<begin> For a triangle with integer sides a,b,c (none over 2000) evaluate the triplet of its medians m_a , m_b , m_c .
Let those three become sides of a new triangle i.e. (a,b,c) =(m_a , m_b , m_c ).
<end>

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

Posted by Ady TZIDON on 2014-12-20 13:13:53