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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.

 See The Solution Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

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 re: extra challenge: | Comment 3 of 6 |
(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

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