<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.
This
is a fascinating problem (if I’m reading it correctly).
Using the cosine rule in triangle ABC and then again in triangle ABM,
where AM is a median, so that both equations feature cos C,
provides simultaneous equations that allow the length of
AM to be found as m_{A} = sqrt(2b^{2} + 2c^{2} – a^{2})/2,
and, by cyclic
rotation of a, b and c, the three medians, m_{A}, m_{B}, m_{C},
are:
sqrt(2b^{2} + 2c^{2} – a^{2})/2, sqrt(2b^{2} +
2c^{2} – a^{2})/2, sqrt(2b^{2} + 2c^{2} – a^{2})/2
Applying the same algebra to these three
lengths as if they were
the sides of a triangle gives the ‘medians’ (3a/4, 3b/4, 3c/4).
 interesting algebra, and a nice surprise! Geometrically this
can be shown by translating the medians to new positions so
that they form a triangle, and constructing its medians.
Clearly, the medians will be integers only if all sides are even, and
whatever power of 2 is common to all sides will be reduced by 1
when considering the medians. Conversely, the medians’ medians
will necessarily have 3 as a common factor and whatever power of
3 they have in common will be reduced by 1 when considering
the original sides. Consequently if a small integer sided triangle is
found with integer medians then, provided they share the factor 3,
a bigger triangle with that property can be found by multiplying
its sides by 4/3. To do this repeatedly will give alternate terms in
the sequence of trios that Ady requires.
Starting with the smallest triangle I can find, (786,948,762) with
integer medians (783,765,612) that all contain 3, and working
backwards gives the longest sequence of integer trios I can find
below 2000:
Sides Medians
(1360, 1088, 1392) (1016, 1048,
1264)
(1016, 1048, 1264) (1020, 816,
1044)
(1020, 816, 1044) (786, 948, 762)
(786, 948,762) (783, 765,
612)
If I’ve misinterpreted the problem, never mind, it’s been fun.

Posted by Harry
on 20141223 14:54:00 