The position of a ladder leaning against a wall and touching a box under it
Let $|FB|=x$. By similarity of triangles we then have $|CE|=1/x$. Pythagoras thus gives $$ 4=\sqrt{x^2+1}+\sqrt{1+(1/x)^2}=\sqrt{x^2+1}(1+\frac1x). $$ Squaring this gives us $$ 16=(x^2+1)(1+\frac2x+\frac1{x^2}), $$ but I prefer to move one factor $x$ from the former factor on r.h.s. to the latter, so $$ 16=(x+\frac1x)(x+2+\frac1x). $$ Getting warmer! Write $u=x+1/x$. We can solve $u$ from the quadratic $$ 16=u(u+2), $$ and then solve for $x$ from the equation $$ x+\frac1x=u. $$
It is clear to discard the negative possibility for $u$. For the positive value of $u$ the two solutions for $x$ are reciprocals of each other. They correspond to "physical" solutions gotten from each other by reflecting the entire picture w.r.t. the diagonal $AD$ at 45 degree with the floor.
EDIT (5.5 years later!)
Had I known my solution would be featured in the video "Ladder and Box Problem" on Presh Talwalkar's "Mind Your Decisions" YouTube channel (thanks, Presh!), I would've made another pass.
Shunting the original solution to this answer's edit history, here's something of a streamlined presentation.
From a length-$b$ ladder resting against a side-$r$ box, we can make the following diagram:
Then we have:
$$\left(\;\color{green}{p + q + r} \;\right)^2 = \color{blue}{b}^2 + \color{red}{r}^2 \qquad\to\qquad p + q + r = \sqrt{b^2 + r^2} \tag{1}$$
(discarding the negative root). The semicircle recalls the classical construction of the geometric mean, and we have $$pq=r^2 \tag{2}$$
Now, $p$ and $q$ are the roots of the quadratic $$0 \;=\; (x-p)(x-q) \;=\;x^2 - (p+q)x + p q \;=\; x^2 - (-r + \sqrt{b^2+r^2})x + r^2 \tag{3}$$
Solving yields
$$\{p,q\} = \frac12 \left(-r + \sqrt{b^2 + r^2} \pm \sqrt{ b^2 - 2 r^2 - 2 r \sqrt{b^2 + r^2}} \right) \tag{$\star$}$$
For $r=1$ and $b=4$, this gives
$$\{p,q\} = \frac12 \left(-1 + \sqrt{17} \pm \sqrt{ 14 - 2\sqrt{17}} \right)$$
Calculating the "bigger cathetus" from this is left as an exercise to the reader.
Consider the equation of the line that the ladder makes:
$$\mathcal{l} = \{ (x, y) ~:~ y = mx + h \}$$
where $h$ is the height of the ladder on the wall and $m$ is the slope of the ladder. We know that $(1, 1)$ is on the line, so
$$1 = m + h$$
And we know that the distance from the x-intersecpt to the y-intersept is $4$. So
$$h^2 + (-h/m)^2 = 4^2$$
So solve the last 2 equations for $h$:
$$h^2 + \left(\frac{h}{1-h}\right)^2 = 4^2$$ $$h^4 - 2h^3 - 14 h^2 + 32h - 16 = 0$$
So the problem is a quartic. It doesn't really have a simple answer (what was this newspaper thinking?), but the one you want is:
$$h = \frac{ \sqrt{14 - 2 \sqrt{17}} + \sqrt{17} + 1 } {2} \approx 3.76$$