Place 15 different positive integers on the vertices of this graph so that the ten products of three numbers in a straight line are all equal.
There are a number of ways to do this. An easy strategy:
To turn this into a puzzle about addition instead of multiplication, we can solve the addition version of the puzzle and take (for example) 4 to the power of everything.
So here, I put a magic square in the middle 3x3, and then just pick numbers for the left and right corners so that the remaining four are all different. (This is why I chose 4 instead of 2 as a base: this way, I can use halves, ensuring no conflict between the wings and the center 3x3.)
The additive solution I found:
And to turn this into a multiplicative solution, all you have to do is raise 4 to the power of each of the circles.
Here is a solution with a constant product of
which I think is the minimum possible:
Some partial progress for a lower bound on the product:
First of all, we need a number with at least $15$ divisors: so, nothing less than $120$ will work. But actually, $15$ is not enough. One of the divisors of $n$ is $n$ itself, and that can only be in a triple with $1$ and $1$, which are not distinct, so we can't use $n$. If $p$ divides $n$, then $n/p$ can only be in a triple with $1$ and $p$; if $p^2$ divides $n$, the $n/p^2$ can only be in a triple with $1$ and $p^2$. Such numbers must be the middles of the sides of a triangle, and since they can only be in a triple with $1$, at most two of them can be used. So we need at least $16$ divisors, plus one for every prime factor past the first two, plus one for every prime square factor past the first two.
This leaves only a few possibilities for improvement:
Of the integers less than $360$, only $240$, $288$, and $336$ pass this test. (Note that if $336$ were possible, then $240$ would be, too: just replace all factors of $7$ with factors of $5$.)
Edit: In an effort to find the minumum, here is a much smaller solution in which the mutual product is
As MKBakker pointed out we could further reduce this by dividing each of the entries 4,8,16,96 and 192 by 2 to get a mutual product of
although they have subsequently improved on this.
I found a solution trying to minimize prime factors. And finding a balance between the minimum value and minimum number of factors.
The product of each three numbers is 720
I noticed that some fields are connected, in that multiplying one of them results in the multiplication of fixed other fields. There are three such patterns:
-C, D, H, L, and M (and any mirror image of that)
-A, E, G, L, M (and any mirror image of that)
-A special case is A,B,C, where multiplication of A can be compensated by division of both B and C (Again, also works for the opposite side and the inverse)
Using this, I was able to assign the minimum number of factors required for a unique solution. I think my solution is the lowest possible, but I'm not 100% confident. Perhaps clever rearrangement could lead to another prime factor being reduced.
Now it's time for weekend!