For a triangle with integer sides a,b,c
(none over 2000) evaluate the triplet of its medians ma
Let those three become sides of a new triangle i.e. (a,b,c) =(ma
, mc )
It is up to you to find a triplet (a,b,c) such that the above procedure can be executed a maximal number of times, creating sets of “medians“ with integer values only.
The answer should include: (a,b,c) and all interim sets of medians.
Rem: Can be solved analytically.
D4: Please solve the puzzle, adding a constraint a<b<c<2001 (or a=b<c<2001 or a<b=c<2001), i.e. not all equal.
If nobody does it - the modified puzzle will reappear in the future as The medians of medians ll.
BTW, Dej Mar's solution is flawless, assuming a=b=c.
Edited on December 20, 2014, 1:37 pm