Previously I posted the Two Ladders Puzzle in which two ladders lean against opposite walls. The length of the ladders is given and so is the height of the point where they intersect. The task was to find the distance between the walls. The problem can be solved using Pythagoras and the law of similar triangles, so really it can classify as an easy problem. Nonetheless, the answer requires the solution of a quartic equation. We begin by naming the lengths of the ladders *a* and *b* and the, as yet unknown heights at which the ladders lean against the walls by *x* and *y*. Also let's divide the distance between the two walls into the parts *m* and *n*. We now have 5 unknowns, so we need 5 equations. The easy one is

*m* + *n* = *d*

Then we have Pythagoras for the two ladders

*a*^{2} = *d*^{2} + *x*^{2}

*b*^{2} = *d*^{2} + *y*^{2}

Finally we have the similar triangles

If we add these two equations, we get the sum (*m* + *n*) / *h* on the right hand side. But *m* + *n* = *d* and thus we can eliminate d to obtain the nice intermediate result

This means that *h* is the **harmonic mean** of *x* and *y*, independent of the values of *a*,*b* or *d*.

The rest of the solution will become a tiny bit messy. Instead of looking for a solution of *d*, we can look first for the values of *x* or *y*. Then we can use Pythagoras to find *d*. To do this we subtract the two Pythagorean equations to eliminate *d*

*x*^{2} − *y*^{2} = *a*^{2} − *b*^{2}

Solving the harmonic equation for *y* we get

With this we can eliminate *y* from the above difference of squares

Obviously this is a fourth order equation for *x*. I know this is not really a nice result so we will have to get some help for solving it. So let's insert the values and stick it into WolframAlpha. Of course there are four possible solutions. We get one positive, one negative and two complex solutions. Only the positive solution of is relevant. Now we can use the first Pythagoras to find the value for *d*

nice puzzle

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