I was reading a newspaper and there was a little math riddle, I thought "how funny, that's gonna be easy, let's do it" and here am I...

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

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 $h=FA, d=AE$: $$ \begin{align} &\frac{h-1}{h}=\frac{1}{d} \\ &h^2+d^2=4^2 \end{align} $$ 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 :$$ \begin{align} &\frac{-h}{d}+h=1 \\ &h^2+d^2=4^2 \end{align} $$ 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: $$ \begin{align} &\frac{-h}{d}+h=1 \\ &\int_0^d \sqrt{1+(f'(x))^2} dx =\int_0^d \sqrt{1+\left(\frac{-h}{d}\right)^2} dx = d \sqrt{1+\left(\frac{-h}{d}\right)^2} \end{align} $$ 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.

What am I missing? Some basic properties maybe? It's really bugging me, not being able to solve this without Wolfram.

  • 3
    possible duplicate of Ladder against a wall. – mvw Jun 30 '15 at 19:42
  • 2
    I am not sure whether the question should be closed as a duplicate. There is a difference between: "I've got this equation from a geometry problem, how do I solve it?" and "How do I solve this geometry problem?" But even if they should be considered duplicates, then the opposite direction would make much more sense in my opinion, since the newer question contains much more detail. – Martin Sleziak Jun 30 '15 at 20:24
  • 1
    I think I should reformulate my question. The question isn't "How to solve this geometric problem", nor "How to solve this equation", but more like : "here are 3 solutions, of which 2 are quartic and one cubic, how could I find an easier solution, possibly a quadratic one?" – Lery Jun 30 '15 at 21:16
  • 2
    The assertion in the title seems to be false; perhaps it should be amended to something like "not obviously solvable by hand" – Glen_b Jul 1 '15 at 2:54
up vote 14 down vote accepted

Consider the following, where we see a full barn (of a curious gravitational nature) with four ladders of length $\color{green}{g}$ leaning against side-$w$ boxes stuck in the corners where floor and ceiling meet walls:

enter image description here

Clearly,

$$\begin{align} (w+\color{violet}{p})^2 &= W + \color{violet}{P} + 2 \color{green}{G} + 2 \color{blue}{B} + 2 \color{red}{R} \\ &= W + \color{violet}{P} + 2 \color{green}{G} + 2 ( \color{blue}{B} + \color{red}{R} ) \\ &= W + \color{violet}{P} + 2 \color{green}{G} + 2 \color{green}{G} \\ &= W + ( \color{violet}{P} + 4 \color{green}{G} ) \\ &= w^2 + \color{green}{g}^2 \end{align}$$ $$\implies \quad w + \color{violet}{p} = \sqrt{w^2+\color{green}{g}^2} \quad\implies\quad \color{violet}{p} = -w + \sqrt{w^2+\color{green}{g}^2}$$

(where we ignore the negative square root). Also, $$\frac{\color{red}{r}}{w} = \frac{w}{\color{blue}{b}} \quad\implies\quad \color{red}{r} = \frac{w^2}{\color{blue}{b}} \quad\implies\quad \color{violet}{p} = \color{blue}{b} + \color{red}{r} = \color{blue}{b} + \frac{w^2}{\color{blue}{b}}$$

so that

$$\color{blue}{b} = \frac12\left(\color{violet}{p}\pm\sqrt{\color{violet}{p}^2-4w^2}\right) \quad\implies\quad \color{blue}{b} = \frac12\left(\color{violet}{p}+\sqrt{\color{violet}{p}^2-4w^2}\right)$$

(where we take the larger root, because the smaller root corresponds to the (presumably) smaller length, $\color{red}{r}$).

Therefore, with $w=1$ and $\color{green}{g}= 4$,

$$\begin{align} \color{violet}{p} &= -1 + \sqrt{17} \\[4pt] \color{blue}{b} &= \frac12\left(-1+\sqrt{17} + \sqrt{14-2\sqrt{17}}\right) \\[4pt] \color{red}{r} &= \frac12\left(-1+\sqrt{17} - \sqrt{14-2\sqrt{17}} \right) \end{align}$$

  • Oh, a really nice, "graphical" one. I think it may be the simpler one (hence the answer to my question). – Lery Jul 1 '15 at 6:12
  • This makes me wonder if a similar graphical construction can be made to demonstrate the general solution to a quartic. I'm guessing not, since it bypasses the cubic component, but it would be nice to see. – DanielV Jul 1 '15 at 15:56
  • @DanielV Actually they happened to do it so, in the "Ancient time". Looks for info about al'Khayyam for the cubic equations, for example. (But it's made using more complicated stuff than "just" squares and triangles) – Lery Jul 1 '15 at 17:03

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.

  • The last paragraph is the key in a way. We do have a quartic, but we also know that $1/x$ is a solution whenever $x$ is. Equivalently, the quartic polynomial is palindromic (=equal to its own reciprocal polynomial). This is a tell-tale sign to write it in terms of $u$. – Jyrki Lahtonen Jul 1 '15 at 5:10
  • Yeah, I saw your answer yesterday, but hadn't time to try and solve it by hand. But the variable's change is clever, I've never got an eye for those. – Lery Jul 1 '15 at 5:59

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:

$$h = \frac{ \sqrt{14 - 2 \sqrt{17}} + \sqrt{17} + 1 } {2} \approx 3.76$$

Expanding on the original diagram (see below), observe that if we draw a circle centred at point $O$ with $\overline{EF}$ as a diameter, then point $A$ is on the circle. We also construct point G such that $\overline{OG} \perp \overline{EF}$. Furthermore, since $\angle{FOG}$ is a right-angle, $\angle{GAF}$ is $45^\circ$ as labeled, so points $A,D,G$ are collinear.

Using the fact that the power of point D relative to the circle is invariant, we have

$(AD)(DG) = (FD)(DE) = (FO + OD)(OE - OD)$

Since $AD = \sqrt{2}$ and $OF=OG=2$, then $DG = 2 \sec{\theta}, OD = 2 \tan{\theta}$, so

$\sqrt{2} (2 \sec{\theta}) = (2 + 2 \tan{\theta})(2 - 2 \tan{\theta}) = 4 - 4 \tan^{2}{\theta} = 8 - 4 \sec^{2}{\theta}$.

Put $w = \sec{\theta}$ to reduce to $4 w^2 + 2\sqrt{2}w - 8 = 0$ and taking the positive root $w = \frac{-\sqrt{2} + \sqrt{34}}{4}$ from which $\cos{\theta} = \frac{1}{w} = \frac{4}{\sqrt{34} - \sqrt{2}}$, so $\theta \approx 0.437896 \text{ rad }$ and $AF = 4 \cos{(\frac{\pi}{4} - \theta)} \approx 3.7609$

Elaboration of original diagram showing additional geometry

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