origami folding problem - nose

Question

There is a square sheet of paper with points $A (0:0:0), B (0:1:0), C (1:1:0)$, and $D (1:0:0)$. The paper is folded along the diagonal, aligning point B with point $D$ (the sheet is divided into two polygons, ABC and ADC). Then, the sheet is folded inward at point C (creating a nose) between the polygons at position $C (1.2:0.5:0)$. Points $A\times C \implies O$, $B\times C \implies E$, and $D\times C \implies F$ are added. This results in the larger polygons $AOEB$ and $AOFD$, and the smaller polygons $OEC$ and $OFC$. It is necessary to determine the position of point $C$ when we raise the $AOEB$ polygon by $30^\circ$ (angle of opening) from the tabletop around the edge $AO$, while symmetrically lowering the AOFD polygon by $30^\circ$ with respect to the tabletop around the edge $AO$. Points $A, O$, and $C$ remain in the tabletop plane. A general formula is needed as the position of point $C$ during the folding of the nose and the angle of opening may vary.

Unfolded sheet picture:

Folded sheet picture:

Answer 1

To take point $C=(1,1)$ to point $P=(a,b)$ the sheet of paper must be folded along the perpendicular bisector $FO$ of segment $PC$ (see figure below; as I'm working in the $(x,y)$ plane it's convenient to drop $z$ coordinates). A straightforward computation gives the equation of line $FO$: $$ y-{1+b\over2}=-{a-1\over b-1}\left(x-{1+a\over2}\right). $$ Substituting $y=x$ into that equation we can find the coordinates of $O$: $$ x_O=y_O={a^2+b^2-2\over2(a+b-2)}. $$ And substituting $x=1$ we can find the coordinates of $F$: $$ x_F=1,\quad y_F={a^2+b^2-2a\over2(b-1)}. $$

When quadrilateral $AOFD$ rotates about line $AC$, point $F$ describes a circle with centre $R$ (the projection of $F$ on line $AC$). If $F'$ is the rotated image of $F$ and $Q$ its projection on $(x,y)$ plane, then $RQ=RF\cos\alpha$, where $\alpha$ is the rotation angle. The coordinates of $Q$ are $(1-\delta,y_F+\delta)$, where: $$ \delta={FQ\over\sqrt2}={RF(1-\cos\alpha)\over\sqrt2} ={CF(1-\cos\alpha)\over2}= {2b+2a-a^2-b^2-2\over4(b-1)}(1-\cos\alpha). $$ As a result of the rotation, point $P$ moves to $C'$, staying on the $(x,y)$ plane. Note that $OC'=OP=OC$ and $F'C'=FP=FC=F'C$. The first equality implies that $C'$ lies on the circle through $C$ of centre $O$. The second equality implies that $C'$ lies on the circle through $C$ of centre $Q$.

As the coordinates of $O$, $C$, $Q$ are known, we can write down the equations of those two circles and find the coordinates of $C'$.

I did the task with some help from Mathematica and got:

$$ x_{C'}=\frac{(a+b-2) (a^2+b^2-a-b) \cos ^2(\alpha ) +(a^2+b^2-2a-2b+2) (a-b) \cos (\alpha)+(a-b)^2} {(a+b-2)^2 \cos ^2(\alpha )+(a-b)^2} \\ \ \\ y_{C'}=\frac{(a+b-2) (a^2+b^2-a-b) \cos ^2(\alpha ) -(a^2+b^2-2a-2b+2) (a-b) \cos (\alpha)+(a-b)^2} {(a+b-2)^2 \cos ^2(\alpha )+(a-b)^2} $$

EDIT.

Here's an animated figure, made with GeoGebra, showing the sheet of paper while folding and unfolding.