Let ABC be a triangle with integral side lengths such that angle A=3 * angle B. Find the minimum value of its perimeter.
(In reply to
re(3): NO DICE.....not a spoiler by Harry)
I could use some help as well.
I got the same equation (3) as you by trisecting the angle 3x to divide ABC into two isosceles triangles, evaluating cos(x) and cos(2x), and substituting into the identity cos(2x) = 2*(cos(x))^2 - 1.
Setting b=n^3 and (a+b)=mn^2 matches the posted solutions but doesn't give a handy form for p.
The ratio (a+b)/b is interesting -- 3/2, 5/3, 7/4, 8/5, 9/5, 10/7, 11/7, 11/6 in order -- but doesn't give me any insight into the general problem.
The only thought I have is to look at area. I don't know if I want to face that algebra. I'll look again later and hope for inspiration.
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Posted by xdog
on 2015-03-06 07:48:35 |
re(4): NO DICE.....not a spoiler
