Turning to Swammerdami's problem, here are all the possible sets of point distances, to within point relabelings:
12 / 13 14 23 24 34
12 13 / 14 23 24 34
12 34 / 13 14 23 24
12 13 14 / 23 24 34
12 13 24 / 14 23 34
I checked that these are the only possible splits using Mathematica, but it should be easy with a program in (say) Python. Some of its standard library is convenient for that, like a permutation generator in itertools and the set() function for unique-member lists
For the first one, triangles 134 and 234 are both equilateral triangles, and they share side 34. Segment 12 thus has length sqrt(3) relative to the others.
For the second one, 234 is an equilateral triangle, and point 1 a little beyond the center of side 23. 12 and 13 have length (sqrt(3)-1)/sqrt(2) times the lengths of the others.
For the third one, 1324 is a square, and 12 and 34 are its diagonals, with length sqrt(2) relative to the others.
For the fourth one, 234 is an equilateral triangle, and 1 is its center, That makes 12 have length 1/sqrt(3) times 23's length.
For the fifth one, the points are in two sequences with equal-length segments: 3124 and 1432. One set of segments will be longer than the other set, with a length ratio (sqrt(5)+1)/2 or the reciprocal of (sqrt(5)-1)/2.
Is this a trick question? I think it's true for ALL integers. This is trivially true when n is a multiple of 6. Step through the other five cases by adding 1 (none of the expressions changes value), then 2 (one increments on the left, another on the right), then 3 (ditto), 4 (ditto), 5 (no change).
Turning to Swammerdami's problem, here are all the possible sets of point distances, to within point relabelings:
...
I checked that these are the only possible splits using Mathematica, but it should be easy with a program in (say) Python. Some of its standard library is convenient for that, like a permutation generator in itertools and the set() function for unique-member lists
The list you defined may be exhaustive, but you've still missed a solution. (I won't say more, and anyway I always prefer solutions in spoilers ... though there's a fair chance lpetrich will be the only contestant. )
Sometimes it's FUN to use Mathematica or Python as a solving tool, but figuring by hand with paper and pencil is also fun and in some cases (including this one) may be much EASIER.
IIUC, the list you developed via Mathematica was a trivial thing that could have been done readily by hand. AND developing that list was an obstacle anyway; did I mention that you missed one of the SIX solutions?
I think at least five of the six solutions can be found readily by hand. Doing so should bring some JOY. ... And help toward the Extra Credit problem ("An interesting comment about difficulty can be applied to the solution").
I used Mathematica because I wasn't sure of my reasoning. I'll go through my reasoning. For points 1, 2, 3, 4: segments 12, 13, 14, 23, 24, 34.
One segment: 12
Two segments: connected to one of 1, 2 or connected to none of them:
12 13
12 23
Three segments: connected to 1 of the first one or 2 of the second one, making the segments radiate out from one point. Or else connected to 2 or 3 of the first one or 1 or 3 of the second one, making a strip. Or else connecting the two that show up only once:
12 13 14 -- radiating from 1
12 13 24 -- strip 4213
12 13 23 -- triangle 123
Let us look at the remaining segments for each one:
12 13 14 / 23 24 34 -- radiation from 1 + triangle
12 13 24 / 14 23 34 -- two strips: 4213, 1432
12 13 23 / 14 24 34 -- triangle + radiation from 4
So the sixth solution is a version of one of the five ones.
Mathematica would be useful for a similar but slightly more complicated puzzle. But for this one, the (3;3) partition is most "difficult" but is readily understood: if there's no equilateral triangle then 3 of the distances must be three sides of a quadrilateral, and the remaining three distances will ALSO be three such sides.
Obviously. The punchline to the puzzle depends on which of the six solutions you find most difficult to find. You might consider the final solution you find to be most difficult, while the "punchline" depends on the hardest-to-find being one you've already found. (Guess which one I mean?)
Two related puzzles:
(a) FIVE points in the plane, with only two distinct point-to-point distances. (b) Five points in the plane, with only three distinct point-to-point distances.
