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I have found a conjecture that there is no triangle whose sides, medians, altitudes and area are all rational. I figure that someone must have already found such a triangle if one existed and yet I saw this conjecture but haven't come across any proof or hints. Is this still an open question? I feel that maybe it can't be proven - is this a question that can't be proven? But how would we show it's logically independent of the standard assumptions in mathematics?

The source where i saw this conjecture is:-https://www.ics.uci.edu/~eppstein/junkyard/open.html

  • While such a statement could conceivably be independent of $\mathsf{ZFC}$ (unless I am mistaken it is $\Pi^0_1$, the same logical complexity as the independent statements in Goedel's incompleteness theorems) this seems very unlikely to me. At least, it seems very unlikely that it could be shown to be independent of $\mathsf{ZFC}$. Finding natural mathematical (as opposed to meta-mathemtical) statements of low logical complexity that are independent of $\mathsf{ZFC}$ is a difficult problem. (Harvey Friedman has some results in this area, but perhaps not quite as natural as this statement.) – Trevor Wilson May 07 '15 at 07:44

3 Answers3

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According to this, it was open when Unsolved Problems in Number Theory was published, which was probably in the eighties, although I can't find the publication date for the life of me. Two rational medians, rational sides, and rational area is possible according to this source. Your problem though in fact was still open in 2004 for the third edition of Unsolved Problems in Number Theory by Guy. It is problem D21 if you can get a copy.

Here are the references listed at the end of the section for it.

Roger C. Alperin, A quartic surface of integer hexahedra, Rocky Mountain J.
    Math., 31(2001) 37-43; MR 2002k:11037.
J. H. J. Almering, Heron problems, thesis, Amsterdam, 1950.
A. S. Anema, Pythagorean triangles with equal perimeters, Scripta Math.,
    15(1949) 89.
Raymond A. Beauregard & E. R. Suryanarayan, Arithmetic triangles, Math.
    Mag., 70(1997) 105-115; MR 98d:11033.
Albert H. Beiler, Recreations in the Theory of Numbers - The Queen of Mathematics
    Entertains, Dover, 1964, pp.131-132.
Ralph Heiner Buchholz, On triangles with rational altitudes, angle bisectors
    or medians, PhD thesis, Univ. of Newcastle, Australia, 1989.
Ralph H. Buchholz, Triangles with three rational medians, J. Number Theory,
    97(2002) 113-131; MR 2003h:11034.

I'm sorry I can't answer your other questions very well, I'm not a logician. However, as I understand it you would have to construct two systems (that are consistent assuming the usual axioms are) in which the usual axioms are satisfied, one in which it is possible to prove this conjecture, and one in which the conjecture is impossible. Anyone with better knowledge please correct me if I'm wrong.

jgon
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A triangle with rational sides and medians is $87,85,68$ (sides), $79,65.5,63.5$ (medians) but irrational area $\sqrt{7207200}$.

A triangle with 3 rational sides, 2 rational medians and a rational area is $73,51,26$ (sides) $48.5,17.5,\sqrt{3796}$ (medians) and area $420$.

And the altitudes will be only rational if both sides and area are rational.

More Info on:

Perfect Rational Triangle
Randall Rathbun
NCR Corporation, Rancho Bernardo

RE60K
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In the abstract, the author mentions that there is no solution to the problem. enter image description here Among the triangles with rational sides no longer than 5000, there are 45 primitive triangles with with rational sides, midians and irrational areas. \begin{align*} (\,\phantom{00}68,\enspace\phantom{00}85,\enspace\phantom{00}87\,)&&(\,\phantom{0}581,\enspace\phantom{0}774,\enspace\phantom{0}907\,)&&(\,1129,\enspace2680,\enspace3161\,)\\ (\,\phantom{0}127,\enspace\phantom{0}131,\enspace\phantom{0}158\,)&&(\,\phantom{0}515,\enspace1223,\enspace1252\,)&&(\,2769,\enspace2848,\enspace3271\,)\\ (\,\phantom{0}113,\enspace\phantom{0}243,\enspace\phantom{0}290\,)&&(\,\phantom{0}810,\enspace1099,\enspace1339\,)&&(\,1999,\enspace3253,\enspace3396\,)\\ (\,\phantom{0}159,\enspace\phantom{0}314,\enspace\phantom{0}325\,)&&(\,1025,\enspace1312,\enspace1583\,)&&(\,1751,\enspace2482,\enspace3481\,)\\ (\,\phantom{0}145,\enspace\phantom{0}207,\enspace\phantom{0}328\,)&&(\,1223,\enspace1524,\enspace1603\,)&&(\,2521,\enspace2924,\enspace3485\,)\\ \\ (\,\phantom{0}327,\enspace\phantom{0}386,\enspace\phantom{0}409\,)&&(\,\,\phantom{0}509,\enspace1323,\enspace1664\,)&&(\,2095,\enspace2151,\enspace3514\,)\\ (\,\phantom{0}233,\enspace\phantom{0}255,\enspace\phantom{0}442\,)&&(\,\phantom{0}877,\enspace1306,\enspace1917\,)&&(\,1516,\enspace2197,\enspace3537\,)\\ (\,\phantom{0}277,\enspace\phantom{0}446,\enspace\phantom{0}477\,)&&(\,\,\phantom{0}839,\enspace1640,\enspace1929\,)&&(\,1266,\enspace2491,\enspace3593\,)\\ (\,\phantom{0}244,\enspace\phantom{0}367,\enspace\phantom{0}523\,)&&(\,1401,\enspace1778,\enspace1973\,)&&(\,1427,\enspace3161,\enspace3684\,)\\ (\,\phantom{0}142,\enspace\phantom{0}463,\enspace\phantom{0}529\,)&&(\,1391,\enspace1921,\enspace2312\,)&&(\,3181,\enspace3444,\enspace4337\,)\\ \\ (\,\phantom{0}377,\enspace\phantom{0}404,\enspace\phantom{0}619\,)&&(\,2547,\enspace2699,\enspace2704\,)&&(\,2047,\enspace2493,\enspace4450\,)\\ (\,\phantom{0}208,\enspace\phantom{0}659,\enspace\phantom{0}683\,)&&(\,1105,\enspace2007,\enspace2822\,)&&(\,1198,\enspace4563,\enspace4639\,)\\ (\,\phantom{0}587,\enspace\phantom{0}632,\enspace\phantom{0}725\,)&&(\,2251,\enspace2410,\enspace2879\,)&&(\,4498,\enspace4507,\enspace4765\,)\\ (\,\phantom{0}466,\enspace\phantom{0}491,\enspace\phantom{0}807\,)&&(\,1118,\enspace2075,\enspace2963\,)&&(\,1592,\enspace3787,\enspace4771\,)\\ (\,\phantom{0}569,\enspace\phantom{0}640,\enspace\phantom{0}881\,)&&(\,1983,\enspace2059,\enspace3158\,)&&(\,2524,\enspace4855,\enspace4949\,)\\ \end{align*}

Eufisky
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