[math-fun] Volume=Integer/3 for Heronian tetrahedra?
Michael Reid: What is the situation in 3 dimensions? Can a tetrahedron with integer edge lengths have a volume that is one twelfth (one sixth, one fourth, ... ) of an integer, without being an integer itself?
WDS: My computer did the following. There are 6 edges of a tetrahedron. It tried all 4^6=2^12=4096 possibilities for the edge lengths (mod 4). A possibility was "unacceptable" if it led to 16*(some triangle area)^2 = nonsquare (mod 1024) or 2*288*(tetrahedron volume)^2 = nonsquare (mod 1024) or GCD(all 6 edge lengths, 4) != 1. Among the acceptable possibilities, it found that 2*288*(tetrahedron volume)^2 was always 0 or 448 (mod 576). If it were always 0 (and if no bugs in my program...) then we would have proven "Any tetrahedron with integer edge lengths and rational face areas and rational volume, automatically has integer volume." But the possibility of 448 means that integer/3 volumes have not been ruled out. I also tried mod-3 analysis instead of mod-4, but that reached the same conclusion: 0 and 448 only. It seems to me possible that such a modular analysis might be able to show nonexistence of Heronian 5-simplices. If that can be done it would solve RK Guy's open problem D22. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
There are 42 primitive Heronian tetrahedra with diameter <= 1250 --- for every one, the volume is an integer, and divisible by 4. WFL On 11/20/11, Warren Smith <warren.wds@gmail.com> wrote:
Michael Reid: What is the situation in 3 dimensions? Can a tetrahedron with integer edge lengths have a volume that is one twelfth (one sixth, one fourth, ... ) of an integer, without being an integer itself?
WDS: My computer did the following. There are 6 edges of a tetrahedron. It tried all 4^6=2^12=4096 possibilities for the edge lengths (mod 4). A possibility was "unacceptable" if it led to 16*(some triangle area)^2 = nonsquare (mod 1024) or 2*288*(tetrahedron volume)^2 = nonsquare (mod 1024) or GCD(all 6 edge lengths, 4) != 1. Among the acceptable possibilities, it found that 2*288*(tetrahedron volume)^2 was always 0 or 448 (mod 576).
If it were always 0 (and if no bugs in my program...) then we would have proven "Any tetrahedron with integer edge lengths and rational face areas and rational volume, automatically has integer volume."
But the possibility of 448 means that integer/3 volumes have not been ruled out.
I also tried mod-3 analysis instead of mod-4, but that reached the same conclusion: 0 and 448 only.
It seems to me possible that such a modular analysis might be able to show nonexistence of Heronian 5-simplices. If that can be done it would solve RK Guy's open problem D22. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Sorry, I should rephrase that: modulo similarity, there are 42 tetrahedra with integer sides and rational face areas and volume --- for every one, face areas are even integers, and volume is an integer divisible by 4. WFL
There are 42 primitive Heronian tetrahedra with diameter <= 1250 --- for every one, the volume is an integer, and divisible by 4. WFL
On 11/20/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
There are 42 primitive Heronian tetrahedra with diameter <= 1250 --- for every one, the volume is an integer, and divisible by 4. WFL
On 11/20/11, Warren Smith <warren.wds@gmail.com> wrote:
Michael Reid: What is the situation in 3 dimensions? Can a tetrahedron with integer edge lengths have a volume that is one twelfth (one sixth, one fourth, ... ) of an integer, without being an integer itself?
WDS: My computer did the following. There are 6 edges of a tetrahedron. It tried all 4^6=2^12=4096 possibilities for the edge lengths (mod 4). A possibility was "unacceptable" if it led to 16*(some triangle area)^2 = nonsquare (mod 1024) or 2*288*(tetrahedron volume)^2 = nonsquare (mod 1024) or GCD(all 6 edge lengths, 4) != 1. Among the acceptable possibilities, it found that 2*288*(tetrahedron volume)^2 was always 0 or 448 (mod 576).
If it were always 0 (and if no bugs in my program...) then we would have proven "Any tetrahedron with integer edge lengths and rational face areas and rational volume, automatically has integer volume."
But the possibility of 448 means that integer/3 volumes have not been ruled out.
I also tried mod-3 analysis instead of mod-4, but that reached the same conclusion: 0 and 448 only.
It seems to me possible that such a modular analysis might be able to show nonexistence of Heronian 5-simplices. If that can be done it would solve RK Guy's open problem D22. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (2)
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Fred lunnon -
Warren Smith