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I saw this question:

What's the fewest weights you need to balance any weight from 1 to 40 pounds?

Suppose you want to create a set of weights so that any object with an
integer weight from 1 to 40 pounds can be balanced on a two-sided
scale by placing a certain combination of these weights onto that
scale.

What is the fewest number of weights you need, and what are their
weights?

Now, the answer was weights of mass in powers of 3 (1, 3, 9, 27, etc.) since they can be placed in either pan. Apparently, this is not really what I conceive of as base-3, but something to the effect of a "-1, 0, 1" and not 0, 1, 2. I am using this reasoning for my guess on this for the triangular cause, but first let me state the question.

What kind of efficient powers of weight suspended from vertices would be required to balance a regular triangle? (The triangle is supported on its centroid (1/3 distance from side to opposite vertex).

And then, what kind of multiples of weights would be required to balance a regular tetrahedron "magically" fastened in space by its centroid (1/4 distance from centroid of triangular face to opposite vertex)? The tetrahedral case has 4 pans (instead of the triangular 3 or the linear 2).

You can add weight to a pan with an unknown weight already in it. Unknown weights can be of any whole number mass, but known weights must be added of mass $x^n$ with $n=0, 1, 2, etc$. What is the optimal $x$ for minimizing the number of weights used? You have an unlimited number of $x^0$ weights, unlimited $x^1$, unlimited number of $x^2$ weights, etc. Find the most efficient $x$.

My guess would be that for a triangle the answer is $4^n$: 1, 4, 16, 64, etc. My guess for the tetrahedral case, if there is only an unknown weight on 1/4 or 3/4 of the vertices, then maybe it could be $5^n$ and possibly $7^n$ for the case of 2/4 vertices with an unknown mass, since the unknown mass is centered on one of the 6 edges rather than the 4 faces/vertices.

You can uniquely weigh items using paired weights, two for every power of $4$:

$begin{matrix}
color{red}{1} & = & 1 & = & 1 \
color{red}{2} + 1 + 1 & = & 4 & = & 4 \
color{red}{3} + 1 & = & 4 & = & 4 \
color{red}{4} & = & 4 & = & 4 \
color{red}{5} & = & 4 + 1 & = & 4 + 1 \
color{red}{6} + 1 + 1 + 4 + 4 & = & 16 & = & 16 \
color{red}{7} + 1 + 4 + 4 & = & 16 & = & 16 \
color{red}{8} + 4 + 4 & = & 16 & = & 16 \
color{red}{9} + 4 + 4 & = & 16 + 1 & = & 16 + 1 \
color{red}{10} + 1 + 1 + 4 & = & 16 & = & 16 \
color{red}{11} + 1 + 4 & = & 16 & = & 16 \
color{red}{12} + 4 & = & 16 & = & 16 \
color{red}{13} + 4 & = & 16 + 1 & = & 16 + 1\
color{red}{14} + 1 + 1 & = & 16 & = & 16 \
color{red}{15} + 1 & = & 16 & = & 16 \
color{red}{16} & = & 16 & = & 16 \
color{red}{17} & = & 16 + 1 & = & 16 + 1\
color{red}{18} + 1 + 1 & = & 16 + 4 & = & 16 + 4 \
color{red}{19} + 1 & = & 16 + 4 & = & 16 + 4\
color{red}{20} & = & 16 + 4 & = & 16 + 4 \
color{red}{21} & = & 16 + 4 + 1& = & 16 + 4 + 1\
color{red}{22} + 1 + 1 + 4 + 4 + 16 + 16 & = & 64 & = & 64
end{matrix}$

This is because every integer has a unique representation in base $4$ using digits valued $-2, -1, 0, 1$.

In order to balance a Triangle every vertice needs to have exactly the same weight.

To show this we can look from a single vertice in the direction of the center and realize that the other two vertices would need to have the same weight as otherwise the triangle would tilt somewhere in the half of the heavier one (depending on the weight of the vertice you look from).

If we name the vertices as A,B,C and look from vertice A that would mean that vertice B and C would need to be equal.

If we now look from one of the other two vertices we can make the same deduction and thereby can see that all three vertices must have the same weight for the triangle to stay in balance.

In order to balance a weight on one vertice that would mean that we can simply use the same strategy as in the original question.

However instead of looking at a single empty vertice we now have two empty vertices which means that we need to place the same weights on each of these two vertices.

therefore we need 2 sets of weights of order $3^n$.

For the tetrahedron we can use the same general idea.

