Is it possible to construct three tetrahedra, each with all six sides being integers and without right angles on any of its faces, such that the volume of one of them is exactly the sum of that of the other two?
Posted Dec 3, 2022, 4:31 AM UTC
1125
Correction: the tetrahedra don't have to be regular
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That's not a correction but just a clarification. I think your question was already perfectly well-stated. Although... Maybe some people define tetrahedra to be regular. I guess you could say "3-simplex in 3-space".
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So, I came up with a way to not only generate such a trio, but in such a way that the two smaller ones result from cutting a regular tetrahedron into two. Except there's one step I need help on.
First, an example. Start with a regular tetrahedron with edge length 8. Choose one edge and mark a point B that divides it into 3 and 5. Now, slice from point B all the way to the edge that is opposite of this starting edge (it will be oriented perpendicular to this starting edge). You will now have two tetrahedra with edges as follows: three edges of length 8, two edges of length 7 and one edge either of length 3 or 5.
This same technique can work for any triple of the form c^2 = a^2 + b^2 - ab for a, b, and c all integers (in the example above 7^2 = 8^2 + (3 or 5)^2 - 8*(3 or 5)). This is the part where I got stuck in how to generate those triples, but they are similar to Pythagorean triples, so I am betting there's a similar trick I'm just not seeing.
How it works: Let's call the original regular tetrahedron's edge length a. When we pick the edge that we are going to divide we notice it is in contact with two faces and not in contact with two other faces. When we cut it the faces that our initial edges is not in contact with will be unaffected, so their three edges remain at length a. We when we choose our point B we are dividing one edge into two smaller edges of length b and a-b. When we slice down through the tetrahedron, we cut the two equilateral faces into smaller triangles. Those smaller triangles still have one edge of length a, then a 60 degree angle, then our new edge of length b or a-b. The new edge that we created opposite the 60 degree angle we can find using the law of cosines: c = sqrt(a^2 + b^2 - 2abcos60) or c=sqrt(a^2 + b^2 - ab) which means the radicand needs to be a perfect square, and thus we have our condition above.
Edit: Some of the triples I found with a spreadsheet (ones not in bold are just multiples of previous solutions)
- 8, 3 or 5, 7
- 15, 7 or 8, 13
- 16, 6 or 10, 14
- 21, 5 or 16, 19
- 24, 9 or 15, 21
- 30, 14 or 16, 26
- 32, 12 or 20, 28
- 35, 11 or 24, 31
- 40, 7 or 33, 37
- 40, 15 or 25, 35
- 42, 10 or 32, 38
- 45, 21 or 24, 39
- 48, 13 or 35, 43
- 48, 18 or 30, 42
- 55, 16 or 39, 49
- 56, 21 or 35, 49
- 60, 28 or 32, 52
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Generating triples satisfying c² = a² + b² - ab...
Pick integers m,n with m > n > 0
a = m² - n²
b = 2mn - n²
c = m² - mn + n²
Copy pasted from https://www.had2know.org/academics/integer-triangles-60-degree-angle.html
Edit: Actually m and n both need to be the same sign in order to have b > 0.
So I added n > 0.
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Wow, that’s a nice algorithm
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Is this an accurate depiction of the technique you described?
https://imgur.com/a/I12QWOl
(what you called B is labeled as Z in the thing I made here)
Assuming yes, we should also check that we don't create any new angles on the faces which are right angles.
Let's stick with a = side length of the starting equilateral tetrahedron.
During the cut three distinct edge lengths are introduced, which are the lengths of
AZ, ZB, ZD.
From the c² = a² + b² - ab equation
- c corresponds to the length of AZ
- WLOG we can say b is the length of ZB
- and a-b is the length of ZD
∠AZB, ∠BZC, ∠AZC, ∠AZD, ∠CZD are the newly introduced angles
∠AZB is congruent to ∠BZC by symmetry
similarly ∠AZD is congruent to ∠CZD
∠AZB and ∠AZD are supplementary, so if one is a right angle then the other is too.
So we only need to check ∠AZB and ∠AZC
Edges CZ and AZ are both length c, so triangle CZA is isosceles, which means that CZA is only a right triangle if c*sqrt(2) = a. This can't happen as a and c are both integers by construction, and rational*irrational ≠ rational.
∠ABZ is 60 degrees, so if triangle ABZ is a right triangle, then we have a 30-60-90 triangle. But in such a scenario, b*sqrt(3) = c. Similar to earlier, this can't happen because rational*irrational ≠ rational.
Another thing. After playing around with it, I realized some 60 degree triangles are not okay, particularly c >= a does not give a valid construction. When c = a, one of the tetrahedrons will have volume 0. When c > a, one of the tetrahedrons will have negative volume.
