Let us take a closer look at (3) by considering the fourth degree polynomium Q(x)=x3(x−c)−1. Note first that c is positive. Next that Q is decreasing for x<0 and 0<x<3c/4 and increasing for x>3c/4 (seen from the derivative Q′(x)=x2(4x−3c)). And since Q(0)=−1 we then know that Qalways has exactly two zeros, one negative and one positive. We are naturally interested in the positive one.
Assume now that we have found the (positive) root x. We then have from (4) and (1),
And that is the width we were looking for. Note, however, that if p>a then no solution is possible. This can happen, for example, if the crossing height h is larger than the shortest ladder.
But how do we solve (3) and find x? It is possible to solve a quartic equation analytically, so let us turn to Abramowitz and Stegun and solve it. By carefully inserting to retrieve the positive root we get
That concludes the solution to the crossed ladders problem. Let us finish by inserting some numbers.
From the initial illustration of the problem, we have a=12, b=10 and h=5. This gives us
There are actually lengths that have an integer solution. Take, for instance, a=119, b=70, h=30 and w=56.
A simple, but tricky, problem.
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