A hallway of width A turns through 90 degrees into a hallway of width B.

geometry/corner

Problem

A hallway of width A turns through 90 degrees into a hallway of width
B. A ladder is to be passed around the corner. If the movement is
within the horizontal plane, what is the maximum length of the ladder?

Show solution

------\---------
B   /  \.......|..B/sin(theta)
   theta\      |
---|-----X     |
         |\    |
         | \...|..A/cos(theta)
         |  \  |
         |   \ |
         | A  \|


Theta is the angle off horizontal.

Minimize length = A/cos(theta) + B/sin(theta)

    d(length)/d(theta)
           = A*sin(theta)/cos(theta)^2 - B*cos(theta)/sin(theta)^2 (?)
           = 0
    A*sin(theta)/cos(theta)^2 = B*cos(theta)/sin(theta)^2

    B/A = sin(theta)^3/cos(theta()^3 = tan(theta)^3

    theta = inverse_tan(cube_root(B/A))

If you use the trigonometric formulas  cos^2 x = 1/(1 + tan^2 x)
and  sin x = tan x cos x, and plug through the algebra, I believe
that the formula for the length reduces to

(A^(2/3) + B^(2/3))^(3/2)

At any rate this is symmetric in A and B as one would expect, and
has the right values at A=B and as either A-->0 or B-->0.