# Find the sum of the possible of values of the least side triangle

1. Oct 17, 2008

### ritwik06

1. The problem statement, all variables and given/known data
Given that the smallest angle of a triangle is half of the largest angle. The perimeter of this triangle is p. Find the sum of the possible of values of the least side.

3. The attempt at a solution
The first thing that I thought was of a quadratic in a(last side) through cosine rule but i realized that I actually don't have an angle.

Here is what I have done:
I assume one of the angles to be A, 2A and 180-3A
I have the inequality A<180-3A<2A
36<A<45
a=2R sin A
b=2R sin 3A
c=2R sin 2A
I thought I might be able to apply the cosine rule converting the other sides in terms of 'a' by converting sin A=a/2R

someone please tell me if R(circumradius) is also fixed if perimeter is set to 'p'?

Last edited: Oct 17, 2008
2. Oct 17, 2008

### jhicks

Re: Triangle

I don't think so. If, for example, you keep one side fixed and then vary the lengths of the other two sides the set of points traces out an ellipse with foci on the endpoints of the fixed side. Sorry I can't be of more help.

3. Oct 18, 2008

### ritwik06

Re: Triangle

Thanks for your contribution. At least you tried to help me!!

4. Oct 18, 2008

### ritwik06

Re: Triangle

One more thing that struck me was that when I a straight line of length 'p'. and then create an angle of A/2 and A on its ends. I get a triangle, I bisect the two created sides to get two points on the base. I join these points with the one at which th lines for angle A and A/2 intersect, to get the triangle. So isnt there an infinite number of triangles possible as 36<A<45????????????