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 The medians of medians (Posted on 2014-12-20)
<begin> For a triangle with integer sides a,b,c (none over 2000) evaluate the triplet of its medians ma , mb , mc .
Let those three become sides of a new triangle i.e. (a,b,c) =(ma , mb , mc ).
<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.

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 Solution | Comment 1 of 6
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) = (an,bn,cn) and n = 0, and each smaller triplet
(ma , mb , mc) being designated (an,bn,cn) 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

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