Messing around in higher dimensions 2

Geometry Level 5

2D: The $3$ lengths $3,4,5$ can be arranged to form a triangle with integer area.

3D: The $6$ lengths $6,7,8,9,10,11$ can be arranged to form a tetrahedron with integer volume (note the arrangement matters - not every tetrahedron with these edge-lengths has integer volume)

4D: Can the $10$ lengths $12,13,\cdots,21$ be arranged to form a pentachoron with integer hypervolume?

If they can, then enter this hypervolume as your answer. If not, find the smallest achievable hypervolume with these edge-lengths, and enter the nearest integer to this minimum hypervolume.

Hint: one approach is to use this formula

The answer is 7.

1 solution

Yuriy Kazakov
Mar 14, 2020

I used combinatorial enumeration of options with Python.
P.S. Fine same problem exist for the 10 lengths $10,11,...,19$ . Try this.

You're right, Kazakov, for 10 lengths $10,11,...,19$ there does exist an unique integer volume. For hot rod fans, it's the same number as one particular Ford V-8 engine block series.

Michael Mendrin - 1 year, 2 months ago

Probably a coincidence...

Are those the shortest consecutive integer edge-lengths that give an integer hypervolume?

Chris Lewis - 1 year, 2 months ago

I went about this the same way (brute-force permutations using the Cayley-Menger determinant). The range of possible hypervolumes is quite surprising (from around $7$ to around $1454$ ).

I wonder if there's a number-theoretical approach to showing no integer hypervolumes exist?

Chris Lewis - 1 year, 2 months ago

Messing around in higher dimensions 2

The answer is 7.

1 solution

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