[bonanova]

More than 200 years ago it was proved that any construction

possible with a compass and straight-edge can be accomplished

with a compass alone.

It is a simple matter to find the midpoint of a line segment using

a compass and a straight-edge. Find the midpoint of a line segment

using a compass alone, without using a straight-edge. Or Google.

To compare methods, consider that you are given the two points

[0, 0] and [0, 1]. You are to locate the point [0, .5].

Bonus for the fewest steps.

[Guest]

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Mascheroni. Lol, It will be nifty to see if anyone gets below seven circles.

Edited April 2, 2009 by Orykle

[Guest]

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After drawing two semicircles whose centers are the end points of the line; an arc from their intersection, tangent to line will touch at midpoint of line.

This is only a suggestion, I think a touch of tangent isn't a valid way to define a point; an intersection of two arcs is always necessary. I posted this to confirm that this tangent method is valid or not?

[Guest]

I believe angles maybe required here as well, like 30deg, 60deg etc.

Just my 2 cents

[Guest]

You can't actually do every construction as you can't draw a straight line, but for most simple cases, the statement is true.

I feel like cheating since I've known this for a long time, but I don't know if it's the shortest steps.

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Let's make two points A and B and we want to find the midpoint of the segment AB. First we draw a circle centered at B with radius AB. Now with the same radius, we put center at A and mark an arc so that it intersects the circle at point C. Repeat this process with C as center to mark D, and D as center to mark E. Now AB is equal to BE and all three points are on the same line (you basically doubled AB).

Now draw a circle centered at A with radius AB. Draw an arc centered at E with radius AE to mark two intersections to circle A. Let's call these two intersection F and G. Now make two arcs, centered at F and G respectively, with radius AF (or AG as they are the same). These two arcs will have an intersection besides A and that is the midpoint of AB.

Edited April 2, 2009 by sihyunie

[bonanova]

  nobody said:

  Reveal hidden contents

After drawing two semicircles whose centers are the end points of the line; an arc from their intersection, tangent to line will touch at midpoint of line.

This is only a suggestion, I think a touch of tangent isn't a valid way to define a point; an intersection of two arcs is always necessary. I posted this to confirm that this tangent method is valid or not?

What if the line is not there - only the two points?

In theory your method works, but its accuracy suffers in practice

[given the line is there] because of tangency rather then intersection.

So if the line is there, I guess a mathematician would say this is a solution,

but an engineer would not like it ...

[bonanova]

  sihyunie said:

You can't actually do every construction as you can't draw a straight line, but for most simple cases, the statement is true.

I feel like cheating since I've known this for a long time, but I don't know if it's the shortest steps.

  Reveal hidden contents

Let's make two points A and B and we want to find the midpoint of the segment AB. First we draw a circle centered at B with radius AB. Now with the same radius, we put center at A and mark an arc so that it intersects the circle at point C. Repeat this process with C as center to mark D, and D as center to mark E. Now AB is equal to BE and all three points are on the same line (you basically doubled AB).

Now draw a circle centered at A with radius AB. Draw an arc centered at E with radius AE to mark two intersections to circle A. Let's call these two intersection F and G. Now make two arcs, centered at F and G respectively, with radius AF (or AG as they are the same). These two arcs will have an intersection besides A and that is the midpoint of AB.

Yeah, that's it.

[Guest]

For those who have never seen it, the WIMS compass-and-ruler program is really interesting. You can find it here. I used it to create...

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1 & 2 are the points that we started with. 9 is the midpoint.

Labeled as sihyunie indicated:

1 = A

2 = B

4 = C

5 = D

6 = E

7 & 8 = F & G

[bonanova]

  jb_riddler said:

For those who have never seen it, the WIMS compass-and-ruler program is really interesting. You can find it here. I used it to create...

  Reveal hidden contents

1 & 2 are the points that we started with. 9 is the midpoint.

Labeled as sihyunie indicated:

1 = A

2 = B

4 = C

5 = D

6 = E

7 & 8 = F & G

Thanks for the sketch - and the drawing program tip.

[HoustonHokie]

  bonanova said:

Thanks for the sketch - and the drawing program tip.

And especially thanks for the final pieces to a puzzle that's bugged me for four months now. Way to go

[Guest]

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Proof for this construction comes from the fact that 2 isosceles triangles, 189 and 168 in the picture, are similar.

If you look at triangle 168, 18 = 12, so the long sides of the triangles are twice as long as the short base.

Same applies to 189 since it's similar to 168, so 18 is twice as long as 19, which also implies that 12 is twice 19 as well, this we have a midpoint.

Edited April 7, 2009 by sihyunie