<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.
(In reply to
extra challenge by Ady TZIDON)
There is little difference for a maximal number of times for an isosceles triangle a<b=c to that of the equilateral a=b=c. The number of triangles is but reduced by one.
For a<b=c<2001, the innermost triangle will have a = 2^{0} and b = c = 2^{1}.
For a<b=c<2001:
(a_{0},b_{0},c_{0}) = (512, 1024, 1024)
(a_{1},b_{1},c_{1}) = (256, 512, 512)
(a_{2},b_{2},c_{2}) = (128, 256, 256)
(a_{3},b_{3},c_{3}) = (64, 128, 128)
(a_{4},b_{4},c_{4}) = (32, 64, 64)
(a_{5},b_{5},c_{5}) = (16, 32, 32)
(a_{6},b_{6},c_{6}) = (8, 16, 16)
(a_{7},b_{7},c_{7}) = (4, 8, 8)
(a_{8},b_{8},c_{8}) = (2, 4, 4)
(a_{9},b_{9},c_{9}) = (1, 2, 2)
For a=b<c<2001, I had initially considered that one could reduce the smallest to a 1,1,2 and for a<b<c<2001 to 1,2,3 but both those triangles would be degenerate, and thus the sides would need be different. For a=b<c<2001, the smallest triangle would be 2,2,3, and for a<b<c<2001, the smallest triangle would need be 2,3,4. Each larger triangle would have sides twice as large as its "medial" triangle. The largest triangle for 2,2,3 (i.e., 2×2^{n} ,2×2^{n} 3×2^{n}) is where n = 9, and for triangle 2,3,4 (i.e., 2×2^{n} ,3×2^{n}, 4×2^{n}) is where n=8.
For a=b<c<2001:
(a_{0},b_{0},c_{0}) = (1024, 1024, 1536)
(a_{1},b_{1},c_{1}) = (512, 512, 768)
(a_{2},b_{2},c_{2}) = (256, 256, 384)
(a_{3},b_{3},c_{3}) = (128, 128, 192)
(a_{4},b_{4},c_{4}) = (64, 64, 96)
(a_{5},b_{5},c_{5}) = (32, 32, 48)
(a_{6},b_{6},c_{6}) = (16, 16, 24)
(a_{7},b_{7},c_{7}) = (8, 8, 12)
(a_{8},b_{8},c_{8}) = (4, 4, 6)
(a_{9},b_{9},c_{9}) = (2, 2, 3)
For a<b<c<2001:
(a_{0},b_{0},c_{0}) = (512, 768, 1024)
(a_{1},b_{1},c_{1}) = (256, 384, 512)
(a_{2},b_{2},c_{2}) = (128, 192, 256)
(a_{3},b_{3},c_{3}) = (64, 96, 128)
(a_{4},b_{4},c_{4}) = (32, 48, 64)
(a_{5},b_{5},c_{5}) = (16, 24, 32)
(a_{6},b_{6},c_{6}) = (8, 12, 16)
(a_{7},b_{7},c_{7}) = (4, 6, 8)
(a_{8},b_{8},c_{8}) = (2, 3, 4)
Edited on December 21, 2014, 4:08 am
Edited on December 21, 2014, 4:11 am

Posted by Dej Mar
on 20141220 16:35:20 