Working through a bit of algebra, we just need a minor tweak to generating the (a,b,c) triples, m > 2n is required rather than m > n.
Edit: typo, and terminology correction
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Yes, that is exactly what I was trying to describe, thank you. Yeah, I had convinced myself no new right angles could be constructed, but good for actually putting it rigorously.
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Just one very minor note. It is possible to get a 90 degree angle if we allow irrational c. This happens when Z is chosen to be the midpoint of BD.
The maximum angle measure for ∠CZA is approximately 70.5 degrees, which is also attained with the same choice of Z, so it can't be 90 degrees no matter the choice of Z in this construction.
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For those who want a tool to just plug edge lengths into and get a volume:
https://keisan.casio.com/exec/system/1329962711
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Interesting, but it would be really tedious to compute the tetrahedra's volumes using this
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I always thought tetrahedra had 4 sides. Check Wikipedia.
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They have four faces, but six edges. Given the context, OP is clearly asking about integer-length ediges.
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Understood. Thank you.
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Ah, I completely missed that I called the edges sides
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No, it’s impossible.
V = s³/(6√2)
Let the edges of the three tetrahedra (in order of their volumes from smallest to largest) be a, b, and c, respectively.
a³/(6√2) + b³/(6√2) = c³/(6√2)
a³ + b³ = c³
From Fermat’s Last Theorem, we know that there are no positive integers that satisfy a³+b³=c³.
-4†
You're assuming regular tetrahedra. OP's question doesn't specify that.
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Thanks, but I meant for the tetrahedra to be irregular, do you know how to prove whether it would be possible if they were?
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It is not possible to construct three tetrahedra with integer side lengths and without right angles on any of their faces such that the volume of one of them is exactly the sum of the volumes of the other two.
To see this, note that the volume of a tetrahedron with side lengths a, b, c, and d is given by the formula V = sqrt(s(s-a)(s-b)(s-c)(s-d)), where s = (a+b+c+d)/2 is the semiperimeter of the tetrahedron. Since s is always a positive integer, it follows that the volume of a tetrahedron with integer side lengths is always a positive integer as well.
Now suppose we have three tetrahedra with integer side lengths and without right angles on any of their faces, such that the volume of one of them is the sum of the volumes of the other two. Let the side lengths of the first tetrahedron be a, b, c, and d, and let the side lengths of the second and third tetrahedra be e, f, g, and h, and i, j, k, and l, respectively. Then we have
V1 = sqrt(s1(s1-a)(s1-b)(s1-c)(s1-d))
V2 = sqrt(s2(s2-e)(s2-f)(s2-g)(s2-h))
V3 = sqrt(s3(s3-i)(s3-j)(s3-k)(s3-l))
where s1, s2, and s3 are the semiperimeters of the first, second, and third tetrahedra, respectively. Since the volumes of the first and second tetrahedra are integers, it follows that the expressions inside the square roots on the right-hand side are perfect squares. But since the tetrahedra do not have right angles on any of their faces, none of the sides of the tetrahedra are perpendicular to each other, which means that the expressions inside the square roots cannot be perfect squares, and therefore the volumes of the tetrahedra cannot be integers.
Therefore, it is not possible to construct three tetrahedra with integer side lengths and without right angles on any of their faces such that the volume of one of them is the sum of the volumes of the other two.
-4
I'm guessing you generated this nonsense with OpenAI.
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I wish you had a good proof of your statement, but i) tetrahedra have six, not four sides; ii) this is the correct formula, as N8CCRG said; iii) there is a counterexample given by Biological_Computer_
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I think I found an example that works.
The two smaller tetrahedrons have side lengths 6 for the base, and 4 for the other sides. The larger tetrahedron has side lengths 12 for the base, and 7 for the other sides.
If I calculated it correctly, the 12 7 tetrahedron has volume 12*sqrt(3),
and each 6 4 tetrahedron has volume 6*sqrt(3)
Edit: Just to be really clear with the side lengths https://imgur.com/a/r4mcSEz
Wow, they work, how did you find them?
I went with the hope that I could find an example which only used tetrahedrons with that sort of pattern of 2 distinct lengths, one length for base edges, and the other for the rest. Because it was easier to get a simple formula for the volume of that sort of tetrahedron.
Then, brute force. Wrote this program: https://pastebin.com/EvUs8P1x which I think is self-explanatory enough.
The program doesn't analytically check if the volumes exactly add up, just sees if they're close enough, and I checked after the fact. The first example it found worked.
If I may ask, what was the motivation for finding such a thing?
No motivation, the problem just popped into my head this morning and I was too lazy to try it