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Math Help - Polygons at a Point

  1. #1
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    Polygons at a Point

    Hey people, i have another problem i'm having trouble with!

    Certain sets of regular polygons fill space around a point without gaps or overlapping. e.g. imagine 3 hexagons joined together and the common vertice is the point they surround. That is classified a [6,6,6] point fill set (3 six sided shapes)

    Although the polygones can be arranged in a different order, we count these as the same.

    We call a set of regular polygons filling space around a point a point-fill set.

    a) Find a point-fill set of 3 polygons containing a 24-gon
    b) Explain why it is not possible to have a point fill set containing a triangle and a pentagon
    c) Find all point-fill sets that contain at least one square.

    Hope i've made it clear

    Any help would be fantastic
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  2. #2
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    Quote Originally Posted by Chuck_3000
    Hey people, i have another problem i'm having trouble with!

    Certain sets of regular polygons fill space around a point without gaps or overlapping. e.g. imagine 3 hexagons joined together and the common vertice is the point they surround. That is classified a [6,6,6] point fill set (3 six sided shapes)

    Although the polygones can be arranged in a different order, we count these as the same.

    We call a set of regular polygons filling space around a point a point-fill set.

    a) Find a point-fill set of 3 polygons containing a 24-gon
    b) Explain why it is not possible to have a point fill set containing a triangle and a pentagon
    c) Find all point-fill sets that contain at least one square.

    Hope i've made it clear

    Any help would be fantastic
    The internal angle at any vertex of a regular n-gon is:

    <br />
\pi(1-2/n) \mbox{ radian}<br />

    If m regular n-gons with n_1,\ n_2,\ \dots\ n_m sides constitute a point-fill set then
    the sum of the internal angles at their vertices sum to 2\pi radians, that is:

    <br />
2\ \pi=\pi(1-2/n_1)+ \dots+\pi(1-2/n_m)<br />

    or:

    <br />
m-2=2(1/n_1+\dots+1/n_m)\ \ \ \dots(1)<br />
.

    Now lets look at part (a).

    m=3, and n_1=24, so plugging these into equation (1) and simplifying:

    <br />
\frac{11}{24}=\frac{1}{n_2}+\frac{1}{n_3}\ \ \ \dots(2)<br />

    then trial and error will allow you to find n_2 and n_3 to complete the point fill set.

    RonL
    Last edited by CaptainBlack; May 30th 2006 at 02:12 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack

    <br />
\frac{11}{24}=\frac{1}{n_2}+\frac{1}{n_3}\ \ \ \dots(2)<br />
    A solution to (2) can also be found by applying the greedy algorithm for
    Egyptian fractions, and techniques for Egyptian fractions can be used
    to find other solutions if they exist.

    RonL
    Last edited by CaptainBlack; May 30th 2006 at 09:41 PM.
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  4. #4
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    thanks a lot

    I follow what you've done in a), but one question - where you've said m=3 and n1 = 24, what does the m=3 refer to?

    And i have no idea what Egyptian fractions are, or how to use them for b)
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  5. #5
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    Quote Originally Posted by Chuck_3000
    thanks a lot

    I follow what you've done in a), but one question - where you've said m=3 and n1 = 24, what does the m=3 refer to?

    And i have no idea what Egyptian fractions are, or how to use them for b)
    The stuff at the beginning is about a general point fill set, with polygons
    of n_1,\ n_2 \dots\ n_m sides, so we see here that m is the number of
    polygons in the point-fill set. In part (a) we are asked for a point-fill set with
    three polygons so we use m=3 for this case.

