Square inscribed in a circle (with a ruler only)

Inscribe a square in a given circle by following the rules of construction with ruler and compass but... without using the compass.
The center point of the circle is given too.

• How can you follow the rules of construction with "ruler and compass" if there is no compass? Commented Aug 20, 2015 at 22:40
• @TomCarpenter Probably by not breaking those rules even if you don't have or use a compass. Simply saying you may have a compass just don't use it, and use only a ruler. Commented Aug 20, 2015 at 22:49
• Is this meant to be exact, or as good as one could regularly do with pencil and paper? Commented Aug 21, 2015 at 0:28
• en.wikipedia.org/wiki/Poncelet%E2%80%93Steiner_theorem
– f''
Commented Aug 21, 2015 at 1:25
• @f'' saw your comment only after I posted. Commented Aug 21, 2015 at 2:36

Name the center of the circle $O$. Draw a diameter $AOB$. Pick another point $C$ on the circle. By Thale's theorem $AC\perp BC$.

Use the method described here to draw lines parallel to $AB$ and $AC$ through $O$. The intersections of these lines with the circle form a square.

• Linking to a solution is suboptimal since the link may be broken at any time. Can you please complete your solution instead of (or in addition to) offering a link? Links to university user space are particularly volatle. Commented Aug 21, 2015 at 18:26
• AB and AC aren't necessarily perpendicular. Did you mean AC and BC in the second part of your construction? Assuming so, could you elaborate, please? Maybe I didn't understand the linked document properly, but it looks like you would need the midpoint of AC to construct the parallel line (through O), likewise with BC. Commented Aug 23, 2015 at 13:01
• An alternate construction that follows the first part of the linked document closely is to start with line AB through O, construct XY parallel to AB (using O the midpoint of AB), then find X' and Y' by continuing the lines XO and YO until they hit the circle. Then AB bisects X'Y at (say) Z. With Z as the midpoint of X'Y, construct the line PQ parallel to X'Y through O, with P and Q on the circle. AB and PQ are perpendicular to each other and APBQ is a square. Commented Aug 23, 2015 at 13:06

We know this is possible

For those that don't know the classic construction with straightedge and compass: Perpendicular through the center, connect the points where the lines and arcs meet.

Poncelet-Steiner shows that straightedge and compass constructions can be done with a straightedge, arbitrary circle and centerpoint:

We can get parallel lines, perpendiculars and transfer lengths. See how here.

Please note, to construct the perpendicular through the center is not directly possible. You need to construct off center through another point first and then transfer it.

Assumption made: Rectangular ruler.

If the ruler is a rectangle, put the corner of the ruler on the center dot,spin around to draw the circle. Then at some arbitrary angle trace along the two sides of the ruler. They are perpindicular, now extend those lines to reach the circle forwards and backwards, and then connect the dots where the lines intersect the circles. That is a square.

• I think the circle already exists; you don't create it. And you might need to explain better how extending two perpendiculars till they hit the circle guarantees you'll get a square. To me, it just seems you'll get a rectangle. Commented Aug 21, 2015 at 2:28