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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    12 Dec '13 07:38
    http://upload.wikimedia.org/wikipedia/commons/thumb/9/97/Pentacle_2.svg/220px-Pentacle_2.svg.png
    http://upload.wikimedia.org/wikipedia/commons/c/ca/DavidStar.png
    http://www.nilesjohnson.net/teaching/9-point-star.png

    All these star shapes have a regular polygon in the middle. David's Star also has a special quality - the hexagon in the middle has exactly the same area as the six little triangles on the rim put together.

    The question is this - is David's Star is the only star shape with a regular polygon in the middle that has that quality, or one of a finite set, or just one of infinite variants?
  2. Subscriber Ponderable
    chemist
    12 Dec '13 13:04
    Originally posted by talzamir
    http://upload.wikimedia.org/wikipedia/commons/thumb/9/97/Pentacle_2.svg/220px-Pentacle_2.svg.png
    http://upload.wikimedia.org/wikipedia/commons/c/ca/DavidStar.png
    http://www.nilesjohnson.net/teaching/9-point-star.png

    All these star shapes have a regular polygon in the middle. David's Star also has a special quality - the hexagon in the middle has exactl ...[text shortened]... n in the middle that has that quality, or one of a finite set, or just one of infinite variants?
    hyothesis: For each regular polygon exists a star where the sum of the area the triangles over each side equate exactly the area of the central polygon.

    The proof is quite trivial : you can always dissect a regular polgon in a set of triangles with the one side as base and the legs as the lines from start and end of these sides to the center. The same triangle can then be taken as the "spikes" of the star. Since they are the same their area has to be the same.
  3. Standard member talzamir
    Art, not a Toil
    14 Dec '13 19:42
    Certainly your method applies. If you take any regular polygon, connect each point to the middle to form triangles, and mirror the triangles over the edge that doesn't touch the middle, you get a star-like shape. Such as the Star of David.

    However, the familiar five-pointed star cannot be generated in this fashion.

    So.. if we think of stars made in this fashion..

    1. start with a circle.
    2. pick n points at regular intervals along the perimeter of the circle.
    3. pick a number above 1 but less than n-1.
    4. connect each point n_x to point n_x+k. If x+k > n, connect to x+k-n.
    5. this usually gives you a star-like shape with a regular n-sided polygon in the middle.

    e.g. with the pentagram n = 5 and k=2, so join 1st to 3rd, 2nd to 4th, 3rd to 5th, 4th to 1st and 5th to 2nd. David's Star is n = 6, k = 2.

    The polygons in the stars generated in this fashion do not always count for exactly half the area of the star. It does so for David's Star. For the pentagram, the middle pentagon looks too small. If you pick n = 1000 and k = 2, the thousand-sided polygon in the middle is huge when compared with the tiny triangles that form the points of the star.

    So, is David's Star unique, one of many, or one of infinite possibilities that can be generated by some method?
  4. Subscriber Ponderable
    chemist
    16 Dec '13 15:21
    you can calculate the respective areas.

    For a pentagramm you find that the area is about 5.568 times a squared. A being the side of the central pentagon. For the area of the pentagon you get roughly 1.72 times a squared. So the area of the triangles around the pentagon is about 3.85 times a squared, which is more than double the area of the central pentagon.

    You can do the claculation (using always triangles) for each star that you can construct and find out that most of those won't be the same area.
  5. 16 Dec '13 19:26
    take any polygon (n sides)
    pick a point inside the polygon
    join all polygon corners to that point, cutting polygon into n triangles.
    reflect each triangle around the relevant polygon side

    You now have a "star" which has the same area as the polygon.

    Since there are an infinite number of triangles with any particular area, there are an infinite number equivalent area stars for every polygon, for regular polygons, only 1 star will have both the same degree of symmetry and the same area.
  6. 16 Dec '13 21:20 / 4 edits
    For the hexagon, the "equivalent area" star has twice the perimeter

    so N/star_perimeter, for a hexagon, is 3

    What is the limit, as N tends to infinity, of N/Star_Perimeter?
  7. 16 Dec '13 22:24
    Sorry, RHP would not let me edit that:

    It should be:

    If a regular hexagon has perimeter 6 then the perimeter of its equal area star is 12

    Let us invent K which is the square of the regular polygon's perimeter, over the perimeter of its equal area star.

    For a hexagon, K = 36/2 = 3

    What is the limit, as the number of sides of the hexagon tend to infinity, of K?
  8. Standard member talzamir
    Art, not a Toil
    17 Dec '13 14:05
    The original.. if you take n points on a circle at regular intervals and join them to form a star.. I wonder if this reasoning is solid? k = 1 is a nicely elegant end result.

    If the lines from the point of a star go to two other points that are adjacent.. as is the case with a pentagram.. then every point on the perimeter of the circle is within the opening of the angle from exactly one point. It's known that if you have a part of the perimeter and join the end points to another point on the perimeter, the angle is exactly half of what you'd get if you connect them to the middle. Thus, the angles at the five points at the ends of pentagram add up to half of 360 degrees, so they are 360 / 2 / 5 = 36 degrees each.

