<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 inner-most triangle will have a = 2⁰ and b = c = 2¹.

For  a<b=c<2001:
(a₀,b₀,c₀) = (512, 1024, 1024)
(a₁,b₁,c₁) = (256, 512, 512)
(a₂,b₂,c₂) = (128, 256, 256)
(a₃,b₃,c₃) = (64, 128, 128)
(a₄,b₄,c₄) = (32, 64, 64)
(a₅,b₅,c₅) = (16, 32, 32)
(a₆,b₆,c₆) = (8, 16, 16)
(a₇,b₇,c₇) = (4, 8, 8)
(a₈,b₈,c₈) = (2, 4, 4)
(a₉,b₉,c₉) = (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., 2×2ⁿ ,2×2ⁿ 3×2ⁿ) is where n = 9, and for triangle 2,3,4 (i.e., 2×2ⁿ ,3×2ⁿ, 4×2ⁿ) is where n=8.

For  a=b<c<2001:
(a₀,b₀,c₀) = (1024, 1024, 1536)
(a₁,b₁,c₁) = (512, 512, 768)
(a₂,b₂,c₂) = (256, 256, 384)
(a₃,b₃,c₃) = (128, 128, 192)
(a₄,b₄,c₄) = (64, 64, 96)
(a₅,b₅,c₅) = (32, 32, 48)
(a₆,b₆,c₆) = (16, 16, 24)
(a₇,b₇,c₇) = (8, 8, 12)
(a₈,b₈,c₈) = (4, 4, 6)
(a₉,b₉,c₉) = (2, 2, 3)

For  a<b<c<2001:
(a₀,b₀,c₀) = (512, 768, 1024)
(a₁,b₁,c₁) = (256, 384, 512)
(a₂,b₂,c₂) = (128, 192, 256)
(a₃,b₃,c₃) = (64, 96, 128)
(a₄,b₄,c₄) = (32, 48, 64)
(a₅,b₅,c₅) = (16, 24, 32)
(a₆,b₆,c₆) = (8, 12, 16)
(a₇,b₇,c₇) = (4, 6, 8)
(a₈,b₈,c₈) = (2, 3, 4)

Edited on December 21, 2014, 4:08 am

Edited on December 21, 2014, 4:11 am

Posted by Dej Mar on 2014-12-20 16:35:20