If we place a single vertice at the bottom we can again see, that all 3 vertices on top need to have exactly the same weight (as long as I understand the question correctly and don't make a deduction based on misunderstood logic).

Next place one of the other vertices at the bottom and we see that all vertices must have the same weight again.

So for the tetrahedron we again use the same strategy only this time with 3 sets of weights of order $3^n$.

Lastly we look at the octahedron.

As an octahedron has 6 vertices consisting of 3 pairs of vertices on exact opposite sides of the center we only need to balance each vertice with its opposite one.

So we use the original strategy 3 times for a total of 3 sets of order $3^n$.

Edit:

I just thought about it again and realised that i made a mistake with the triangle (not in the logic but in the number of sets required).

We effectively need to use our weights to make every digit of our base 3 weight numbers equal.

So if we have our 3 weights and $mod 3$ them we would need to distribute our 1-weights on the 3 vertices to get the same result on each of them.

After that we take $mod 9$ and distribute our 3-weights and so on until we balanced all 3 vertices.

We reach a problem with the triangle when we have a $mod 3^n$ result of 0,1 and 2 as it is impossible to make them equal with only 2 sets of weights.

Therefore we will need 3 sets of weights for the triangle as well.

Triangle case:
If the triangle has three weights, we can subtract the smallest value from each side. That reduces to the case of only two unknown weights and a zero in the third pan.

The original problem defines the maximum weight as 40. Rather than examine this exact case, I assume an arbitrarily large MAX weight and solved for that. This will give an optimal solution for arbitrarily large MAX, but for specific values (such as 40) it may merely give a very good solution rather than an optimal one.

There's a particularly good solution I want to mention, which doesn't quite fit the rules. You can balance the two unknown weights using a single set of 3^n weights as in the solution for the original problem, then fill the empty pan using a single set of 2^n weights. This efficiently uses fewer weights in total, but requires two differing sets of weights.

To solve this problem we are going to be working in some base x, and assume each weight is a number written in base x. When solving any particular case we will work on it one digit at a time, starting from the smallest digit. If we can balance the triangle for the smallest digit, then we simply repeat the same process to solve each larger digit. Note that because the process is the same for each digit, we only need to ensure that our process works for the smallest digit. If it works for one digit it will work for every digit.

This means that for x=2, we only need to examine the cases where the weights are 0,0 0,1 1,0 and 1,1. For X=3 we only need to examine the cases where the weights are 0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 and 2,2. Similarly for x=4 we only need to examine the 16 cases with weights ranging up to 3,3. If that works we can solve the lowest digit and the process can repeat to solve every digit.

When x=2 the worst case is when the unknown weights are 0,1. We can add 1 to the second pan - this changes it into a 0 and a carry into the next digit. This results in 0,0,0 for the lowest digit in each pan. (We don't care about the carry, it gets absorbed into the next digit and solved when we handle that digit.) We only need 1 of each size weight.

When x=3 the worst case is when the unknown weights are 1,2. We will need 2,2,2 to balance the triangle, meaning we will need 3 of each size weight.

When x=4 the worst case is when the unknown weights are 1,3. The most efficient solution is to add 3 to the first pan and 1 to the second pan. This means we'll have 0,0,0 as the lowest digit in each pan, and the weights generate a carry into the next digit. This means we need 4 of each size weight.

When x=5 the worst case is when the unknown weights are 1,3. The most efficient solution is to add 2 to the first pan and 3 to the empty pan. This means we need 5 of each size weight. And things just get worse for larger x.

So which is the most efficient x?

For x=2 we will need 1 weight of each size, times LogBase2(MAX) sized-weights.

For x=3 we will need 3 weights of each size, times LogBase3(MAX) sized-weights.

For x=4 we will need 4 weights of each size, times LogBase4(MAX) sized-weights.

For x=5 we will need 5 weights of each size, times LogBase5(MAX) sized-weights.

For an arbitrarily large MAX value, lower is better:

x=2 has an efficiency of 1/Ln(2)=1.4427

x=3 has an efficiency of 3/Ln(3)=2.7307

x=4 has an efficiency of 4/Ln(4)=2.8854

x=5 has an efficiency of 5/Ln(4)=3.1067

For arbitrarily large MAX weight best solution is x=2, with one copy of each weight. I have not checked the tetrahedron yet but the reasoning will be the same. You'll have three unknown weights, and you only need to find how many unit-weights are needed to solve a single digit in the worst case.