Posers and Puzzles

Posers and Puzzles

  1. Standard membertalzamir
    Art, not a Toil
    60.13N / 25.01E
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    14 Dec '11 17:31
    Christmas is coming, and one of the ornaments I see here often looks like a nine-pointed star, a nonagram. How wide is the angle at the sharp points?

    [img]http://thesaurus.maths.org/mmkb/media/png/Nonagram.png[/img]
  2. SubscriberSuzianne
    Misfit Queen
    Isle of Misfit Toys
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    15 Dec '11 20:091 edit
    Originally posted by talzamir
    Christmas is coming, and one of the ornaments I see here often looks like a nine-pointed star, a nonagram. How wide is the angle at the sharp points?

    [img]http://thesaurus.maths.org/mmkb/media/png/Nonagram.png[/img]
    40 degrees?

    Edit: Oh, wait, no the angles don't originate at the center, so it would be more... how much more, I'm not sure.
  3. Joined
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    15 Dec '11 20:53
    Is that three overlayed equilateral triangles I see?
  4. Joined
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    15 Dec '11 23:102 edits
    Originally posted by talzamir
    Christmas is coming, and one of the ornaments I see here often looks like a nine-pointed star, a nonagram. How wide is the angle at the sharp points?

    [img]http://thesaurus.maths.org/mmkb/media/png/Nonagram.png[/img]
    One can visualize two circles; one that intersects the "outer (so called sharp) points" and one that intersects the "inner" points (the points near the center, between the outer points). Each circle has a radius; call them ro and ri. The inner circle radius can approach zero as a minimum, and can approach ro as its maximum (beyond this maximum, the inner points become outer points). It does not seem to me to be possible that the angle the puzzle asks for is some specific value; it is a function of ri/ro. Maybe the puzzle is to define this function.

    At the limit of ri=0, the angle of the sharp outer points is zero (9 line segments radiating from a center) and the angle of each "sharp" inner point is 360/9 = 40 degrees.

    At the limit of ri=ro, we have an 18-sided regular polygon with a rather unsharp interior angle of 160 degrees. (http://www.mathopenref.com/polygoninteriorangles.html)

    That's as far as I am going with this before someone weighs in on my thinking.
  5. Joined
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    15 Dec '11 23:271 edit
    Ah, I see, it is sharper than two equilateral triangles.

    Assuming that If you "walk" from a point along a side, and continue in that same direction across the middle then you get exactly to the opposite side (certainly looks like that), then:

    We can "walk" the nonagram as follows:
    Start on the bottom point

    Walk up the left side, carrying on across the middle and getting to a the top lefthand point
    Turn clockwise by angle A

    If you repeat those two moves 9 times you have walked the star.

    Looking at the diagram we can see that we will then have made 4 complete revolutions through 360

    So angle A = 360*4/9 = 160 degrees

    The angle at each point is 180-A

    So the angle at each point is 20 degrees.
  6. Standard membertalzamir
    Art, not a Toil
    60.13N / 25.01E
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    16 Dec '11 11:58
    Indeed it is. There are many ways to find it of course. Here are two more -

    draw a circle around the nonagram, and linking two points of the star to the center of the circle as well the farthest point of the star. By symmetry, the arc between the points is one ninth of the whole, as is the angle at the center of the circle, so that angle needs to be 360 / 9 = 40 degrees. It is known that the angle at the center of the circle is twice the angle at the perimeter if the sides of the angle touch the same arc; so the angle needs to be 40 / 2 = 20 degrees.

    Alternatively, without using the knowledge that center angle = 2 x perimeter angle, link all 9 outer points A1 .. A9, and all 9 inner points, B1 .. B9 of the star to the center point =, dividing the full 360 degrees into 18 equal parts of 20 degrees each. Draw a triangle with points A1, A5, and O. The angle A5OA1 consists of eight parts of 20 degrees each for a total of 160 degrees. The other two are thus 10 degrees each. And since they split the angle at the point of a star in half, that angle is 2 x 10 = 20 degrees.
  7. Joined
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    17 Dec '11 01:38
    Originally posted by talzamir
    Christmas is coming, and one of the ornaments I see here often looks like a nine-pointed star, a nonagram. How wide is the angle at the sharp points?

    [img]http://thesaurus.maths.org/mmkb/media/png/Nonagram.png[/img]
    Are you guys talking about the specific star at the link?

    BTW, it's now 404 not found, when I look.
  8. Joined
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    17 Dec '11 10:00
    The link works for me if I paste it into the address bar in without the [img] bits at the start and the end.
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