    That can be expanded so that if you have an n-gram, the angle is 360 / 2 / n. Further, if there are points in the opening of the angle, the angles together cover the perimeter multiple times - so if you have n points to a star, and there are k points in the opening of each angle, the angle is

    360 * (k + 1) / (2n)

    degrees. Now, cutting the n-gon in the middle into triangles as Ponderable suggested, you get a quadrangle. Due to symmetry, the part closer to the point and the part within the n-gon are equal if and only if the quadrangle is a parallelogram. Obviously there are n triangles so each degree is 360 / n. Which gives the equation

    360 * (k + 1) / (2n) = 360 / n

    which simplifies into k = 1.

    So, if the star opens so that exactly one point of the star is in the opening of the angle, the area of the polygon in the middle equals the total area of the triangles near the points. If the star is drawn so that the two lines from each point are draw to two adjacent points.. like a pentagram, from point 1 to points 3 and 4.. the polygon is too small. If there are 2 or more points in the opening of the angle, the polygon is too large.
  9. 17 Dec '13 22:35 / 1 edit
    hmm, not what I got...Were you thinking of areas rather than perimeters?

    Although it is very neat (but perhaps a bit obvious (if area ratios) as one is a dissection of the other).
  10. Standard member talzamir
    Art, not a Toil
    20 Dec '13 09:03
    areas, yes
  11. 20 Dec '13 13:05
    ok, but my problem was perimeters:

    If we have a regular polygon, and we form its "equal area star", with the same symmetry as the original polygon, then was is the value, as number of sides tends towards infinity, of k if:
    k = polygon_perimeter^2/star_perimeter
  12. 20 Dec '13 15:06
    Originally posted by iamatiger
    ok, but my problem was perimeters:

    If we have a regular polygon, and we form its "equal area star", with the same symmetry as the original polygon, then was is the value, as number of sides tends towards infinity, of k if:
    k = polygon_perimeter^2/star_perimeter
    I think that there has to be another constraint to make it converge properly, namely that the polygon edge size is 1, so that the polygon perimeter is n

    (sorry)
  13. 23 Dec '13 16:02 / 1 edit
    anyone thinking about this, or do you want the answer.

    Put completely in one post:
    Start with a regular polygon, sides one unit long, N sides. Draw lines from each corner of the polygon in towards the middle to divide it into N triangles. Now reflect each triangle in the relevant side and you have a star. Take the perimeter of this star, P.

    What is the value, as N tends towards infinity, of:
    N^2/P

    For bonus points:
    Why is it necessary for the sides to be one unit long?
  14. 27 Dec '13 20:45
    ok, nobody is thinking about this, so here is the answer (I had to work it out myself, as all I had initially was an excel sheet showing that as N got large the ratio became ever closer to pi)

    If we join the points of a polygon with N sides (length 1) to the centre, the angles at the centre are all 2*pi/N

    a = 2*pi/N

    The cosine rule is:
    A^2 = B^2 + C^2 - 2BC.cos(a)

    we can use this to derive the length from the centre of the polygon to its corners (r):

    r = 1/sqrt(2(1- cos(a))

    As the perimeter of the star is 2N sides, each of this length:

    perimeter_of_star = 2N/sqrt(2(1- cos(a))

    the perimeter of the polygon itsefl is N, so, if we call our ratio x, which is:
    polygon_perimeter^2/star_perimeter
    then we can construct the equation:

    x = N.sqrt(2(1-cos(a))/2

    we want to work out what the value of x is, as N tends towards infinity.

    we can simplify our subsequent work, if we observe that 1-cos(a) is a small positive number, for very large N, so we can square both sides of this equation, and later take the positive square root of them, and this will still be unambiguously the value of x

    This gives us the equation:

    x^2 = N^2(1-cos(a))/2

    the problem here is that N => infinity, (1-cos(a)) => 0, so the value of x^2 as N=>infinity is not obvious.

    We can use l'Hopitals rule to proceed, basically this states that if we have a fraction, and the top and bottom of the fraction both tend to zero (or both tend to +- infinity) then we can take the differential of the top and the bottom, and the resultant new fraction will tend to the same thing as the original fraction.

    If we recast our expression so that it is a fraction:
    x^2 = (1-cos(a)) / (2/N^2)

    we can see that, as required, both the top and the bottom tend to O as N=> infinity, so we can differentiate them:

    d/dn { 2/N^2 } = -4/N^3

    d/dN { 1 - cos(a) } = -a.sin(a)/N

    so:

    lim {N=>inf} x^2 = lim {N=>inf} N.pi.sin(a)/2

    using lim, to stand for lim {N=>inf}

    lim x^2 = (pi/2) lim sin(a)/(1/n)

    once again, the top and the bottom of the fraction tend to 0 as N=> inf, so we can again apply L'hopital:

    d/dn {sin(a)) = -2.pi.cos(a)/N^2

    d/dn {1/n} = -1/N^2

    so:

    lim x^2 = (pi/2) lim 2.pi.cos(a)

    as, N => inf, clearly cos(a) => 1, so this evaluates to:

    lim x^2 = pi^2

    and taking square roots:

    lim x = lim {N=>inf} N^2(1-cos(a))/2 = pi

    so our ratio tends to pi, as the number of the sides of the polygon tend to infinity.