I am NOT masochistic enough to attempt (b). I might add it to my already-overlong To-Do list.
ETA: In fact, unless you have a VERY clever program to let a machine do the work, (b) might be ridiculously cumbersome. In an effort to make amends I've crossed it out.
Since there are 5 points, there are 5*(5-1)/2 = 10 line segments connecting them. That means that we must consider sets of 1, 2, 3, 4, and 5 segments.
I used Mathematica to sort out the possibilities. Here are all the distinct network topologies for each number of segments and number of segments in its complement (total segments - those segments). When both a set of segments and its complement have the same number, then we have (number of all that have distinct flipped versions, number of all with the same flipped version)
2: 1
0 - 1: 1
3: 3
0 - 3: 1
1 - 2: 1
4: 6
0 - 6: 1
1 - 5: 1
2 - 4: 2
3 - 3: 3 (1,1)
5: 10
0 - 10: 1
1 - 9: 1
2 - 8: 2
3 - 7: 4
4 - 6: 6
5 - 5: 6 (2,2)
For 5 points, many of the topologies can easily be ruled out because of the impossibility of creating equilateral triangles with distinct vertices that satisfy that. That rules out all those that contain squares with both diagonals and squares with center points connected to 3 or 4 vertices. Those are all of (0,10), (1,9), and (2,8), and this criterion leaves 1 of (3,7), 4 of (4,6), and all of (5,5).
Of these, all but one of them can be ruled out by taking the complement segments and finding the possible lengths of the original segments.
The remaining one is a pentagonal topology for both the original segments and their complement.
This leads to the only solution: a regular pentagon and a pentagram.
Instead of 5 points with 3 different lengths, I will try 4 points with 3 different lengths. The 3-point case is trivial.
Let us first see how the edges divide up:
1 - 1 - 4
1 - 2 - 3
2 - 2 - 2
There are solutions for all three cases, solutions with one free parameter.
I've considered 6 points. For two segment lengths, there are no solutions. For three, there is at least one: a regular hexagon with a hexagram and the three diagonals.
For an n-gon, this kind of solution will always exist, and will contain floor(n/2) different segment lengths.
Still waiting for someone to list the six solutions to the (4,2) problem — 4 points in the plane which have only 2 unique distances — and explain what's "paradoxical" about the difficulty of finding the (allegedly) most difficult-to-find of the six.
...
Two related puzzles:
(a) FIVE points in the plane, with only two distinct point-to-point distances. (b) Five points in the plane, with only three distinct point-to-point distances.
I deleted the suggestion about the (5,3) investigation for fear of being arraigned by the Society for the Prevention of Cruelty by Sadists. However masochism is not a crime, and I have pursued the problem a bit.
I will list some of the (5,3) solutions. The list will not be exhaustive. Would it be a fun contest to see who can find more solutions?
I wrote a program with the same goal as lpetrich's: to find all classes of distance assignments. There are 142 classes, but 108 of those derive simply: start with one of the six (4,2) solutions, and add a 5th point. I'll refer to those 108 classes as "derived classes."
This leaves 34 classes for which any 4-sized subset uses three distances. I'll refer to these as "non-derived classes."
Note that a class, whether "derived" or "non-derived," may or may not lead to a satisfactory geometry, and might lead to two or more such geometries.
Let's start with some of the "derived" solutions.
From the square (a (4,2) solution with distances 1 and 1.414), a 5th point can be added in five places, all along a lateral bisector. An isosceles triangle can be added to a side, either pointing in or pointing out, and with side either 1 or 1.414. This is four solutions; a fifth solution is to place the 5th point in the center of the square. Is that it for the square?
Another (4,2) solution is the four-fifths of a pentagon with distances 1 and 1.618. Now there seem to be eight derived solutions: add an isosceles triangle to either of the two parallel sides, pointing either out or in, and with side either 1 or 1.618. However these eight solutions are just seven because the (5,2) solution -- the perfect 5-star -- is derived with EITHER an in-pointing 1.618 isosceles triangle from the short side OR an out-pointing 1.0 isosceles triangle from the long side.
