The center of a circle


by pixel01
Tags: circle
pixel01
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#1
Feb19-07, 01:30 AM
P: 691
Hi everybody,

I ve got a question : is it possible to identify the center of a given cirle with only a compass?

Thanks for reading
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ObsessiveMathsFreak
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#2
Feb19-07, 03:51 AM
P: 406
Quote Quote by pixel01 View Post
I ve got a question : is it possible to identify the center of a given cirle with only a compass?
No you need a ruler as well.

Pick a point A on the circle. Draw a circle with center A and diameter less than that of the given circle. Mark off the two points B and C where this new circle intersects the given one. Join B and C with the line BC.

Bisect BC. The midpoint is D. Draw a straight line through A and D and extend it until it meets the given circle again at E. AE is a diameter.

Repeat this process for a second point on the given circle. Where the two diameters meet is the center.
pixel01
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#3
Feb19-07, 04:51 AM
P: 691
Thank you for answering.
But the problem is without a ruler !!! It seems impossible. I have tried many times but failed.

Edgardo
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#4
Feb20-07, 02:24 AM
P: 685

The center of a circle


It works without a ruler, though only for a circle that's on a sheet of paper.
Instead of using a ruler fold the paper to make a line visible.
pixel01
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#5
Feb20-07, 02:56 AM
P: 691
Quote Quote by Edgardo View Post
It works without a ruler, though only for a circle that's on a sheet of paper.
Instead of using a ruler fold the paper to make a line visible.
Well, it seems like a trick ! Let's say the circle is on a table. May be we need to prove it impossible to do so.
jeroen
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#6
Feb20-07, 03:56 AM
P: 25
http://steiner.math.nthu.edu.tw/disk...13/center.html
pixel01
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#7
Feb20-07, 05:30 AM
P: 691
It's great !. Thank you Jeroen.
Anyway, how can we prove that friends?
jeroen
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#8
Feb21-07, 04:05 AM
P: 25
The method works by drawing a circle, picking two points on it, then draw a series of circles based on where the existing circles.
The final answer is the intersection of the last two circles.

All you need to do is write a big formula containing all these circles and resulting in the coordinates of the final intersection and then prove that "the_big_function(center_of_circle, radius_of_circle, point_1_on_circle, point_2_on_circle) = center_of_circle" is true for any input.


This would be a start:
point
  float x 
  float y 

circle
  point center 
  float radius 

point intersect (circle ca, circle cb, point pn)
  returns intersection between ca and cb that is not pn

float distance (point pa, point pb)
  returns distance between points pa and pb

point pointoncircle (circle c, float angle)
  returns a point on the circle going the given angle clockwise from the top

input:
  circle c0
    c0.center = variable
    c0.radius = variable
  float a1 
    a1 = variable
  float a2
    a2 = variable

output:
  point pf

process:
  point p1
    p1 = pointoncircle(c0,a1)
  point p2
    p2 = pointoncircle(c0,a2)
  circle c1
    c1.center = p1
    c1.radius = distance(p1,p2)
  circle c2
    c2.center = p2
    c2.radius = distance(p1,p2)
  circle c3
    c3.center = intersect(c0,c1,p2)
    c3.radius = distance(p1,p3)
  circle c4
    c4.center = intersect(c2,c3,p1)
    c4.radius = distance(intersect(c2,c3,p1),p1)
  circle c5
    c5.center = intersect(c4,c3,p1)
    c5.radius = distance(intersect(c4,c3,p1),p1)
  circle c6
    c6.center = intersect(c2,c4,p1)
    c6.radius = distance(intersect(c2,c4,,p1),p1)
  pf = intersect(c5,c6,p1)
HallsofIvy
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#9
Feb21-07, 06:35 AM
Math
Emeritus
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Thanks
PF Gold
P: 38,890
I may be wrong but I seem to remember a theorem that any construction that could be done with compasses and straight edge could be done with compasses alone. Of course "drawing a line" has to be interpreted as constructing two points on the line.

Check:
http://thesaurus.maths.org/mmkb/entr...ryById&id=4066


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