In triangle ABC, it is known that AB = 9 and BC/AC = 40/41.
Find the maximum possible area of the triangle ABC.
(In reply to
re: Nice problem by Bractals)
To be quite honest I started out with a 'Heron' solution, and only in the course of that saw the key features of the problem, leading to the 'simpler' solution.
When an expression like:
1/4((-(2n-1)^8+2(2n-1)^4(2(n^2-n))^4+4(2n-1)^4(2(n^2-n))^2(2(n^2-n)+1)^2+2(2n-1)^4(2(n^2-n)+1)^4-(2(n^2-n))^8+2(2(n^2-n))^4(2(n^2-n)+1)^4-(2(n^2-n)+1)^8)/(2n-1)^4)^(1/2)
simplifies to n(n-1)(2n^2-2n+1) (i.e.) bc/2, there just has to be an easier way!
|
|
Posted by broll
on 2014-10-25 09:04:35 |
re(2): Nice problem
