Sequence of folds for finding intersection of two circles, given centers/radii I know that any ratio that can be constructed by use of a straightedge and compass (and some which cannot) can be constructed by folding paper. I am not certain whether or not the same is true of points or lines.
Given four points, $A_1$, $B_1$, $A_2$ and $B_2$, so that $A_1B_1$ and $A_2B_2$ are parallel and the circles centered on $A_1$ and $A_2$ that pass through $B_1$ and $B_2$ intersect at some point $C$, is it possible to construct point $C$ using paper folding? If so, how?
 A: EDIT: I've found a much simpler solution with fewer branches. I'm leaving the old image up until I can build a new one.
Without loss of generality, assume that $A_1A_2$ has length 1, and say $A_1B_1$ and $A_2B_2$ have lengths $a$ and $b$ respectively with $b \leq a$. The triangle $A_1A_2C$ has altitude (from point $C$) of length $h$:
$$h=\frac{1}{2}\sqrt{(a+b+1)(a+b-1)(a-b+1)(b-a+1)}$$
Note that:
$$(a+b+1)(a+b-1)(a-b+1)(b-a+1)
\\= ((a+b)^2-1)(1-(a-b)^2)
\\= (a^2+b^2+2ab) + (a^2+b^2-2ab) - ((a-b)(a+b))^2 - 1
\\= 4a^2 - (2(a^2-b^2) + (a^2-b^2)^2 + 1)
\\= (2a)^2 - (a^2-b^2+1)^2$$
So
$$h=\frac{1}{2}\sqrt{(2a)^2-(a^2-b^2+1)^2}$$
Say that $a^2 - b^2 = c$. A right triangle thus exists with hypotenuse $2a$ and legs $c+1$ and $2h$, so another exists with side lengths $a$, $\frac{c+1}{2}$, and $h$.
Also note that a right triangle exists with hypotenuse $a$ and legs $b$ and $\sqrt{c}$.
Getting back to the construction, to find $C$, follow these steps:


*

*Mark the line through $B_2$ perpendicular to $A_2B_2$

*Fold $B_1$ onto that line, and mark the line through $B_1$'s folded image perpendicular to $A_1A_2$ before unfolding. Mark point $D$ where the new line and $A_1A_2$ intersect.

*Note that $A_1D$ has length $\sqrt{c}$. Fold along $A_1B_1$ and mark the point $E$ at the folded image of whichever of $A_2$ and $D$ is farther from $A_1$ before unfolding.
4a. If $A_2$ was farther from $A_1$ than $D$ (so $1>\sqrt{c-1}$) fold the line through $A_1$ that places $E$ onto the perpendicular line from step 2, then mark the line perpendicular to the folded image of $A_1E$ through $D$ and mark its intersection $F$ with $A_1E$ before unfolding.
4b. Otherwise, fold the line through $A_1 that places $E$ onto $A_2B_2$ and mark point $F$ where the folded image of $A_1E$ intersects the perpendicular line from step 2 before unfolding.

*In both cases, $A_1F$ has length $c$, so $A_2F$ has length $c+1$. Fold $A_2$ to $A_1$, then fold $F$ to $A_1$ and $A_2$ and mark the line perpendicular to $A_1A_2$ through the point $G$ where the last fold intersects the folded image of $A_1A_2$ opposite $A_2$ from $A_1$ before unfolding everything.

*Now $A_1G$ has length $\frac{c+1}{2}$. Fold $B_1$ onto the perpendicular line through $G$ and mark its folded image $H$ before unfolding.
The right triangle $A_1GH$ has hypotenuse $a$ and one side length $\frac{c+1}{2}$, so the height must be $h$ and $H$ is the point $C$ we were searching for.
I have an image of the entire process (EDIT: currently using an older,less efficient process)

