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 One of a kind (Posted on 2014-07-27)
Find a triangle with integer sides and medians.

There exists only one generic solution!

 See The Solution Submitted by Ady TZIDON No Rating

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 re: contradictions? Yes, Sir | Comment 2 of 3 |

Apparently, there is more than one generic solution.

Ch,

I have checked the compliance of your triplets with the formula:
mc=.5*sqrt(2*a^2+2*b^2-c^2) and found the following:
(68, 85, 87)=>(136, 170, 174) with medians  158, 131 and 127
(127, 131, 158) a wrong triplet (medians of the previous)
(113, 243, 290)=>(226,486,580) with medians 523,367 and 244
(145, 207, 328)=>(290,414,656) with medians 529,463, and 142
(327, 386, 409)=>(654,772,818) with medians  725,632, & 587.
The original question (from a Russian contest) was neither solved
nor altered by me. The claim of uniqueness was there.
I believe that the only way to solve it is checking  all triplets, say
below 1000, s.t. a<b<c<(a+b), rotating the values of the triplet,
applying the formula up to 3 times, and saving the "all integers" results.
Once again both of us learn not to be misguided by some web authority.

 Posted by Ady TZIDON on 2014-07-28 02:17:41

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