This Geometry problem seems like it should be easier.

Question

Two ladders are located in and alley between two buildings. They are leaned up agains the buildings such that they take up the full width of the alleyway and cross at a point 12 ft above the ground. One ladder is 40 ft long. The other ladder is 30 ft long.

How wide is the alley?

I was able to find a quartic equation that gave the height above the ground where the 30' ladder meets the building.

$x^4 - 24x^3 + 700x^2 - 16800 x + 100800 = 0$

I did not see an easy way to solve this, and plugged it into wolfram alpha. WA gives a solution to this quartic of 19.000. But, 19 can't be a root by the rational root theorem... The root is very nearly, but not exactly 19.

And then we can use the Pythagorean theorem to find that $w\approx \sqrt {539}$

On the face of it this problem seems like it should not require quartic equations. I am curious to know if anyone has a simpler solution, and even better an exact solution.

Answer 1

You say that you don't think that this problem should require quartic equations, but you also seem to be saying that the answer is a quartic number (i.e. an algebraic number of degree four over the rationals): You say it's not exactly equal to a rational number, and also that it's not the square root of any rational number.

I didn't check your equations, but if the previous assertions are true then how could the answer not involve a quartic equation?

Of course potentially you might be able to write the answer in terms of some trig functions instead of in terms of polynomial equations, but typically we consider an expression in radicals to be "better" than an expression in terms of transcendental functions.

Answer 2

This is the old crossed ladders problem, and the resolution of a quartic is unavoidable in the generic case.

$\frac w{10}$ solves $$x^8-\frac{1106}{25}x^6+697x^4-\frac{114264}{25}x^2+\frac{6543504}{625}=0$$ which is irreducible and has generic Galois group ($S_4$ with respect to $x^2$), so it cannot be simplified beyond what the quartic formulas provide.

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More generally, suppose the ladder lengths are $a$ and $b$, they cross at height $h$ and the alley width is $w$. Let $A=a^2$, $B=b^2$, $H=h^2$, $W=w^2$, $S=A+B$ and $P=AB$. Then $W$ satisfies the (homogeneous in $A,B,H,W$ and symmetric in $A,B$) quartic $$\begin{align} &W^4\\ +(4H-2S)&W^3\\ +(S^2+2P-6HS)&W^2\\ +(2H(S^2+2P)-2SP)&W\\ +(H^2(S^2-4P)-2HSP+P^2)&=0 \end{align}$$ For $a=30,b=40,h=12$ this yields $w=23.21673746900\dots$