be equal to or larger than the sum of the two equal sides. As 2015 is odd it automatically is not the sum of two equal integers; we just need to make sure it's not larger. It can't be larger than half of 2015; i.e., not larger than 1007.5. The largest it can be is 1007.

The base must be odd so that what's left of 2015 can be evenly split into the two equal integer sides.

How many odd integers lie between 835 and 1007 inclusive? (1007 - 833) / 2 = 87, the numerator coming from the fact that we need to count the even/odd pairs before dividing by 2 and we need to exclude the even/odd pairs through 832 & 833.

The answer is 87.

  Posted by Charlie on 2015-04-14 10:42:14
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