The Bi-Fold Door Problem

The following is a well-traveled problem that has appeared in both the Mathematics Magazine (June 1993, p193) and The Mathematics Teacher (October 1993 article and November 1993 Letters):

A bi-fold closet door consists of two one-foot-wide panels, hinged at point P. One of the panels is fixed at the point O. Assume that as the endpoint Q moves to the right, the door rubs against a thick carpet. What shape will be swept out on the carpet?


Figure 1

Figure 1 shows that when the door first starts to close, the path of the boundary curve follows the hinge of the door, and therefore the curve is part of a circle of radius 1. After some point, however, the hinge of the door lies below the curve as the carpet is brushed by the right-hand side of the door. In particular, this means that along this part of the curve, the edge of the door is tangent to the curve; that is, the boundary of the brushed carpet is determined by the envelope of tangent line segments represented by the right-hand side of the door.


Figure 2

Consider the position of the door when it has been opened to an angle theta as illustrated in figure 2. The right side of the door is tangent to the curve we seek at some point (x(theta),y(theta)). In the figure we have t = 2*sin(theta) and k = cos(theta) and so the slope of the right-hand side of the door is -2*k/t = -cot(theta). The equation of the line represented by this side of the door is therefore

formula for y for theta

Next consider what happens to the door when it is opened by an additional angle h as shown in figure 2. The equation of the line representing this new door position is

formula for y for theta+h

The intersection of these two lines occurs at the x-value

formula for x(h)

This intersection point will converge to the point (x(theta),y(theta)) as h approaches 0. Thus

formula for x(theta)

(An application of the definition of the derivative!) Of course, this is only valid if the two line segments intersect at an x-value to the right of the first door hinge, which means we must have 2(sin(theta))^3 .gt. sin(theta). Thus this parameterization of the curve is valid for theta >= pi/4, or when x .gt. 1/sqrt(8).

Finally, we get from the equation of the line found earlier that

formula for y(theta)

For 1/sqrt(8) .le. x .le. 2 the curve is therefore given by

(x/2)^(2/3) + (y/2)^(2/3) = 1

or

y = (2^(2/3) - x^(2/3))^(3/2)

For 0 .le. x .le. 1/sqrt(8), the curve is just part of a circle of radius 1 centered at the origin, and so

final solution


Larry Riddle, November, 1993

(Converted to HTML, February 1996)