Falling Ladder

Let the length of the ladder be $w$ , the angle the ladder makes with the ground $\theta$ , and $(x, y)$ be a coordinate on the ladder:

Then by similar triangles, $\frac{y}{w \cos \theta - x} = \frac{w \sin \theta}{w \cos \theta}$ , which simplifies to $y = w \sin \theta - x \tan \theta$ .

For a fixed $x$ , the maximum height $y$ with respect to $\theta$ is when $\frac{dy}{d\theta} = 0$ , so $\frac{dy}{d\theta} = w \cos \theta - x \sec^2 \theta = 0$ , which simplifies to $\cos \theta = \sqrt[3]{\frac{x}{8}}$ .

Then using the identities $\sin^2 \theta + \cos^2 \theta = 1$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$ we have $y = w \sqrt{1 - (\sqrt[3]{\frac{x}{8}})^2} - x \frac{\sqrt{1 - (\sqrt[3]{\frac{x}{8}})^2}}{\sqrt[3]{\frac{x}{8}}}$ which can be rearranged to $x^{\frac{2}{3}} + y^{\frac{2}{3}} = w^{\frac{2}{3}}$ , which is the equation of an astroid .

Therefore, the area $A$ swept by a falling $8\text{-ft}$ ladder is the same as the area of an astroid in the first quadrant, which is $A = \frac{1}{4} \frac{3}{8} \pi 8^2 = 6 \pi$ , and so $n = \boxed{6}$ .

This is nearly identical to my "Squeegees" problem that I posted last year! Here is a link to that problem, which has a few relevant solutions...