A 2x4 in a doorway This problem has puzzled me for years.  I've asked students, teachers and engineers and none of them have solved it.  It looks simple, but as of yet, no solution.  I can come up with several equations, but I can't reduce them down to a solution.
Imagine a doorway of height "A" and width "B".  You have a board of width "w".  You want to place the board in the door frame and run it corner to corner.  The board must just fit (too long and it won't fit inside the door way, too short and it doesn't run corner to corner).  I'm looking for the general solution so that it could be used for any door size and a board of any width.  You can introduce any number of angles to solve this, but your final solution must be purely in terms of "A", "B", and "w".
I would love to see a worked solution (or a layman's explanation as to why there isn't one).
Have fun!
I had a diagram drawn, but I'm too new to this site so it couldn't be posted
 A: If you let $t$ be the angle created by the door in degrees, the diagram below shows the start.  There are lots of $90-t-(90-t)$ triangles about.  Now you have $B-w\sin t=L \cos t, A-w \cos t = L \sin t, \sin^2t+\cos^2 t=1$, two equations in $\sin t, L$.
The practical answer is to say $w \ll A,B$, in which case you can take $\tan t=\frac AB$ You can't cut accurately enough to care about the corrections.    Then $\sin t=\frac {\frac AB} {\sqrt{1+(\frac AB )^2}}$ $\cos t= \frac 1 {\sqrt{1+(\frac AB )^2}}$  Now divide the equations to get $L \tan t$ and you are there.

A: Using Ross's symbols, I wrote 
$$  x = w \sin t, \; \; y = w \cos t.   $$
Now, there is a solution in radicals for $x,$ but it is very unpleasant because we arrive at
$$ 4 x^4  - 4 B x^3 + (A^2 + B^2 - 4 w^2) x^2 + 2 B w^2 x + (w^2 - A^2) w^2 = 0.   $$
Oh, note
$$  x^2 + y^2 = w^2  $$ and
$$  L^2 = (A-y)^2 + (B-x)^2 = A^2 + B^2 + w^2 - 2 A y - 2 B x.  $$
So, while there is a method for writing the value of $x$ using radicals, see QUARTIC, it is just not going to do you any good. I would solve for $x$ numerically, then find $y$ from $y = \sqrt{w^2 - x^2},$ then find $L.$
