Question for fun

[Johnny5]

I give you a circle, but I don't tell you where the center is. The only tools you have, are a compass, and a straightedge. How do you locate the center of the circle?

[Callipygous]

i would put the point of the compass on the edge of the circle, and then widen it to the farthest point on the other side of the circle. (just make sure its wide enough that if you were to draw a cirlce with it the other circle would have a tangent point, but never outside) i would use that to make a line elsewhere (pretty hard to pinpoint the tangent point) the same length as the diameter. bisect it, set the compass to that length. put it back on the edge of the circle in two different places and draw circles, where they intersect is the center.

 

(this maybe a very round-about and possibly inaccurate method, i havent done much construction besides the basics in geometry class 4 years ago : P)

[Johnny5]

i would put the point of the compass on the edge of the circle, and then widen it to the farthest point on the other side of the circle.

 

How would you determine which point is the farthest point?

[ydoaPs]

it has been a long time since i have done constructions, so this may be really wrong.

 

if you pick three points on the circle and draw tangents from those points, the lines intersect to create a triangle. if you find the center of the triange, won't it be the center of the circle as well?

[Callipygous]

How would you determine which point is the farthest point?

 

 

"just make sure its wide enough that if you were to draw a cirlce with it the other circle would have a tangent point, but never outside"

 

you move it along toward the circle, when you hit a point that is going to be outside of your new line, widen it until you just barely draw over a point but dont cross it.

[Cap'n Refsmmat]

I've seen this before.

 

Try doing a search on SFN. I think someone already asked.

[Johnny5]

it has been a long time since i have done constructions' date=' so this may be really wrong.

 

if you pick three points on the circle and draw tangents from those points, the lines intersect to create a triangle. if you find the center of the triange, won't it be the center of the circle as well?[/quote']

 

 

 

Suppose two of the points you pick are diametrically opposed, meaning they are endpoints of a diameter of the circle. Thus, the tangent lines through those two points will be parallel to each other. No matter what other point on the circle you pick, there will be no triangle constructed. So no.

[Johnny5]

"just make sure its wide enough that if you were to draw a cirlce with it the other circle would have a tangent point' date=' but never outside"

[/quote']

 

This just makes no sense to me.

 

Make sure (whats) wide enough?

If you were to draw a circle with (what) the other circle (what other circle?)

[5614]

Draw two tangents to the circle so that they meet up at some point.. you then have two sides of a triangle... then bisect the angle made between the two tangents and draw a line.

 

Do this again with different tangents... where the "bisect lines" cross will be the center of the circle.

[ydoaPs]

Suppose you two of the points you pick are diametrically opposed, meaning they are endpoints of a diameter of the circle. Thus, the tangent lines through those two points will be parallel to each other. No matter what other point on the circle you pick, there will be no triangle constructed. So no.

 

well, connect those two points and find the midpoint.

[Johnny5]

Draw two tangents to the circle so that they meet up at some point.. you then have two sides of a triangle... then bisect the angle made between the two tangents and draw a line.

 

Do this again with different tangents... where the "bisect lines" cross will be the center of the circle.

 

Yes.

 

Euclid's proof is rather interesting too.

 

Euclid Book III Proposition 1

 

Incidentally' date=' his proposition on how to bisect a given angle was already covered in book one, so Euclid could have used your proof in book III, or his own.

 

Euclid Book I Proposition 9

[Johnny5]

well, connect those two points and find the midpoint.

 

Interesting.

 

Yes of course that would work.

 

Here is Euclid's proposition on how to bisect a given finite straight line.

 

Euclid Book I Proposition 10

 

 

But issues about how you would have randomly picked two diametrically opposed points arise. If you really did choose at random, you could never find the diametrically opposed point by luck (the probability you randomly chose the diametrically opposed point would by 1/infinity = 0), SO you would need to know where the center is and then draw the diameter, which pre-supposes that you already knew where the center was. So I suppose I objected unfairly, but of course now we are analyzing Euclidean geometry with a tinge of probability theory, and I don't think Euclid knew about probability theory at all. Call it a hunch.

 

I see your point.

 

if you pick three points on the circle and draw tangents from those points' date=' the lines intersect to create a triangle. if you find the center of the triange, won't it be the center of the circle as well? [/quote']

 

Yes.

 

Proving it symbolically might take you a bit of time though. For what it's worth, how do you find the center of a triangle?

[bloodhound]

yeah. this been answered before.

http://www.scienceforums.net/forums/showthread.php?t=3903

 

heres my solution.

[Johnny5]

 

if you pick three points on the circle and draw tangents from those points' date=' the lines intersect to create a triangle. if you find the center of the triange, won't it be the center of the circle as well?[/quote']

 

You didn't specify how to locate the center of a triangle, so I will. Connect a vertex to the midpoint of the opposite side, then do that again with another vertex. Where the two lines intersect is the midpoint of the triangle. I can prove this if you want.

