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Can one determine the surf/circumf with a geometrical figure

  1. Oct 6, 2012 #1
    Exists in mathematics a geometric figure, with straight lines (rectangle, parallelogram, or so), allowing you to determine the circumference and area of a circle by drawing / calculating / approximating?
    Without pi or formulas. So figure with no curved lines (and no polygons).
    for instance using sin, cos, tan.

    In other words: with pen, paper and a protractor drawing a figure to determine the circumference and surface(area) of a circle.


    PS.
    from other forums I heard it isn't possible...
    BUT IT IS!!

    I've found one model myself! It's a kite! With angle 144.686

    For more detail, drawings and calculations view my blogsite:
    http://quarks-divided.over-blog.fr
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2

    Simon Bridge

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    Welcome to PF;
    http://a10.idata.over-blog.com/500x375/5/88/58/73/Kite-Circle-1.JPG [Broken]

    That figure shows that when the angle at C is a particular value, then the area of the kite is the area of the circle.

    The calculation of the angle needs pi (and formulas).
     
    Last edited by a moderator: May 6, 2017
  4. Oct 7, 2012 #3

    micromass

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    Right. Of course it is possible.

    But the idea is to construct that figure with compass and straightedge from the original circle. THAT is what is impossible.
     
  5. Oct 9, 2012 #4
    Simon thanks for welcoming me!

    I have never seen a circle being kited!

    I added some new drawings, (logically):
    "rectangling the circle"
    and
    "boxing the circle".

    The big question is: why this angle of 17.65678715141......???????????????
     
  6. Oct 9, 2012 #5

    Simon Bridge

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    The big question is: why this angle of 17.65678715141......???????????????

    because the angles inside the triangle ABC must sum to 180 degrees.

    and those are the angles that make the area of the kite the same as the area of the circle.

    You get this kind of construction where you want to know (say) what the lowest point of reflection can be to reflect a signal between two places on the surface of the Earth.
     
  7. Oct 10, 2012 #6
    because the angles inside the triangle ABC must sum to 180 degrees

    I understand yes.
    But why 17.686 and not 15.3 or 26.88
    What's the relation of 17.6...to a circle or pi or so?
     
  8. Oct 10, 2012 #7

    HallsofIvy

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    The angle is NOT necessarily 17.686. There are an infinite number of kites, of different dimensions and different angles that will work. You happen to have found one.
     
  9. Oct 10, 2012 #8
    Thank you HallsofIvy for answering.

    But I can't believe you, sorry.
    I think there's just 1 kite:

    The kite which has
    2 arms equal r
    and 2 arms equal πr
    and an area equal the area of a circle.

    Other angles can't give that I think.(I tried several)

    If you can give me an example of another kite with those 3 arguments, please!

    If not, why 17.686...?
     
  10. Oct 10, 2012 #9

    Simon Bridge

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    Because none of those angles will make the sum come to 180 degrees.

    The key to the figure is the angle at C.
    Here is how the kite is determined:
    A and B are on the circle, C is in the center, Point D is the intersection between the tangents to the circle at A and B.

    Angles: A+B+C+D=360, A=B=90, so D=180-C
    ... so setting angle C will determine the entire geometry of the kite. [1]

    You want to find the kite whose area is also the area of the circle ... namely ##\frac{1}{2}|CD||AB|=\pi r^2##. To find the angle C that does this, you need to express |AB| and |CD| in terms of sines and cosines of C.

    ##|AB| = 2r\sin(C/2)## .. that was easy! The other one takes a bit more work but you should be able to do it OK.

    But you needn't bother - ##C/2## is the angle ##\theta## in a right-angle triangle whose adjacent side is length ##r## and opposite side length ##\pi r##
    The area of the kite is two of these.

    It is easy to find shapes whose significant feature is also the area of a circle
    ... eg: the area of an oblong width pi and length r2 is also the area of a circle radius r;
    ... the volume of a box whose base is a square with sides length r, and of height pi, is the area of a circle radius r too.
    ... as above, the area of a box length ##\pi r## and width ##r##: cut that box corner-to-corner to make two triangles - flip one and put the hypotenuses back together and you have the above kite.

    ---------------------

    [1] per your question ... if a = ∠CAB = ∠CBA, then 2a= 180-C or a=90-(C/2)
    i.e. the angle is 17.686degrees because of what C has to be (by above).
     
    Last edited: Oct 10, 2012
  11. Oct 11, 2012 #10
    Hello Simon, thanks for the extended answer.

    I think you should look at my blogsite with the drawings and calculations to see what I ment.
    We're talking 2 directions.
    It's not a box with edges pi*r and r but those with cos and cos/tan edges.
    I think my representation of the kite is the only one.
    Or not?
    Thanks

    see: quarks-divided.over-blog.fr and to the "kyte-ing the circle"page / "rectangling the circle"page
     
  12. Oct 11, 2012 #11

    Simon Bridge

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    I did see your blog site... from the first post right?

    Your kite shape - cut it along the CD line to get two triangles, flip one of them over, put them back together.
    See? Each triangle of the kite is just a rectangle cut corner-to-corner.

    Using the figure on http://quarks-divided.over-blog.fr/
    The area of the kite is the area of the two triangles CAD and CBD added together, which is twice the area of one of them since they are mirror images.

    The triangles have base length ##|AD|=|BD|=\pi r## (half the circumference) and height ##|CA|=|CB|=r## for area ##\frac{1}{2}\pi r^2## and twice that is???

    The kite you describe is not the only one that can be drawn by those rules, but it is the only one whose area is the same as the area of the circle.
    You cannot construct it using a straight-edge and a protractor without also knowing the angle at C ... which you have to compute from knowing the value of pi and the tangent function. It has to be the one that makes the sides ##|AD|=\pi|CA|## and that is a tangent ratio.
     
