Easy peasy:
0
(Confused? Just simplify the last three items in the polynomial)
The following is more of a Play With Dots problem than a Brain Teaser, but I may as well bump the thread.
Describe all the ways to place four points in the plane such that only two distinct values occur among the inter-point distances. (That is, the six different distances will have values like {a, a, a, a, b, b}.)
(There’s a particular reason why I find this puzzle interesting, but I’ll wait until any Play-with-Dotters have had their play.)
There are exactly four non-degenerate ones:
A square
A 60° rhombus
An equilateral triangle with a fourth point in the center
An equilateral triangle with a fourth point “above” one vertex by one side-length
There are arguably three more, and they’re probably why you find the puzzle interesting, but their validity is debatable:
Three points superimposed on each other, with one anywhere else
Two pairs of points superimposed on each other
An equilateral triangle with a fourth point superimposed on one vertex
ignore this post
No one one has answered this yet.
But I probably should amend the riddle to say that last year *many people would have said *that the word I am thinking of did not share the common factor. Certain people might have argued that it did. In any case, nobody can now deny the commonality.
I say there’s five, and that in your last case, you have forgotten its non-isomorphic partner:
A square
A 60° rhombus (i.e., two equilateral triangles joined base-to-base)
An equilateral triangle with a fourth point in the center
An equilateral triangle with a fourth point “above” one vertex by one side-length (where “above” denotes the direction from the midpoint of the opposite side to the chosen vertex)
An equilateral triangle with a fourth point “below” one vertex by one side-length (where “below” of course denotes the opposite direction from “above”)
I say there’s SIX solutions, the five Indistinguishable mentions and one other. What I find interesting about the puzzle is something related to the difficulty of noticing the sixth solution.
Ah, I should say, I was trusting Chronos to have generally carried out a correct enumeration, and did not actually think about the problem from scratch or devise a proof. There is at least one more configuration, as you note:
Take the points to be any four among the five corners of a regular pentagon
I haven’t actually proven this is a complete enumeration either, but now I will switch to trusting septimus instead of Chronos. 
It is interesting that was psychologically so difficult to summon. Particularly when
If you had asked us to place 5 points rather than 4, it would surely have jumped out as the very first thought!
Precisely why I find the puzzle to be an excellent example of an interesting theme.
OK, not looking at those spoiler boxes yet. I’ve got to find a way to think about this more systematically. Fortunately I’ll have a fair bit of time for that in the next couple of days.
For what it’s worth, the systematic thing I was too lazy to do but trusted Chronos and then septimus to do was this:
Enumerate every 2-coloring of the complete graph on 4-vertices. I originally trusted that Chronos had done this and found that each one would have either a monochromatic 3-cycle (thus, an equilateral triangle, and it is easy to show geometrically that the only solutions for these are as noted) or a monochromatic 4-cycle (thus, a rhombus [or an isosceles triange with one point doubled, but that won’t do], and it is easy to show geometrically that the only solutions of this form are either the square or the 60 degree rhombus (also containing an equilateral triangle), as noted].
Now I trust instead that septimus has done this and found that each one contains either a monochromatic cycle OR is isomorphic to the graph consisting of two complementary monochromatic Hamiltonian paths of length three [which can be shown geometrically to correspond to 4 corners of a regular pentagon].
I still haven’t done this enumeration myself, mind you. It’s only tens of cases, but I’m lazy…
Proof that there are no solutions beyond the six shapes (and Chronos’ 4 degenerate cases) is easier than this, especially since geometry has to be considered anyway on top of a graph-coloring approach.
[SPOILER]Either an equilateral triangle is present or it isn’t. In the former case, the 4th point must be equidistant to two other points, so lies on a bisector of the triangle. From here it is easy to find all solutions.
If no equilaternal triangle is present, all four triangles will be isoceles. Wlog the points can be relabeled so that AB = BC = CD. In other words, BC forms the base of two similar triangles, with the remaining length AD equal to either AB or AC. Straightforward exhaustion finds the remaining two shapes. (Actually, you may find four cases: square and partial pentagon can each be derived two ways from the ABCD formation.[/SPOILER]
Great! That approach isn’t different to the one I was noting; it can be thought of in precisely the terms I set out:
Either a 2-coloring of the complete graph on 4 vertices contains a monochromatic 3-cycle or it doesn’t. In the latter case, all four 3-cycles will contain a 2-1 color split. WLOG, the vertices can be relabelled so that the edges AB, BC, and CD have the same color (which is different from BD’s since we don’t have a monochromatic 3-cycle)), with the edge AD equal either to AB (so that we have a monochromatic 4-cycle) or not to AB but to AC (in which case, our coloring consists of two complementary monochromatic Hamiltonian paths).
