Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world — including you! You’ll find this week’s puzzle below.
Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the link below. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!
Before we get to the new puzzle, let’s return to last week’s. Congratulations to ÑÑâÐ Lewis Kershaw ÑÑâÐ of London, our big winner. You can find a solution to the previous Riddler at the bottom of this post.
Now here’s this week’s Riddler, a dungeons-and-dragons puzzle that comes to us from Brandon Hensley, a scientist at NASA’s Jet Propulsion Laboratory.
A video game requires you to slay monsters to collect gems. Every time you slay a monster, it drops one of three types of gems: a common gem, an uncommon gem or a rare gem. The probabilities of these gems being dropped are in the ratio of 3:2:1 — three common gems for every two uncommon gems for every one rare gem, on average. If you slay monsters until you have at least one of each of the three types of gems, how many of the most common gems will you end up with, on average?
And here’s the solution to last week’s Riddler, concerning a nearly overflowing martini, which came to us from an anonymous philosophy professor. I’ve adapted the solution the professor provided here. (There are undoubtedly many other ways to get there. Oh, and here’s an interactive virtual martini.)
The outline of the liquid’s surface, which is the intersection of the cone and the plane of the drink’s surface, is an ellipse. So the liquid forms an oblique elliptical cone (a cone whose base is an ellipse and whose apex is off-center). Like any cone, whatever the shape of its base and however oblique it might be, it has a volume 1/3 times the area of its base, here \(\pi ab\), where \(a\) and \(b\) are the semi-major and semi-minor axes, times its height \(h\).
Without loss of generality,1 assume the glass cone has height and radius 1. A right circular cone of any other size and shape can be obtained from ours by two transformations: vertical scaling to match the shape, and three-dimensional scaling to achieve the size. Both of these affect the volumes of all objects by a linear multiple, and so the volume of the liquid will still be in the same ratio to that of the cone. And both affect distances along corresponding lines by a linear multiple, so that the ratio of distances that is the answer to our problem will remain constant.
The volume of the upright liquid is \(1/3\cdot \pi p^3\), and so \(abh = p^3\). Let \(x\) be the height of the upright liquid (and hence the fraction that is our answer). The area of the triangle in the diagram formed by the liquid is \(ah\), and so:
Where r is the radius of the glass cone at the center of the ellipse:
Phew! And this is what it all looks like:
Understandably, some readers went with a more … hands-on approach.
Slaying monsters, doing math and drinking martinis — y’all must be exhausted. Have a relaxing weekend.