The other four (4,2) solutions are described by an equilateral triangle and a 4th point on its bisector. For these cases I've given up (temporarily?) after deriving just one (5,3) solution: Add a 4th point in the center of the triangle and a 5th point that forms a triangle congruent to the little triangles just formed.
So far we're up to 13 solutions total. Now let's look at "non-derived" solutions.
I've listed the 34 non-derived cases in a Spoiler; I describe the notation with this example (first in the list): . . . . . . . . AAAA // ABC // BA // C . #Equ= 2 #Iso= 6
Reading left to right, point #5 has distance A to points #1,#2,#3,#4; point #4 has distances A,B,C to points #1,#2,#3 respectively; point #3 has distances B,A to points #1,#2 and the distance from #2 to #1 is C. (The software reports, redundantly, that the case has 2 equilateral triangles and 6 isosceles triangles. (Equilaterals are included in the Isosceles count.)
If you think about this with a pencil, you find one solution: a 1-by-1.732 rectangle with its center. (1.732 is the square root of 3.)
Here's the 2nd non-derived class . . . . . . . . AAAA // ABC // BB // C . #Equ= 1 #Iso= 8
Again, the AAAA to start means point #5 is the center of a circle, with the other 4 points somewhere on the circumference. Think of angles rather than distances; the central angle (call it t) for points 1-3 equals the angle for 3-2 which equals the angle for 2-4; and the 4-1 angle is 60 degrees. This means t is either 100 degrees or 140 degrees; either leads to solution.
I've not looked at the other 32 non-derived cases. Any other masochists out there?
AAAA // ABC // BA // C . #Equ= 2 #Iso= 6
AAAA // ABC // BB // C . #Equ= 1 #Iso= 8
AAAB // AAC // AB // C . #Equ= 1 #Iso= 7
AAAB // AAC // BB // C . #Equ= 0 #Iso= 7
AAAB // AAC // BC // C . #Equ= 0 #Iso= 8
AAAB // ABC // AB // C . #Equ= 1 #Iso= 7
AAAB // ABC // BA // C . #Equ= 1 #Iso= 5
AAAB // ABC // BB // C . #Equ= 0 #Iso= 7
AAAB // ABC // BC // C . #Equ= 0 #Iso= 7
AAAB // ABC // CB // C . #Equ= 0 #Iso= 8
AAAB // ACC // AC // B . #Equ= 2 #Iso= 6
AAAB // ACC // BA // C . #Equ= 1 #Iso= 6
AAAB // ACC // BB // C . #Equ= 0 #Iso= 7
AAAB // ACC // BC // B . #Equ= 1 #Iso= 6
AAAB // ACC // BC // C . #Equ= 1 #Iso= 7
AAAB // ACC // CB // C . #Equ= 0 #Iso= 8
AAAB // BCC // AB // C . #Equ= 1 #Iso= 6
AAAB // BCC // AC // B . #Equ= 2 #Iso= 6
AAAB // BCC // BA // C . #Equ= 1 #Iso= 7
AAAB // CCC // AB // C . #Equ= 2 #Iso= 6
AABB // AAB // CC // C . #Equ= 2 #Iso= 6
AABB // AAC // BC // C . #Equ= 0 #Iso= 8
AABB // ABA // CC // C . #Equ= 1 #Iso= 6
AABB // ABC // BA // C . #Equ= 0 #Iso= 6
AABB // ABC // BC // C . #Equ= 0 #Iso= 7
AABB // ACA // CA // B . #Equ= 0 #Iso= 6
AABB // ACA // CB // B . #Equ= 0 #Iso= 6
AABB // ACA // CC // B . #Equ= 0 #Iso= 6
AABB // ACB // CA // B . #Equ= 1 #Iso= 4
AABB // ACB // CC // B . #Equ= 1 #Iso= 5
AABB // ACC // CA // B . #Equ= 0 #Iso= 6
AABB // CCA // CC // B . #Equ= 0 #Iso= 6
AABC // ABA // BC // C . #Equ= 0 #Iso= 5
AABC // BBA // CC // A . #Equ= 1 #Iso= 3
12 points on a 7-pointed star shape (heptagram), making both 4 equilateral triangles and 3 squares. The heptagram has each vertex connected to those nearest the opposite side from it.