For what it’s worth, though I was only proposing enumerating colorings at the above level of granularity originally before switching to geometry, turns out the full enumeration of colorings up to isomorphism isn’t so hard anyway; I just wasn’t thinking about it appropriately:
Since there are 6 edges in the complete graph on 4 vertices, every way of splitting these into at most 2 sets is either a 0-6, 1-5, 2-4, or 3-3 split.
[ul]
[li]A 0-6 split means a regular tetrahedron; we can’t get this to lie flat, except degenerately by having all 4 points coincide. (Even then, we get only one distinct distance, not two, if that is our goal)[/li]
[li]A 1-5 split means just 1 odd edge; there’s nothing more to say about the coloring. We must have two equilateral triangles joined base-to-base; making this lie flat, we get [LIST][/li][li]a 60 degree rhombus, or [/li]li an equilateral triangle with one point doubled up.[/li][/ul]
[li]For a 2-4 split, either the 2 odd edges meet at a vertex or they don’t. [ul][/li][li]If the 2 odd edges meet, we have an equilateral triangle, plus attached to the apex of this triangle we have another point another side-length away, equidistant from the other two corners; thus, making this lie flat, we have [LIST][/li][li]an equilateral triangle with extra point one side-length “above” the apex, or[/li][li]an equilateral triangle with extra point one side-length “below” the apex.[/li][/ul]
[li]On the other hand, if the 2 odd edges don’t meet, we have a loop of 4 edges of the same length; thus, two isosceles triangles joined base-to-base. Making this lie flat, we get a rhombus, or an isosceles triangle with odd point doubled up. Ensuring that the 2 odd edges have the same length, these amount to [ul][/li][li]a rhombus with equal diagonals (i.e., a square), or[/li]li an isosceles triangle with odd point doubled up, and odd side of length zero (i.e., a line segment with both ends doubled up)[/li][/ul]
[/LIST]
[li]Finally, for a 3-3 split, we should note that any 3 edges of the complete graph on 4 vertices either all meet at a common point, or comprise a 3-cycle, or comprise a Hamiltonian path of length 3. The first two of these complement each other in a split, while the latter complements itself. These correspond to either [ul][/li][li]An equilateral triangle with 4th point equidistant from all its corners. Thus, a pyramid with equilateral triangle base. Making this lie flat, we can have either [LIST][/li][li]an equilaterial triangle with 4th point in its center, or[/li]li a line segment with one end tripled up[/li][/ul]
[li]Or, finally, we may have three line segments back-to-back AB, BC, and CD of the same length as each other, with CA, AD, and DB also all of the same length as each other. In 3d, this can be done in infinitely many ways: specifically, thinking of our points as given by the three vectors u, v, and w such that B = A + u, C = B + v, and D = C + w, our conditions are that u, v, and w all have the same length, while u + v, v + w, and u + v + w all have another same length. What remains is only to know the angle between these vectors, or just as well, the cosines of the angles between these vectors; in 3d, we have a bevy of choices. We may pick a value e whose square does not exceed 5, then take the two angles x and y whose cosines are (-1 plus or minus e)/4, and take the angle from u to v and from v to w to be x while the angle from u to w is y. But the only way to get this to lie flat is by making e maximally large, resulting in[ul][/li][li]four corners of a regular pentagon[/li][/ul]
[/LIST]
[/LIST]
and so now we’re done. Hooray!
Alright, to keep the flow of puzzles from dying off, here’s one that’s an oldie but a goodie: come up with two nonstandard dice (6-sided objects with, let’s say, a positive whole number of dots on each side) such that, when you roll them and take the sum, the chances of any particular outcome are exactly the same as with standard dice.
Is it required that the faces of each die occur with equal probability? Either way, did you mean 6-sided dice, or N-sided for N of your choice? In the latter case, note that
If zeros are permitted on a face, a single 6-sided die can be emulated as the sum of a 2-sided die (i.e. a coin!) and a 3-sided die. Thus the sum of two ordinary dice can be emulated as the sum of 2-, 2-, 3- and 3-sided dice. Combine these four dice in a different way to get emulation as the sum of 4- and 9-sided dice.
I’m assuming you’re looking for the numbers themselves to change, as opposed to just the positioning of them, yes?
With standard dice, the opposing faces always add up to seven. (Six and one on opposite sides, three and four, etc…) Meaning I could make nonstandard dice by having, say, six and two on opposite faces instead of six and one.
But that’s not where you’re going with that, right?
Need the six-sided objects be cubes?
Right, sorry, to clarify: I’m looking for two cubes (thus, each is 6-sided), each with each side equiprobable, which are nonstandard by more than just changing the positions of the values of standard dice (that is, the dice do not both have each of the values 1 through 6 once).