 

At any rate, you should now be able to figure out that the center of the triangle you construct will not coincide with the center of the circle. Suppose that you have two points which are almost but not quite diametrically opposed. Therefore, they intersect far far away from the center of the circle, and they can be made to intersect arbitrarily far away.

 

Now, connect both vertices at the base of the triangle to the midpoints of the opposite sides, where they intersect will be the center of the triangle. Since the sides can be made arbitrarily long, the center of the triangle can be made to be arbitrarily far from the center of the circle.

 

Hence, the center of the triangle depends upon which points on the circle you pick, so that the method you have suggested will not work.

[Johnny5]

yeah. this been answered before.

http://www.scienceforums.net/forums/showthread.php?t=3903

 

heres my solution.

 

 

I don't understand it' date=' can you explain it to me? I read your post there, and didn't follow.

 

ust take four points ABCD. use the normal process to bisect AB and CD. join up the two bisectors and the point where they meet is the centre

[Dave]

Pick two points on the circle fairly close to each other. Draw a line between the two to form a chord. If you can draw a perpendicular line through the centre of your chord, then you know that this line will pass through the centre of the circle. Then you do this again to get the centre.

 

(That's my interpretation anyway).

[syntax252]

If the circle is drawn on paper, carefully fold the paper so that the circumfrence line matches on both sides and run your fingers back and forth along the fold. Then carefully fold it the other way and do the same. Upon unfolding the paper, the creases will cross in the center.

[syntax252]

Actually, the easiest way to do this is to lay the protractor's edge so that it is just tangent to the circle and so that the center mark is coincident to the circle also. Make a dot at this center mark and another dot at the 90 degree point. using the straight edge, draw a line that connects those 2 dots and that passes through the circle.

 

Adjust the compass so that it encompasses more that 1/2 the diameter of the circle and putting the point at one end of the line discribe an arc above and below the line. Moving to the other side of the circle do the same.

 

Using the straight edge, draw a line that connects the points where these arcs cross. Where these 2 lines cross is the center of the circle. (or close enough)

[syntax252]

I misread the opening post. I thought that we were allowed a protractor. My mistake. Therefore.......

 

[Johnny5]

Pick two points on the circle fairly close to each other. Draw a line between the two to form a chord. If you can draw a perpendicular line through the centre of your chord' date=' then you know that this line will pass through the centre of the circle. Then you do this again to get the centre.

 

(That's my interpretation anyway).[/quote']

 

That's almost like what 5614 said. He said draw two tangents, then bisect them, and they will intersect at the center, which will work.

 

Your approach is more like Euclid's, in that he just drew a chord, then a perpendicular to it (through the midpoint), and then bisected a diameter. Obviously, it's easier to draw a chord than a tangent line, which explains why Euclid did it that way.

 

I knew there would be more than one answer to this question. It's nice to see different approaches, and figure out what assumptions go into them.

 

A couple of questions Dave.

 

1. Suppose you already know how to construct a perpendicular to your chord, from a point on the chord. Suppose you already know how to bisect your chord. How can you infer that the infinite straight line through the midpoint of the chord will contain the center of the circle?

 

2. Do you know how to construct a tangent to a circle, from a given point on it?

[Johnny5]

If the circle is drawn on paper, carefully fold the paper so that the circumfrence line matches on both sides and run your fingers back and forth along the fold. Then carefully fold it the other way and do the same. Upon unfolding the paper, the creases will cross in the center.

 

 

Excellent.

 

And this is the point of the thread, that there are many different ways to solve one and the same problem.

 

I didn't think of this one, but then again I was assuming my circles were drawn in the vacuum, traced out by particles, not circles that are drawn on paper. But that was my assumption, not yours.

[Johnny5]

I misread the opening post. I thought that we were allowed a protractor. My mistake. Therefore.......

 

 

This was fantastic. It looks like a sphere. You have arcs intersecting.

 

Can you explain your solution in a bit more detail. Why did you have to ensure that you increased the radius of the compass, so that it would exceed half the distance between any two points?

 

I also like how you referred to a straight line that you didnt draw, as an imaginary one.

[syntax252]

Glad you liked it.

 

Give us another one.....

[Johnny5]

Glad you liked it.

 

Give us another one.....

 

 

Aye

[Algebracus]

When the task is to find the center of the circle using only a compass and an unmarked ruler, is it then really allowed to base oneself upon approximations??

 

A method not based on approximations is to draw two lines, both dividing the circle in two parts. Then you find the midpoints of the two circles, and construct a right angle there. The two new lines cross eachothers in the center of the circle, as can easily be proved.