# Folding a rectangular paper such that one corner moves along an opposite side

A rectangular paper is folded such that one corner moves along the opposite side. Prove that all the creases formed are tangent to a parabola.

Attempt:

Let the paper be oriented such that its in the first quadrant and has one corner as the origin and sides along the axis. After folding like so:

Let one side of the paper be $a$.

The equation of the crease is $$y=(x-h)\tan\theta$$ Also, in $\Delta LOH$, $$\cos(\pi-2\theta)=-\cos(2\theta)=\frac{h}{a-h}$$ $$-\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{h}{a-h}$$ Substituting the value of $\tan\theta$ from equation of crease, $$ah^2+2(y^2-ax)h+a(x^2-y^2)=0$$

I am not getting anywhere close to proving the statement given.

• Good question truly Dec 2, 2015 at 16:07

This is a basic question to be solved using Mathematical Origami, which was initially developed to serve as a tangible key to mathematical comprehension of geometry shapes.

Let $L_1$ be the line forming the bottom edge of our paper. Let $P_1$ be a point towards the middle, fairly close to $L_1$, and $P_2$ be a point on the left or right edge of our paper. Fold the paper and call the creased line $L_2$.

From the point $P_1$ construct a line which is $\perp$ to the folded portion of $L_1$. Let $X$ be the point where this line intersects $L_2$.

By opening our paper, we observe that the line segment $XP_1$ and the line segment from $X$ to $L_1$, call it $\overline{XA}$, are equal.

Therefore, $X$ is the point on $L_2$ which is equidistant to both $P_1$ and $L_1$. By deﬁnition, this point is on the parabola with focus $P_1$ and directrix $L_1$. $L_2$ is also the $\perp$ bisector of $\angle AXP_1$. Therefore, any point on $L_2$ is equidistant to $A$ and $P_1$.

Choose a point $Y$, on $L_2$ between $P_2$ and $X$. Construct the $\perp$ line to $L_1$ passing through $Y$, call it $\overline{YB}$. Note, that $\triangle YBA$ is right, thus $\overline{YB}<\overline{YA} = \overline{YP_1}$. Since all points of the parabola must be equidistant to both the directrix, $L1$, and the focus, $P1$, we know the parabola lies above $L_2$ at this point.

Similarly, we can show this is true for any point on $L_2$ from $X$ to $L_1$. Thus, $L_2$ is the tangent line to the parabola.[Source]