|
47
I never saw the solution given by CAML, but you can solve this problem quite easily just from intuition. Since it's a right triangle with sides that are reciprocals of integers, we can compare it to a right triangle with integer lengths by multiplying the sides by its common multiple. So working backwords, we take the smallest right triangle with integer lengths (3,4,5) and divide through by the least common mulitple (60) to get (1/20, 1/15,1/12). Adding them up gives 47.
Now here comes the hard part, proving that it is the minimum. I'm sure there are other ways to do it too, but here's the one I came up with. One way to represent the three lengths of a right triangle is (2mn,m^2-n^2,m^2+n^2), where m and n are two different positive integers, with m>n>0. This will ensure a right triangle since (m^2+n^2)^2=(2mn)^2+(m^2-n^2)^2. I'm sure there are exceptions where the lengths may not be represented by that but the common ones that we know of are.
So, to get the lengths to be integral values, we divide through by the least common multiple to get (1/(m^4-n^4),1/(2mn*(m^2+n^2)),1/(2mn*(m^2-n^2))). Adding up the denominators will yield m^4-n^4+4m^3*n.
By inspection, we can see that the sum increases as m increases. So the minimum sum is found by the minimum m=2 ==> n=1. That makes the sum, 16-1+4*8*1=47.
|