The first step is constructing that heptagram. For that purpose, we need to consider constructing a regular polygon. For a regular n-gon, the construction involves trigonometric functions of the angle (pi/n) or (180d/n).
That may seem very difficult, but trigonometric identities yield polynomial equations. Like cos(2x) = cos(x)2 - sin(x)2 and sin(2x) = 2*cos(x)*sin(x) along with cos(x)2 + sin(x)2 = 1.
In general, for nonnegative integer n, cos(n*x) = T(n,cos(x)) and sin(n*x) = U(n-1,cos(x))*sin(x) where T and U are the Chebyshev polynomials of the first and second kinds. These definitions give a variety of features of them -- recurrence relations, differential equations, special values, generating functions, etc. -- including having integer coefficients.
So T(n,cos(pi/n)) = cos(pi) = -1, so one has a polynomial equation to solve for cos(pi/n). At first sight it seems very difficult to do this solution, but there are a variety of simplifications that one can use.
Before continuing, note that this equation's solutions are not unique. All the solutions can be expressed as cos((2k+1)*pi/n), for k = 1 to (n-1), though that means that if one has cos(pi/n), one can find all the others using Chebyshev polynomials of it. Furthermore, replacing k with (n-k-1) gives the same solution, and if n is odd, then k = (n-1)/2 gives -1. This means that for n even, this becomes the square of an equation with degree (n/2), and for n odd, the square of an equation with degree ((n-1)/2) multiplied by (variable + 1). So the effective degree of this polynomial equation is reduced by a factor of 2.
For n even, one can use the half-angle formulas to get cos(pi/(2n)) in terms of cos(pi/n) -- x becomes sqrt((x+1)/2). Likewise, for n a multiple of 3, one can do something similar with a cubic equation, though that is much more complicated.
So let us consider n odd. This degree reduction explains why cos(pi/3) = 1/2 with no square root and cos(pi/5) = (sqrt(5)+1)/4 and not some more complicated equation. Because (3-1)/2 = 1 and (5-1)/2.
For a heptagon, one thus has a cubic equation to solve, and that is very complicated.
for some integers k, where n has prime factors p with powers m. So we must consider prime powers.
One can do prime powers by chaining degree-p polynomial equations, but for power 1, one gets some simplifications. In particular, if the prime is a Fermat prime, one can solve with a chain of quadratic equations, and if a Pierpoint prime, with a chain of quadratic and cubic equations.
The Fermat numbers F(n) = 22^n + 1. The only known prime ones are for n = 0, 1, 2, 3, 4 -- 3, 5, 17, 257, 65535
For n = 5 to 32, and also for some larger values of n, the Fermat numbers are known to be composite, and there are some handwaving arguments from the frequency of prime numbers that suggest that every other Fermat number is composite. But there is still no proof that the five known Fermat primes are the only ones, or else that there are any others.
The Pierpoint numbers P(n,m) = 2n*3m + 1 for nonnegative integer n, m
I watched that video, and the Mathologer pointed out that the points being on a heptagon was something of a cheat. They are on a hypotrochoid, a kind of Spirograph pattern, for a circle rolling inside another circle. The other kind, an epitrochoid, is a circle rolling outside of another circle.
For independent variable t, frequencies m and n, initial phases q and p, and relative size a: coordiantes x and y are:
x = cos(n*t+p) + a*cos(m*t+q)
y = sin(n*t+p) + a*sin(m*t+q)
To get an approximate heptagram, one uses n = 3, m = -4, a = 3/4.
One places the points with parameter t = t0 + (2pi)*k/12 where k is 0 to 11.
