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Posers and Puzzles

Posers and Puzzles

  1. 09 Oct '10 08:51
    I have a regular decagon, and inside it I have drawn the largest square that will fit.

    What it the ratio:

    Area_Of_Square / Area_of_Decagon
  2. 09 Oct '10 12:15
    0,6639
  3. 09 Oct '10 13:27
    Originally posted by Thomaster
    0,6639
    No, that must be wrong.
  4. 10 Oct '10 21:30
    Originally posted by Thomaster
    No, that must be wrong.
    Yes, I think it is wrong too, but quite close.
  5. 10 Oct '10 21:36
    Originally posted by iamatiger
    Yes, I think it is wrong too, but quite close.
    That is because my square was a diamont.
  6. 11 Oct '10 07:34
    Originally posted by Thomaster
    That is because my square was a diamont.
    Ah, yes. I got a diamond at first too.
  7. 11 Oct '10 11:48 / 1 edit
    Are you sure? I'm probably missing something, but I got the same answer with something that I'm pretty sure is a square. (Though I'm not certain it's the largest square, admittedly).

    Presumably that 0.6639 is an approximation of (cos 9)^2/(5 sin 18 cos 18)
  8. 11 Oct '10 13:49
    To provide more detail, my "quadrilateral" is constructed as follows:

    If the decagon has a vertex at 0 degrees (and 36, 72, etc), then the square has vertices at 9, 99, 189, 279 degrees.

    If the answer isn't right, where I went wrong must be:
    - that isn't a square
    - that isn't the largest square
    - miscalculating the areas

    Which is it?
  9. 12 Oct '10 00:51 / 5 edits
    Originally posted by mtthw
    If the answer isn't right, where I went wrong must be:
    - that isn't a square
    - that isn't the largest square
    - miscalculating the areas

    Which is it?
    I think you miscalculated the areas, unless I'm wrong (The fourth possibility!).
  10. 12 Oct '10 09:33
    Originally posted by iamatiger
    I think you miscalculated the areas, unless I'm wrong (The fourth possibility!).
    OK, I'll have another look at it later. Ta.
  11. 12 Oct '10 10:53 / 1 edit
    <slaps head/>

    Silly mistake .

    0.6310 any closer?
  12. 12 Oct '10 20:16
    Originally posted by mtthw
    <slaps head/>

    Silly mistake .

    0.6310 any closer?
    Now only the 4th decimal place is wrong I think.

    Can you give your expression for the answer?
  13. 13 Oct '10 08:42 / 1 edit
    Originally posted by iamatiger
    Now only the 4th decimal place is wrong I think.

    Can you give your expression for the answer?
    I get:

    cos 18/(5 sin 18 cos^2 9)

    = 0.6309778... = 0.6310 to 4 d.p. (according to this calculator )
  14. 19 Oct '10 14:57 / 1 edit
    1 / (20 × sqrt( 5 + 2sqrt(5)) × sin² 9) = 0,6639

    Same answer.
  15. 02 Dec '10 21:23 / 24 edits
    Sorry I forgot about this

    you are both still slightly wrong:

    From the wikipedia entry on decagons:

    Area of a regular decagon = 2.5dt where:

    t = length of a side

    d = distance between two parallel sides

    they are related:

    d = 2t(cos(54) + cos(18)) {angles in degrees}

    so area_of_decagon = 5t^2(cos(54) + cos(18))


    lets say the side is length one, then:

    area 5(cos(54) + cos(18))


    Now a decagon has 10 sides, and a square has 4, so there should be 2.5 decagagon sides between any two square corners. Draw a decagon.


    Try setting a square corner to be a decagon corner,

    now drawing corners at

    2.5 side spacings we see that even though the sides are the

    same length, the diagonals of the "square" are different lengths


    so it isnt actually a square


    Shunt every square corner round

    by a quarter of a side, and we can see that all the diagonals

    will now be the same, and the sides are the same, so its a square.


    By pythagoras its easy to see that the length of a diagonal

    of this square is

    sqrt(d^2 + (t/2)^2)


    where d and t are as above


    the area of a square, in terms of its diagonal length is:

    diagonal_length^2/2


    so the area of the square is:

    d^2/2 + t^2/8


    recalling that d = 2t(cos(54) + cos(18)) {angles in degrees} and that t = 1


    area_of_square = 2(cos(54) + cos(18))^2 + 1/8


    so the ratio:

    square_area/decagon_area


    (2(cos(54) + cos(18))^2 + 1/8)/5(cos(54) + cos(18))


    so the answer is:
    0.6317826922466960