    You don't really need to know about Egyptian fractions to solve the problem,
    in fact I did it by trial and error. If you do want to know about Egyptian
    fractions though there is another thread discussing them here

    RonL
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  6. #6
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    does anyone know how one would go about question c) ?
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  7. #7
    lozer
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    need help

    hi, am stuck with the same question. does anyone know how to do part b and c. thanks
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  8. #8
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    Quote Originally Posted by lozer
    hi, am stuck with the same question. does anyone know how to do part b and c. thanks
    Part b) can be done by looking at the angle left after you put a triangle and
    pentagon together at a vertex. Then try to fill this angle with a combination
    of other polygons. You will quickly find that there is no combination of
    polygons that will perfectly fill this angle.

    RonL
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  9. #9
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    Part b) can be done by looking at the angle left after you put a triangle and
    pentagon together at a vertex. Then try to fill this angle with a combination
    of other polygons. You will quickly find that there is no combination of
    polygons that will perfectly fill this angle.

    This is not true as if u use 3 triangles and a pentagon, u can get a point-fill set of (3,3,3,5).
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  10. #10
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    wat i don't get is a)
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  11. #11
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    Quote Originally Posted by confused
    Part b) can be done by looking at the angle left after you put a triangle and
    pentagon together at a vertex. Then try to fill this angle with a combination
    of other polygons. You will quickly find that there is no combination of
    polygons that will perfectly fill this angle.

    This is not true as if u use 3 triangles and a pentagon, u can get a point-fill set of (3,3,3,5).
    The internal angle of a vertex of a 3-gon is \pi/3, the corresponding angle
    for a 5-gon is \frac{3}{5}\pi so the sum of these angles for three 3-gons and a 5-gon is \frac{8}{5}\pi.

    But if [3,3,3,5] where a point fill set the angle sum would be 2\pi

    RonL
    Last edited by CaptainBlack; May 27th 2006 at 11:20 PM.
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  12. #12
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    Quote Originally Posted by CaptainBlack
    The internal angle of a vertex of a 3-gon is \pi/3, the corresponding angle
    for a 5-gon is \frac{3}{5}\pi so the sum of these angles for three 3-gons and a 5-gon is \frac{8}{5}\pi.

    But if (3,3,3,5) where a point fill set the angle sum would be 2\pi

    RonL
    who cares about that technical stuff?? u try drawing it on paper - a pentagon, and three triangles. it worxs
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  13. #13
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    Quote Originally Posted by confused
    who cares about that technical stuff?? u try drawing it on paper - a pentagon, and three triangles. it worxs
    For three triangles, since they have to be equilateral, the internal angles a
    vertex are \pi/3 or 60 degrees, three times sixty is 180 degrees.
    so for a regular pentagon to complete a point fill set with three triangles the
    internal angle at a vertex would have to be 180 degrees, which it is not. So
    [3,3,3,5] is not a point fill set (not even nearly!).

    So check what you have done, it may be wrong, then check the
    wording of the problem you may have misunderstood it.

    RonL
    Last edited by CaptainBlack; May 27th 2006 at 11:21 PM.
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  14. #14
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    Quote Originally Posted by CaptainBlack
    For three triangles, since they have to be equilateral, the internal angles a
    vertex are \pi/3 or 60 degrees, three times sixty is 180 degrees.
    so for a regular pentagon to complete a point fill set with three triangles the
    internal angle at a vertex would have to be 180 degrees, which it is not. So
    [3,3,3,5] is not a point fill set (not even nearly!).

    So check what you have done, it may be wrong, then check the
    wording of the problem you may have misunderstood it.

    RonL
    plz look at my attachment. i'm not the best of drawers on computer so excuse my graphics but you can get the general idea.
    p.s. who said anything about the triangles being equilateral??
    Attached Thumbnails Attached Thumbnails Polygons at a Point-maths-thing.bmp  
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  15. #15
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    Quote Originally Posted by confused
    plz look at my attachment. i'm not the best of drawers on computer so excuse my graphics but you can get the general idea.
    p.s. who said anything about the triangles being equilateral??
    The problem asks for regular polygons. Regular means all sides and angles
    are equal, so a regular 3-gon is a equilateral triangle. Also the pentagon
    has to be a regular pentagon.

    RonL
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