A magic hexagon of order n is an arrangement of close-packed hexagons containing the numbers 1, 2,…, H(n-1)$, where Hn is the nth hex number. The magic hexagon can help us remember that by going clockwise around any of these three triangles.
The magic hexagon can be used to find the sine, cosine, and tangent of angles by finding the point on the hexagon that corresponds to the angle. The lines from the point to the center of the hexagon can also be found using trigonometric functions sine, cosine, and tangent.
A normal magic hexagon contains the consecutive integers from 1 to 3n2 − 3n + 1, while an abnormal magic hexagon starts with a number other than 1. The solutions to the Magic Hexagon Problem use appropriate systems of linear algebra.
A unique solution has been discovered independently several times, with all rows and columns adding up to 38. A normal magic hexagon contains the consecutive integers from 1 to 3n² – 3n + 1, whereas an abnormal one begins with a number other than one.
In conclusion, trigonometric identities are equalities that include trigonometric functions and are true for every value of variables that occur for which both sides of equalities are defined. A magic hexagon of order n can help us remember that by going clockwise around any of these three triangles, we can find the magic number of the hexagon.
📹 Magic Hexagon – Numberphile
Dr James Grime talking Magic Hexagons (and magic squares). More links & stuff in full description below ↓↓↓ Support us on …
📹 The Magic Hexagon!
Today’s math puzzle is about magic hexagons! New Puzzles every Sunday and Thursday. I post math puzzles and their solutions …
Thank you Brady:) It’s always great hearing Dr Grime talk about math. I did, however, notice a distinct lack of prime numbers in this article, and was wondering if there were any interesting mathematical things going on with geometric shapes that have a prime number of sides. I find it hard to imagine that there isn’t.
In the article you never mentioned that the Magic Hexagon must be made of consecutive numbers. Since you can just multiply all of the Numbers in the Hexagon shown in this article by 2 and get a new Hexagon that Works. (MAGIC NUMBER: 76) If you want a Magic Hexagon in it’s simplest form, you can take the Hexagon shown in the article and Add 8 to the Pink, then Add 16 to the Blue and Center. This will give you a new Magic Hexagon in it’s simplest form. (MAGIC NUMBER: 70)
For a moment, when James started adding the numbers 1 + 2 + … + nn, I was lost. But then I remembered that a magic square must be made of all the numbers up to and including nn. Also, are there any other magic hexagons if we remove the constraint that the numbers have to be from the set {1, 2, 3, …, 3n(n -1) + 1} only? Like, could we use distinct whole numbers not necessarily from that set to fill out the honeycomb, and still get this effect? It seems to me that that problem would have a much more involved and non-elementary solution than this one.
At the end, it said that that were 120 possible combinations that were rejected, with one solution. Those add up to 121 with is 11^2. The only other magic hexagon is the 1 hexagon, with has only one combination, which is 1^2. Is there any pattern there with hexagons of different sizes where the total number of combinations is a square number?
Floyd’s triangle is easy enough for anyone to understand, and the numbers on the left edge of the triangle are the lazy caterer’s sequence, which kind of looks awful, with tiny itsy bitsy pieces of the pizza and the numbers along the right edge are triangular numbers and the nth row sums to the constant of an n x n magic square
I found a mistake! 🙂 While only hexagons of 1 and 3 work the math done to prove this was incorrect. M = ( (3n^2 – 3n + 2) • ((3n^2 – 3n + 1) / 2) ) / (2n – 1) This is what is originally shown, when the polynomials are divided out you get this: M = (9/4)n^3 – (27/8)n^2 + (45/16)n + (27/32) + (5/(4n – 2)) This equation is not shown but instead the fractions are removed by multiplying both sides by 32 to get this: 32M = 72n^3 – 108n^2 + 90n + 27 + (80/(2n – 1)) In the article 5 is shown instead of 80 in the numerator of the last fraction. The answer of only hexagon sizes 1 and 3 still works because they are they only whole numbers that can be plugged in for n and result in a whole number for M. (Please let me know if I am right or missing something.) -Aspiring high school mathematician.
