Generalizing the Crease Length Problem
Let be a convex region in the plane and a point. Thinking of as a piece of paper, fold a boundary point to to form a crease. Given a closed subset of the boundary of , the crease length problem is to determine the shortest and longest creases for .[more]
This generalizes the "classic" crease length problem, where is a rectangular piece of paper and a vertex is folded to the longer opposite edge. (The creases are the same, whether a boundary point is folded to a point or the point is folded to a boundary point.)
Given a parametrization , , of the boundary of , the crease length function for is the length of the crease , .
The boundary set is colored purple, where . You can vary the endpoints and with sliders.
You can vary , the blue point, either with a 2D slider or, on the boundary, using a slider for the boundary parameter. Initially, . You can move the black point , which is folded to , around the boundary with the slider for . Select "fold" to see the fold making the crease and select "crease" to see the crease as a dashed line. The top of the paper is yellow and its underside is red. You can draw up to 100 creases to reveal the pattern of creases for in . Check "show envelope" to show the envelope of the creases, for in . Move the slider all the way to the left and the slider all the way to the right to see the complete envelope, for in (the boundary of ).
Click "graph" to see the graph of the corresponding crease length function and click "both" to see both the region and the graph. A vertical bar shows the crease length when . Note how the crease length function varies as you vary and the dimensions and .
The crease length problem is solved by letting and finding the location of the minimum and maximum values of for in .
Can you eyeball the minimum and maximum crease lengths? Show the crease length graph and use the slider for to approximate the creases for which is a minimum or a maximum for in .
Select "yes" to solve the crease length problem when . The shortest and longest creases are drawn as dashed red and blue segments. When the crease length graph is shown, the values of the minimum and maximum of for in and their parameter locations are displayed and indicated in red and blue.
This Demonstration lets you study the creases, the correspondence between creases and the crease length function, and the solutions to crease length problems when is the interior and boundary of:
(1) the rectangle with vertices , , ,
(2) the triangle with vertices , ,
(3) a "constructible polygon", initially the triangle with vertices , , and
(4) the ellipse
The "classic" crease length problem is the rectangular case with , and . It is studied in detail in the Demonstration "Exploring the Crease Length Problem". The solution depends in a surprising way on . You can use the Demonstration to check that the shortest and longest creases when and are the "opposite corner" crease with and the "upper-left corner" crease with an endpoint at , while when and , the shortest and longest creases are the "vertical" crease with and the 45° crease with an endpoint at .
Choose "constructible polygon" to study crease length functions for a general polygon. Initially, and the polygon is the triangle with vertices , , and . When "crease length function" is not checked, and just the polygon is shown, you can drag the vertices. The vertices are , , and . Ctrl-click to add a new vertex between and , which reparameterizes the boundary so , , and the new point is . When there are vertices, the new vertex becomes . Position the vertices however you want, but make convex. Then click "crease length function" to see the crease length function.
Choose "ellipse" to study creases and crease length functions for . The envelope of the creases is well-known when the ellipse is a circle. See Details.
What can be said about crease length functions? When is a polygon, the local maxima of can occur only where an endpoint of is a vertex, and does not exist at these points. Local minima do occur where and can occur when is a vertex of . Thus, for the crease length problem where is a polygon and , the minimum occurs either when , , where is a vertex of , or where . The maximum occurs either when , , or where an endpoint of is a vertex.[less]
The endpoints of the crease formed by folding to are equidistant from and and therefore satisfy or , which is an equation for the perpendicular bisector of the segment joining and . The endpoints of the crease are found by determining the points where this line intersects the curves (lines for a polygon) defining the boundary of and choosing the ones on the boundary of .
Some problems are easy to solve "by hand". For example, suppose has and for boundary segments. Let and . Using the equation for the perpendicular bisector of and , the crease with endpoints and satisfies
, , .
When is a polygon and is on a linear segment of the boundary, the envelope of the creases is part of the parabola whose focus is and whose directrix is the line containing the linear segment.
When is a circle and is interior to the circle, the envelope of the creases is the ellipse with foci and . When is outside the circle, the envelope is part of the hyperbola with foci and .
Parametric equations of the perpendicular bisector of and are given by
where is the vector rotated counterclockwise 90°. Parametric equations for the envelope of the perpendicular bisectors therefore have the form
and is parallel to .
These are linear equations in and with matrix the transpose of , which can be solved for at nonsingular points of the envelope.
Let denote the ellipse whose center is and whose axes end at and . Then the ellipse has , , and . If is the flipping function, mapping a point to the point symmetric with respect to the perpendicular bisector of and , the folded part of the ellipse is mapped interior to ; is defined by
 L. W. Berman, "Folding Beauties," The College Mathematics Journal, 37(3), 2006 pp. 176–186.
 S. Ellermeyer, "A Closer Look at the Crease Length Problem," Mathematics Magazine, 81(2), 2008 pp. 138–145. www.jstor.org/discover/10.2307/27643095?uid=3739656&uid=2&uid=4&uid=3739256&sid=21102015257303.
 Mathematical Art Galleries. "Sharol Nau." (May 15, 2013) gallery.bridgesmathart.org/exhibitions/2012-joint-mathematics-meetings/sharol-nau.