<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.

For a maximal number of times, it would be where each formed triangle had each side of length 2ⁿ, i.e., the triangle would be an equilateral triangle with the inner-most triangle having sides of length 2⁰, with vertices at the medial points of an equilateral triangle with sides of length 2¹, and so on until the vertices of the triangle are the medial points of a triangles with sides of length 2¹⁰. (2¹¹ = 2024 and exceeds the maximum limit of 2000.)

Where (a,b,c) = (a_n,b_n,c_n) and n = 0, and each smaller triplet
(m_a , m_b , m_c) being designated (a_n,b_n,c_n) and n++:

(a₀,b₀,c₀) = (1024, 1024, 1024)
(a₁,b₁,c₁) = (512, 512, 512)
(a₂,b₂,c₂) = (256, 256, 256)
(a₃,b₃,c₃) = (128, 128, 128)
(a₄,b₄,c₄) = (64, 64, 64)
(a₅,b₅,c₅) = (32, 32, 32)
(a₆,b₆,c₆) = (16, 16, 16)
(a₇,b₇,c₇) = (8, 8, 8)
(a₈,b₈,c₈) = (4, 4, 4)
(a₉,b₉,c₉) = (2, 2, 2)
(a₁₀,b₁₀,c₁₀) = (1, 1, 1)

Posted by Dej Mar on 2014-12-20 10:46:01