I am missing something. I see at 12:24 that P=57+y/2 that number is smaller than when I calculate the pinks together… I end up with 71 for P and with the equation on 12:24 I see that it is 68. I am missing three somewhere and so on for B and C (C too since it was 5 and the equation tells me its 8) please enlighten me what I missed
Numberphile If I get this right, the definition of a magic hexagon is to use each number once. If you allowed that rule, you could create infinite magic hexagons by simply adding 2 to all outer numbers, 1 to all middle ring numbers and 0 to the central number (in this example they all add up to 44 then (you take 3x the number you added to the outer ring)), however in my workings you do get the number 5 three times.
Here’s a very different kind of number hexagon, but as usual inspired by Dr Grime years later, and has a strange magic. Do an Ulam spiral but on a hexagon lattice, starting with a 7 in the middle and winding round indefinitely in the natural number sequence. Eventually you’ll see every odd number except 7 itself starts a row which includes its square. For example 11 21 37 59 87 121 . . . So far it only seems to work with 7 in the middle. Anything in the literature?
2x-1… 2n-1….. That is an increadibly popular value by itself and is seen in almost every numberphile article. Obviously it’s important. Could we see a article on why that, though? There has to be an explanation scientifically/mathematically that explains why it is used so frequently more than others.
TL;DR: The proof is correct, assuming that the numbers used must be 1 … X – but this is not a fair assumption for hexagon mesh. In magic squares, it is convenient to define that the numbers must start at 1, because you could just easily add the same number X to all the numbers and get a (in a sense isometric) square from number (X + 1). In the case of hexagons is the situation is different, since not each “line” (column/diagonal/row…) has the same number of elements. Therefore it makes sense to “allow” to start at any positive integer number. Or even negative. Or let’s allow even fractions with the same denominator. Or any real number 🙂
By only magic hexagon he means magic hexagon filled with the natural numbers from 1 to one of the hexagonal numbers. Technically it’s not the only but the other one is the single hexagon with a 1 in it. There are plenty more though if it’s not just the natural numbers from 1. There’s one from like, -89 to 89 or something that works. And of course if you just use any natural numbers you like with repeats then it’s easy to make one.
32M = 72n^3 – 108n^2 + 90n -27 + 5 / (2n – 1), as mentioned in the article. Now, if n = (1, 3), then M = (1, 38). Boring/obvious details below. 32 M = 72 – 108 + 90 – 27 + 1 = 72 – 108 + 90 – 27 + 5 = 32 M = 32 /32 = 1 32M = 72 – 108 + 90 – 27 + 1 = 1944 – 972 + 270 – 27 + 1 = 1216 M = 1216 / 32 = 38 Obvious, I know, but I haven’t seen anyone mention it yet, so I did. Did I miss its previous mention? If so, I don’t want to take credit for being first.
Notice that 3, 5 & 7 are the only triplet primes, using those numbers with 3 at a vertex, 5 as the center and 7 between them leads immediately to 15 as the opposite vertex with 8 opposite the 7. This leads quickly to the solution but doesn’t prove uniqueness. I wish heartily that James had at least sketched the uniqueness proof. The only ones I’ve found online are mind-bending use of combinatorics.
From a pure number theory stance there are a couple more solutions, but they don’t really make sense in practice as you have to allow n outside the natural numbers. n=0 gives the trivial hexagon, n=-2 also give a whole number solution but it makes even less sense to talk about a hexagon of size -2 than it does to talk about one of size 0
So… by my calculations, a magic soccer ball (where each ‘row’ is a center pentagon and the 5 hexagons surrounding it) should be possible, with numbers 1-32 and the sum of each ‘row’ being 44. (sum of available numbers) = (# of rows) * (Magic Number) 1+2+3+…+32 = 12M 528 = 12M 44 = M I just don’t have the maths skill to take the extra step of solving it or testing to see whether that it is actually possible. 🙁
Quite Fascinuberant (exuberant and fascinating) and Entertollicking (entertaining and rollicking). Proof of uniqueness is not easy, but it is highly revealing. But of what? What can this magic hexagon insight lead to? Best Regards! –The Slowcounters’ Association and Mius Eternus (The Eternal Monster in the Unit Square)
Question, you have a 8 1 6 | 3 5 7 | 4 9 2 magic square as mentioned in the article, it would seem to me that if the magic hexagon has to be able to work from any angle with any number of hexagons then why doesn’t the magic square have to say add up to the magic number when you take say 1 from the top row middle column and 7 from the middle row right column?
