Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Quick announcement: Have you enjoyed the puzzles in this column? If so, I’m pleased to tell you that we’ve collected many of the best, along with some that have never been seen before, in a real live book! It’s called “The Riddler,” and it will be released in October — just in time for loads of great holidays. It’s a physical testament to the mathematical collaboration that you, Riddler Nation, have helped build here, which in my estimation is the best of its kind. So I hope you’ll check out the book, devour the puzzles anew, and keep adding to our nation by sharing the book with loved ones.
And now, to this week’s puzzles!
Riddler Express
From Steven Pratt, some nail-biting calculation:
This year’s World Cup has been chock full of exciting penalty shootouts. Historically, about 75 percent of soccer penalty kicks are successful. Given that number, what are the chances that a shootout goes past its fifth kick for each team and into the even more exciting sudden-death portion of the penalty-kick period?
Riddler Classic
From Adam Wagner, a puzzling bisection. Adam writes that this novel challenge is “my all-time favorite puzzle, and I like the exclusivity of knowing that nobody I tell it to will ever have encountered it before. But as Chris McCandless said, ‘Happiness [is] only real when shared,’ so I’ve decided at long last to let it leave the nest and be enjoyed by all of Riddler Nation.”
Say you have an “L” shape formed by two rectangles touching each other. These two rectangles could have any dimensions and they don’t have to be equal to each other in any way. (A few examples are shown below.)
Using only a straightedge and a pencil (no rulers, protractors or compasses), how can you draw a single straight line that cuts the L into two halves of exactly equal area, no matter what the dimensions of the L are? You can draw as many lines as you want to get to the solution, but the bisector itself can only be one single straight line.
Solution to last week’s Riddler Express
Congratulations to 👏 Joe Lindell 👏 of St. Louis Park, Minnesota, winner of last week’s Riddler Express!
Last week you were employed as a delivery driver for a peanut butter and jelly restaurant in Riddler City. That booming metropolis was built on a grid of east-west streets (numbered from 1st to 61st) and north-south avenues (named from A to U), with each block 0.1 miles long. Your restaurant was at 20th and F. The speed limit was 20 mph everywhere — except on Avenue U, aka the Ultra-Speed Trafficway, where the speed limit was 200 mph. For deliveries to what parts of the city was it helpful to use Avenue U?
Essentially, the trafficway helps for deliveries to regions in the city’s north and southeast — because Avenue U runs north-south up the east side and you start closer to the south.2
As you may have noticed, Riddler City’s layout looks an awful lot like an Excel spreadsheet — a tidy series of locations laid out in a tidy rectangular grid. Therefore, one handy approach was to map Riddler City itself in Excel, as a 60-by-20 spreadsheet, and calculate for each cell the amount of time that taking Avenue U saved you.
For example, let’s suppose you’re considering a delivery to 61st and F — 41 blocks, or 4.1 miles, directly north of you at the edge of the city. You have two options: drive straight north, or drive 1.5 miles east to Avenue U, north on the trafficway, and then back west to your destination. The first option, at 20 mph, will take you 12.3 minutes. The second option will take you 4.5 minutes to get the Avenue U, 1.23 minutes on Avenue U, and 4.5 minutes to backtrack west, for a total of 10.23 minutes. So, for that address, you should take the high-speed trafficway!
In Excel, the time for the first route option could be calculated with a formula like this: 0.3*(ABS(ROW()-20)+ABS(COLUMN()-6)). It takes 0.3 minutes to travel a block at 20 mph, and you are traveling some distance away from the 20th street and the sixth avenue. The time for the second route option could be calculated with a formula like this: 0.3*(0.1*(ABS(ROW()-20))+ ABS(COLUMN()-6)+ IF(ROW()<>20, 2*(21-MAX(6, COLUMN())), 0)). Basically, it takes 0.3 minutes to travel each block toward Avenue U, the 20th “column” in the city spreadsheet, and 0.1 times that to travel on Avenue U, since you can drive 10 times faster.
If taking Avenue U saved a positive amount of time, great, take it and gun that engine! If not, take the slower surface streets. Solver James Barton published a nice explanation of exactly how he used Excel to solve this problem, including his breakdown of the formulas above. You can also explore Nick Harper’s solution spreadsheet.
But this is a solution best expressed graphically — with a map of Riddler City!
Solver Miguel Restrepo plotted the optimal PB&J delivery times to every address starting from your restaurant (the red dot). The toughest deliveries are to the northwest corner of the city, while Avenue U makes things much faster on the east side:
Given these optimal delivery times, we can also draw another map, simply showing those addresses for which we should use Avenue U and those for which we should not. Miguel provided that as well:
Again, the far north, northeast and southeast quadrants are well served by the Ultra-Speed Trafficway. And please step on it — I’m getting hungry!
Solution to last week’s Riddler Classic
Congratulations to 👏 Madeline Barnicle 👏 of Los Angeles, winner of last week’s Riddler Classic!
Last week found you in the desert, carefully examining its shifting populations of meerkats and scorpions. Meerkats feed on scorpions, and you knew the following things about how the two populations interact: 1) If there were no meerkats, the population of scorpions would double every month. 2) If there were no scorpions, the population of meerkats would halve every month. 3) If you had exactly 20 scorpions and five meerkats, both populations would not change at all. The following questions relied on the Lotka-Volterra model of predators and prey.
First, what is the number of meerkats when the desert has as many scorpions as possible? It is five meerkats. Adorable! What is the number of meerkats and the number of scorpions when the meerkat population is increasing by four meerkats per month and the scorpion population is decreasing by two scorpions per month? It is about five meerkats and 42 scorpions.3 Less adorable!
How do we get to these numbers? Call the population of scorpions (the prey) x, the population of meerkats (the predators) y, and time (in months) t. The model equations that describe the populations’ changes (dx and dy) over time (dt) are
\begin{equation*} \frac{dx}{dt} = a x – b xy \end{equation*}
\begin{equation*}\frac{dy}{dt} = d xy – c y \end{equation*}
So the first thing to do now is to solve for the model’s parameters: a, b, c and d. We know, for example, that when there are no meerkats (y = 0) the population of scorpions doubles each month. That gives us \(a=\ln(2)\). We also know that when there are no scorpions (x = 0) the population of meerkats halves each month. That, similarly, gives us \(c=\ln(2)\). Finally, we know that if we have five meerkats and 20 scorpions, the populations don’t change at all. That means \(c/d = 20\) and \(a/b = 5\), and therefore that \(b = \ln(2)/5\) and \(d = \ln(2)/20\).
With those parameter values in hand, we can answer the questions. When there are as many scorpions as possible, it must be the case that \(dx/dt = 0\), that is, the population is maximized when its derivative is equal to zero. Therefore, from the first equation above, \(y=a/b=5\) — five meerkats.
When the meerkat population is increasing by four meerkats per month and the scorpion population is decreasing by two scorpions per month, we know that \(dx/dt = -2\) and that \(dy/dt = 4\). We can turn to an online solver for this one, and after plugging in our parameter values, we find that the number of meerkats is about 5.35 and the number of scorpions is about 41.59. (I won’t spend too much time ruminating on what exactly 35 hundredths of a meerkat looks like.)
Thanks for spending this time in the desert. And always remember to stay hydrated while Riddling.
Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.