Putnam 2024, Problem A-5

Consider a circle \( \Omega \) with radius \( 9 \) and center at the origin \( (0,0) \), and a disc \( \Delta \) with radius \( 1 \) and center at \( (r,0) \), where \( 0 \leq r \leq 8 \). Two points \( P \) and \( Q \) are chosen independently and uniformly at random on \( \Omega \). Which value(s) of \( r \) minimize the probability that the chord \( \overline{PQ} \) intersects \( \Delta \)?

Solution:
We will show that $r = 0$ (and no other value of $r$) minimizes the stated probability. Note that $P$ and $Q$ coincide with probability $0$; thus we can assume that $P \neq Q$.

**First solution.** First restrict $P,Q$ to points on $\Omega$ such that the segment $\overline{PQ}$ makes an angle of $\theta$ with the $y$ axis, where $\theta$ is a fixed number with $-\pi/2 < \theta \leq \pi/2$. By rotating the diagram by $-\theta$ around the origin, we move $\overline{PQ}$ to be a vertical line and move $\Delta$ to be centered at $(r\cos\theta,-r\sin\theta)$. In this rotated picture, $P$ and $Q$ are at $(9\cos\phi,\pm9\sin\phi)$ where $\phi$ is chosen uniformly at random in $(0,\pi)$. Now the vertical tangent lines to the boundary of $\Delta$, $x = r\cos\theta \pm 1$, intersect the $y > 0$ semicircle of $\Omega$ at $(9\cos\phi,9\sin\phi)$ where $\phi = \cos^{-1}\left(\frac{r\cos\theta \pm 1}{9}\right)$. Thus the probability that $\overline{PQ}$ intersects $\Delta$ for a specific value of $\theta$ is $\frac{1}{\pi}f(r,\theta)$, where we define
$$f(r,\theta) = \cos^{-1}\left(\frac{r\cos\theta - 1}{9}\right) - \cos^{-1}\left(\frac{r\cos\theta + 1}{9}\right).$$

If we now allow $\theta$ to vary (uniformly) in $(-\pi/2,\pi/2]$, we find that the overall probability that $\overline{PQ}$ intersects $\Delta$ is
$$P(r) = \frac{1}{\pi^2} \int_{-\pi/2}^{\pi/2} f(r,\theta) d\theta.$$

The function $P(r)$ is differentiable with
$$P'(r) = \frac{1}{\pi^2} \int_{-\pi/2}^{\pi/2} \frac{\partial f(r,\theta)}{\partial r} d\theta.$$

Now
$$\frac{\partial f(r,\theta)}{\partial r} = (\cos t)\left((80 - 2r\cos t - r^2\cos^2 t)^{-1/2} - (80 + 2r\cos t - r^2\cos^2 t)^{-1/2}\right),$$
which, for $t \in (-\pi/2,\pi/2)$, is zero for $r = 0$ and strictly positive for $r > 0$. It follows that $P'(0) = 0$ and $P'(r) < 0$ for $r \in (0,8]$, whence $P(r)$ is minimized when $r = 0$.

**Second solution.** (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang) We interpret the first paragraph of the first solution as reducing the original problem to the following assertion: given two parallel lines at distance $2$, both of which intersect a circle of radius $9$, the length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.

To see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above.