Amnesia's Dream: June 2026

An infinitesimal change in length \(ab\) produce a similar infinitesimal change at intercepts \(a'\) and \(b'\) and such a change is reversible, change in length \(a'b'\) is continuous as length \(ab\) changes continuously. The change in length between the red and blue centers is smooth as the closed curve is smooth.

Can the two centers be made to meet by changing length \(ab\) or points \(a\) and \(b\)? Can you drive the red center into the blue center by manipulating point \(a\) and \(b\)?

Note: Only when \(ab=a'b'\) guarantees a square; when blue and red are the same circle.

Erratic Hopping Circle On a Close Curve

From the previous post "Circle And Circle In A Closed Curve" dated 11 Jun 2026, the phrase

"The blue center will move erratically, as \(a\) and \(b\) hop along the closed curve."

If it is still possible to make incremental changes at the intercepts \(a'\) and \(b'\) although the change moving points \(a\) and \(b\) are amplified by an angle of rotation, then it is possible to make incremental changes to the position of the blue center. The blue center will not hop, but changes position continuously; a smooth locus is possible.

It is then possible to adjust for \(ab=a'b'\) and so \(aa'bb'\) a square by only adjusting length \(ab\).

One variable (length \(ab\)) for one change (\(ab=a'b'\)).

Can this happen for all closed curve?

Note: Concentric circles do not produce a square.

Circle And Circle In A Closed Curve

Consider,

where \(a'\) and \(b'\) are two points along the perpendicular bisector of the diameter \(ab\), intersecting with the arbitrary closed curve. A blue circle is drawn using \(a'b'\) as diameter. When the lengths \(ab=a'b'\) and the red center of the circle defined by diameter \(ab\) will then coincide with the blue center of the circle define by diameter \(a'b'\), by adjusting point \(a\) and \(b\), we have one circle along the circumference which points \(aa'bb'\) forms a square.

Both red center and blue center will move as \(a\) and \(b\) are moved.

The blue center will move erratically, as \(a'\) and \(b'\) hop along the closed curve. This is because line \(ab\) changes length (thus a new center) and rotates as points \(a\) and \(b\) move. The red center will trace out a smooth locus as \(a\) and \(b\) move along the continuous curve.

Does this method of drawing a square in the closed curve guarantee a solution? That a square is possible always?

Move the red center towards the blue center always, util both centers meet and \(ab=a'b'\).

Plenty of fun moving all over the closed curve.

Note: In this example, as the diagram indicates, move point \(b\) upwards rotates the perpendicular bisector \(a'b'\) and may result in \(ab=a'b'\).

Or move point \(b\) downwards to rotate the perpendicular bisector \(a'b'\) clockwise may result in \(ab=a'b'\).

Wednesday, June 10, 2026

Not A Square

Noooooo...

The need to pass the chord \(a'b'\) through the center of the circle as \(ab\) is adjusted to rotate \(a'b'\) means that the closed curve's continuity does not guarantee smooth operation. This is a case of two changing variables; center and angle.

The 'center' was hidden behind the word 'bisect'.

In this particular case, it maybe possible to move point \(b\) closer to \(a'\) or \(b'\) to form a smaller circle that makes \(ab\) and \(a'b'\) bisect (both set of chord new intersections between circle and the closed curve).

No proof.

Fitting A Square Into A Closed Curve

Given two points \(a\) and \(b\), on a closed curve, it is always possible to draw a circle using length \(ab\) as diameter. This circle will intersect the closed curve at, at least two points. It is possible to move \(a\) and \(b\) along the curve so that the center of the circle is inside the closed curve and the circle insects more points on the curve.

Since the center of the circle is inside the close curve there can be pairs of intersection points (\(a'b'\)) on opposite side of the diameter \(ab\), between the circle and the curve. Make it so.

Then it is possible to rotate/move/adjust the diameter \(ab\), with both points still on the closed curve, that \(a'b'\) is rotated to perpendicularly bisect the diameter \(ab\). (Changing one variable at a time, in this case the angle between the lines \(ab\) and \(a'b'\).)

When \(ab\) is perpendicular to \(a'b'\), \(aa'bb'\) forms a square.

Good night.