    Last edited: Oct 11, 2012
  13. Oct 11, 2012 #12
    to Simon:

    but it is the only one whose area is the same as the area of the circle !!!!

    That has been my statement all allong!

    The question is: why is that so and why does it give that kite with angle 17.656...(or angle C 144.686...)

    What's the relationship between that (those) angle(s) and a circle (or pi, or...).

    i.o.w. where does this number 17.65678715141....come from??


    http://quarks-divided.over-blog.fr/pages/Kiteing_the_Circle_-8348481.html [Broken]

    http://quarks-divided.over-blog.fr/pages/Rectangling_and_boxing_the_circle-8354341.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  14. Oct 11, 2012 #13

    Simon Bridge

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    I'm sorry I keep telling you - because the geometry dictates it. The angles have those (approximate) values for the same reason that 3-4-5 triangle has angles (approximately) 36.86° and 53.13°; it is for the same reason your legs are exactly long enough for your feet to touch the ground!

    One... last... time... and since you seem to be fixated on the 17thingy number I'll do it that way this time:

    17.65678715141...... is the half-angle at D, so ##D/2=17.65678715141......## (Using my convention of labeling interior angles with the point name.)

    So it is the angle of a right-triangle whose adjacent side is length ##|DA| = \pi r## and whose opposite side is length ##|AC|=r## .... so ##\tan(D/2) = 1/\pi## because of the definition of the tangent function.

    The sides have to have those lengths because one of them is defined to be the radius (so that's fixed) and the other side has to be long enough that the product of the sides is ##\pi r^2##.

    Perhaps this will help:
    If you hadn't set the angles at A and B to be 90°, then all these numbers would be different.
    eg. if you had set the angle at C to 60°, then ##|AB|=r##, and ##|DC|=2\pi r##
    ... and the half-angle at D would be

    ##\tan(\frac{D}{2})=1/(4\pi-\sqrt{3})\Rightarrow D/2 \approx 5.2734^\circ##.

    This method is more useful in terms of construction because you do the following:
    1. draw a circle with radius r - mark the center
    2. construct an equilateral triangle sides length r with one corner on the center - mark the other two corners
    3. bisect the angle at the center - draw a line radially along the bisector
    4. cut out a circle radius r to make a wheel
    5. roll the wheel along the radial line until it turns exactly once and mark that point
    6. join each of the marked points to form a kite-shape.
    ... the area of the kite is the area of the circle.
     
    Last edited: Oct 11, 2012
  15. Oct 11, 2012 #14
    Well, to make a long story short:

    the answer to why 17.656...
    is
    because

    atan (1/pi) = 17.65678715141288..

    (thanks to Simon.)
     
  16. Oct 11, 2012 #15

    Simon Bridge

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    ... and 144.6868-etc is because atan(pi)=72.343=C/2
    ... both because you cut a box with sides r and pi*r in half diagonally to make the kite.

    the kite is unique for angles A=B=90, but not unique if that restraint is relaxed.
    (if this restraint is relaxed, then the kite comes from cutting a parallelogram)
    setting the angle at C determines |CD| and the angles at A and so the entire kite... easier than setting the half-angle at D.
    yadda yadda yadda... you actually had a lot of restraints in the setup - which is why you only get the one solution.

    another version would be to put C=180. Then C and D would be centers of sides of a rectangle.
    then the length |CD| would have to be half-circumference to make it fit.
    (angle D would be 180 while A=B=90 again(!) ...as well, the side going through D would be exactly pi away from the circumference.)

    convinced of how simple this is yet?
    are you going to update your web-page?

    Note: you measure an angle with a protractor ... by
    1. measuring a distance around the circumference
    2. dividing by the circumference (2*pi*r)
    3. multiplying by 360.
    The manufacturer usually does this for you when they mark out the scale around the edge.

    For radiens, the process is
    1. measure the distance around the circumference
    2. divide by the radius.
    ... this method is more natural because it does not require you to be able to make a circle with radius (some multiple of) 1/2pi to make a protractor ;)

    You should be using radiens in preference to degrees.
    If you use radiens then A=B=pi/2 is the main restraint but you don't introduce pi in the angles as well because this way does not require that you measure out pi units to get an angle where the degrees one does.
    (However atan(pi) = 1.2626 does not come to a simple ratio of pi)


    Bonus:
    If you want to make a building to excite mystical-pi enthusiasts, make a wheel 1m in diameter and roll it along the ground to mark out the sides. Now all the sides of the building have lengths some multiple of pi meters. If you mark out your right-angles using 3-4-5 triangles (in the same units!), you'll get another "mystical" ratio and you can tell people you had extra-terrestrial help with the construction :D
     
    Last edited: Oct 11, 2012
  17. Oct 12, 2012 #16
    Well, it seems like Simon is getting ironic....
    I don't know why that's necessary.

    I just wondered why the angle was 17.686... and not another one.
    Just like pi = 3.1415... and no-one knows why.

    Nothing mystical about, but just interesting.

    Point is I DID found the only kite with the 3 arguments
    - 2 arms length r
    - 2 arms length πr
    - area kite = area circle

    And that in an unconventional way:

    with a flashlight and a piece of paper.

    We can't be all wonderkids of mathematics....

    (I wonder what he has to say about my found relationship between Pi e and Phi...also mystical nonsense too perhaps)
     
    Last edited: Oct 12, 2012
  18. Oct 12, 2012 #17

    Simon Bridge

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    But we do know how - it is because the geometry supports it.
    Like with pi - if it were any other number, the object would not be a circle in Euclidean space.
    Science does not do "why".

    No worries - I thought you were getting it so I just had some fun at the end.
    Time for me to bow out.
    Have fun.
     
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