What I wonder is whether the fact that the magic number is an integer guarantees that there is exactly one solution. Intuitively, just because a solution is possible doesn’t mean it exists or is unique. If we take a different range of numbers and a different size hexagon, does the fact that the magic number for that set of parameters is an integer guarantee that there exists a unique solution?
With Magic Squares, you’re only measuring rows, columns and diagonals of length 3, you’re not measuring the diagonals of 2. Are there more magic hexagons where all rows of length x are the same? For instance, where all the rows of 5 are the same, all the rows of 4 are the same (but not the same as the rows of 5), &c.?
I mostly enjoyed this article, but the way you did that “add the first and last” formula was lacking. I think you should have first applied that to a generic sequence of numbers from 1 to x to get the formula of x(x+1)/2, and then applied that formula to the two sequences you did. By doing it from scratch both times, I think people might miss over that there’s an underlying pattern there; looking over that x(x+1)/2, referencing triangular numbers, explaining why it’s always going to produce a whole number, etc, would make this article far more interesting and useful as a learning tool (even if it would be a bit of a tangent, and I know enough that I wouldn’t really need such explanation).
The comparison between the magic squares and magic hexagons as you describe them is kind of incorrect. In your hexagon the line 9-6-5-2-16 goes from corner to corner, so it is really a diagonal. All your lines are parallel to one of these diagonals. This also explain why the lines are of differing lengths. The true corresponding magic square to this hexagon would be one where all the lines parallel to a diagonal add up, and some of these lines would be the corners with 1 number in them. Constructing such a square would be impossible. The hexagon corresponding to the square you present would be one where lines are going perpendicular to the hexagons sides. In your example, lines such as 9-8-10 and 11-5-12. All these lines would have lenght 3. The diagonals are harder to pinpoint, but they could be lines of length 5, or they could be similar skipping lines, such as 9-5-16. Would such a hexagon be easier to construct?
There’s actually one more magic hexagon. I think. Answer? Just one hexagon with a one number in it. It’s made only of hexagons and each number appears once and any direction adds up to the same number. So technically… EDIT: Oh. I really should watch the whole article first, but still! You yourself said you were being sneaky! You cheated!
Tried to see if the major “diagonals” added up to the magic sum, defining that as the lines through the center that are intermediate between the grid directions; if you only count the cells those lines pass through the centers of surprisingly one of them does work (14 + 5 + 19 = 38). However the other two do not, making this imo at best a Parker hexagon
Compared to squares, with hexagons you can’t really fit a whole number of hexagons inside a bigger one. That’s frustrating me. Hexagons have such nice properties compared to squares, but hierarchical structures aren’t as elegant as being able to contain 4 squares into a larger square, etc. It’s tempting to just go with dividing up space into squares, like with the yee-lattice for the purpose of electromagnetic simulations. But what if there’s a holographic principle, where information contained in the lines define what’s in the surface, can i just take the information contained in the lines to make up a larger hexagons? That might regain some elegance as the line segments can be seen as individual parts, compared to cutting the hexagon surfaces up and assigning parts to larger hexagons. This has nothing to do with this article, but thanks for making me think of hexagons.
Actually, the size one hexagon doesn’t check out. If you plug one into the equation you get 32M=248, which means the magic number M has to be 31/4. As you mentioned earlier, the magic number is a whole number, and since 31/4 is not a whole number, the answer “one” is an extraneous solution to the problem.