I was recently reading a newspaper in the train and there was a little math riddle, I thought "how funny, that's gonna be easy, let's do it" and yet...

The problem goes as follow : in a barn, there is a 1 meter cubic box against a wall and a 4 meter ladder is leaning against the wall, touching the box at its corner. Here is a picture :

Illustration of the situation
Illustration

So, the big triangle has a hypotenuse \(FE\) of \(4\), the square \(ABDC\) has sides of length \(1\) and is basically "insquared" at the right angle, i.e. \(D\in \overline{FE}\). The question is "what is the length of the biggest cathetus", here \(AF\).

So far, no problem.

Now here are my solutions:

  • By Thales' intercept theorem, \(\frac{FB}{FA}=\frac{BD}{AE}\), by hypothesis, \(FB=FA-1\) and \(BD=1\). Now by Pythagoras, \(FA^2+AE^2=FE^2\); by hypothesis, \(FE=4\), so we end up with a system of equations, letting \(h=FA, d=AE\):

\[ \begin{aligned} \frac{h-1}{h}&=\frac{1}{d} \\[2ex] h^2+d^2&=4^2 \end{aligned} \]

Which solves (removing 3 non-relevant solutions) into \(d \cong 1.3622\) and \(h \cong 3.76091\).

  • Now, if I consider the "function" of the line : \(f(x)=\frac{-h}{d}x+h\), I know that \(f(1)=1\) and I end up with Pythagoras with the system :

\[ \begin{aligned} \frac{-h}{d}+h&=1 \\[2ex] h^2+d^2&=4^2 \end{aligned} \]

it solves again into the same, again removing 3 non-relevant solutions

Okay, this means that using Pythagoras is no good since it ends up giving a quartic equation (4 answers, of which 3 are "non-relevant").

  • Now if I consider the length of the arc \(f(x)\) between \(0\) and \(d\) it has to be \(4\) and again \(f(1)=1\) I end up with the system:

\[ \begin{aligned} \frac{-h}{d}+h&=1 \\ \int_0^d \sqrt{1+(f'(x))^2} dx &=\int_0^d \sqrt{1+\left(\frac{-h}{d}\right)^2} dx = d \sqrt{1+\left(\frac{-h}{d}\right)^2} \end{aligned} \]

Which solves again into the same answers, but this time removing only 2 non-relevant solutions (i.e. it gives a cubic equation instead of a quartic).

I tried also using the areas and the smaller trangles \(FAD\) and \(AED\) for example :

\[\frac{h \cdot d}{2} = \frac{h\cdot 1}{2}+\frac{d\cdot 1}{2}\]

Yet I wasn't able to get to any "hand solvable" solution : if I were able to bring it down to some quadratic equation, that would be nice, since it is a common assumption, here, that everybody has seen the "general formula for solving quadratic equations" in school and so would be able to solve this, I may then see how it is seen as a funny riddle in the newspaper.

My best trial, with "just" a cubic equation, is way too complicated for the normal readers of this newspaper, so it's bugging me.

So, I asked around! And somebody proposed me the following solution, using a clever variable change:

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 he preferred to move one factor \(x\) from the former factor on the right hand side to the latter, so

\[ 16=(x+\frac1x)(x+2+\frac1x). \]

Almost there: write \(u=x+1/x\). We can solve \(u\) from the quadratic

\[ 16=u(u+2), \]

and then solve for \(x\) from the equation

\[ x+\frac1x=u. \]