The triangles are for points 0,4,8 - 1,5,9 - 2,6,10 - 3,7,11
The squares are for points 0,3,6,9 - 1,4,7,10 - 2,5,8,11
I've verified with Mathematica that this produces the right shapes.
Still waiting for someone to list the six solutions to the (4,2) problem — 4 points in the plane which have only 2 unique distances — and explain what's "paradoxical" about the difficulty of finding the (allegedly) most difficult-to-find of the six.
An alternate arrangement of the points in solution 2, with point 1 reflected across point 4, so that edges 12 and 13 are longer than the others instead of shorter.
As for which is most difficult, that's in the eye of the beholder. I didn't attack it systematically but just tried to visualize solutions, and only came up with four. I missed lpetrich's solutions to 2 and 5, so obviously those are the most difficult. Not seeing a paradox on either, though, just a failure of imagination.
The (5,2) problem has only one solution -- the vertices of the regular pentagon. It is obvious at once that it solves; and I think easy to find. (Proving it to be ONLY solution may be more difficult.)
Obviously removing any point from this (5,2) solution immediately yields a (4,2) solution. Since the (5,2) is easy to find and gives a(4,2) at once, should that (4,2) be easy to find? But in fact that (4,2) -- 4/5 of a star is HARDEST to find!
Paradoxical? Sort of. Sometimes the HARDER version of a math problem is easier to work with. (I'm trying to write this on my phone but will give one ...) Example: Didn't Gauss solve FLT N=3 elegantly by solving a harder complex case?
Here's another Michael Penn problem. Solve for x, y, z in
\( \displaystyle{ \frac{1}{x} + y + z = 3 } \\ \displaystyle{ x + \frac{1}{y} + z = 3 } \\ \displaystyle{ x + y + \frac{1}{z} = 3 } \)
Here is a nice LaTeX preview service:
QuickLaTeX - "QuickLaTeX.com is a free web service aimed to help people to include mathematical formulas&graphics into the web pages easily and without sacrificing flexibility and high quality of LaTeX system."
Subtract one equation from another and rearrange to get
\( \displaystyle{ x - y = \frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy}} \)
This has two solutions:
\( \displaystyle{ x = y } \\ \displaystyle{ x = \frac{-1}{y} } \)
Similar deductions relate x to z and y to z. This leads to four solutions:
\( \displaystyle{ (x,y,z) = (1,1,1) } \\ \displaystyle{ (x,y,z) = (3,3,\frac{-1}{3}) } \\ \displaystyle{ (x,y,z) = (3,\frac{-1}{3},3) } \\ \displaystyle{ (x,y,z) = (\frac{-1}{3},3,3) }\)
These are some of the solutions, but not all of them. Let us rearrange the equations and combine their terms:
\( \frac{1}{x} (x y + x z - 3 x + 1) = 0 \\ \frac{1}{y} (x y + y z - 3 y + 1) = 0 \\ \frac{1}{z} (x z + y z - 3 z + 1) = 0 \)
These are three equations in three variables, so we try to eliminate all but one of the variables. There is a trick for doing so, a trick that is an expansion on Euclid's algorithm. Find the greatest common denominator of positive integers 75 and 840, and do that without factoring either number.
Make a sequence of numbers starting with those as the first two, then each new sequence member is the remainder of dividing the second one before it by the first one.
75
840
75 - 0*840 = 75
840 - 11*75 = 15
75 - 5*15 = 0
So the two numbers' gcd is 15, which can easily be verified: 75/15 = 5, 840/15 = 56.
One can do the same for two polynomials and some variable, where one uses remainders after dividing by that variable. I like to take the numerators after each division, because the denominators are not relevant to the results. Related to this is the "resultant" of two polynomials by that variable.
Equations 1 and 2 by z:
\( -((x-y) (x y+1)) \)
(equals the resultant)
Equations 1 and 3 by z:
\( (y-3) \left(x^2+x y-3 x+1\right) \)
(equals the resultant)
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