A horizontal plane is ruled by parallel lines at distance 2a…
A horizontal plane is ruled by parallel lines at distance $2a$ from each other. A convex contour whose greatest diameter is less than $2a$ is thrown on the plane at random. Find the probability that the contour intersects one of the lines.
Explanation of Method
The position of the contour is described by its orientation $\theta$ (uniform) and the position of a reference point relative to the lines (uniform across a strip of width $2a$). For a fixed orientation the contour has a certain width $w(\theta)$ in the direction perpendicular to the lines, and it meets a line with probability $w(\theta)/(2a)$. Averaging over $\theta$ gives the mean width, and Cauchy's formula says that the mean width of a convex figure equals its perimeter divided by $\pi$. So the answer is $s/(2\pi a)$, where $s$ is the perimeter.
Step 1: The random position
Because the lines repeat every $2a$, only the position within one strip matters. "Thrown at random" means:
- the orientation angle $\theta$ of the contour is uniform on $[0,\pi)$;
- for a fixed orientation, the distance $y$ from the lowest point of the contour to the line just below it is uniform on $[0,2a)$.
Step 2: A fixed orientation
For orientation $\theta$ let $w(\theta)$ be the width of the contour perpendicular to the lines (the distance between the two supporting lines parallel to the ruling). Since the diameter is less than $2a$, $w(\theta)\lt2a$, so the contour meets at most one line. It meets the line above it exactly when $y+w(\theta)\ge 2a$, i.e. $y\ge 2a-w(\theta)$:
\[ P(\text{intersection}\mid\theta) = \frac{w(\theta)}{2a}. \]
Step 3: Average over the orientation
\[ P = \frac1\pi\int_0^\pi\frac{w(\theta)}{2a}\,d\theta = \frac{\bar w}{2a}, \qquad \bar w = \frac1\pi\int_0^\pi w(\theta)\,d\theta \ \ (\text{mean width}). \]
Step 4: Cauchy's formula $\bar w = s/\pi$
For a convex polygon with sides $s_1,\dots,s_m$ (perimeter $s=\sum s_i$): projecting the polygon onto a line perpendicular to the ruling, the boundary covers the projection exactly twice (once going "up", once "down"), so
\[ w(\theta) = \frac12\sum_{i=1}^m s_i\,|\cos(\theta-\varphi_i)|, \]
where $\varphi_i$ are fixed angles depending on the directions of the sides. The average of $|\cos|$ over a period is $\frac1\pi\int_0^\pi|\cos t|\,dt = \frac2\pi$. Therefore
\[ \bar w = \frac12\sum_i s_i\cdot\frac2\pi = \frac{s}{\pi}. \]
For a smooth convex contour, approximate it by inscribed convex polygons; widths and perimeters converge, so $\bar w=s/\pi$ as well.
Step 5: The answer
\[ \boxed{P = \frac{\bar w}{2a} = \frac{s}{2\pi a},} \]
where $s$ is the perimeter of the contour. Remarkably, only the perimeter matters, not the shape.
Step 6: Examples
| Contour | Perimeter $s$ | $P$ |
|---|---|---|
| circle of radius $\rho$ ($\rho\lt a$) | $2\pi\rho$ | $\rho/a$ (width is always $2\rho$) |
| square of side $q$ | $4q$ | $2q/(\pi a)$ |
| "needle" of length $\ell$ (a flat contour traced twice) | $2\ell$ | $\ell/(\pi a)$ — Buffon's needle |
With the lines $2a$ apart, Buffon's classical needle result $\frac{2\ell}{\pi\cdot 2a}$ is exactly the case of a contour of perimeter $2\ell$.
Result
| Answer | |
|---|---|
| Given orientation | $P=w(\theta)/(2a)$ |
| Mean width (Cauchy) | $\bar w=s/\pi$ |
| Probability | $P=\dfrac{s}{2\pi a}$, $s$ = perimeter of the contour |