This week’s Riddler Express asks:

The other day I was playing around in my kitchen with a tall glass and a smaller disk, when, to my surprise, I was able to balance the disk neatly atop the rim of the glass.

First of all, try this at home!

Now suppose you have a disk of radius \(R\) and a glass with a circular rim of radius \(2R\). If you randomly place the disk so that its center lies within the glassâ€™s rim, what is the probability that the disk will balance atop the glass? (Assume the distribution is uniformly spread across the circular area inside the rim.)

### Solution

To start with, we need to build a geometric model for this problem. Since the center of the disk will lie within the glass’s rim, we don’t need to worry about the disk falling *outside* the glass, only in. To determine if the disk will fall into the glass, we need to figure out the threshold for how close to the center of the glass the center of the disk can lie. My approach to determining where that threshold is follows:

The relevant bifurcator of the disk is the chord that cuts through the disk and ends at the points of intersection between the circumference of the disk and the circumference of the glass (dashed lines below). When more than half of the area of the disk falls on the interior side of the chord (closer to the center of the glass), the disk will fall in. But when more than half of the area of the disk falls on the exterior side (away from the center of the glass), the disk can balance on the glass.

We can figure out how close to the center of the glass the limit case is by recognizing that at the limit case, the two points of intersection and the center of the glass form an equilateral triangle with sides of length \(2R\) (gray line below). Then the distance that the center of the disk lies from the center of the glass is \(R\sqrt{3}\) at the limit case (dashed line below).

We can use this to figure out the area of the donut between the outside edge of the glass and this interior boundary.

The area of this region is

\[A=\pi(2R)^2-\pi(R\sqrt{3})^2=\pi R^2.\]

The portion of the glass that this represents is

\[p=\frac{\pi R^2}{4\pi R^2}=1/4.\]

So the probability of balancing the disk is **25%.**

Here’s a simulation: