How Many Turkey Trotters Can You Pass?
Nov. 18, 2022, at 8:00 AM
.share
.has-bugs
Illustration by Guillaume Kurkdjian
.single-featured-image
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,<a class="espn-footnote-link" data-footnote-id="1" href="#fn-1" data-footnote-content="
Important small print: In order to 
win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!
“>1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.
Due to the holidays, the next column will appear on Dec. 2. See you then!
Riddler Express
I recently competed in a 5-kilometer “turkey trot” race. Before the race began, all the runners (including me) gathered behind the starting line in a random order. Once the race began, everyone started running at their own fixed pace.
I hadn’t run in several years, so I wasn’t sure how my pace would compare to that of the other racers. Nevertheless, once the race began, I found myself passing quite a few other runners — and being passed myself.
On average, what fraction of the other runners could I expect to pass during the race? (Assume that if my pace is faster than that of another runner’s, I will pass them at some point during the race.)
Riddler Classic
From Michael Branicky comes a challenge involve many, many dice:
I have five kinds of fair Platonic dice: tetrahedra (whose faces are numbered 1-4), cubes (numbered 1-6), octahedra (numbered 1-8), dodecahedra (numbered 1-12) and icosahedra (numbered 1-20).<a class="espn-footnote-link" data-footnote-id="2" href="#fn-2" data-footnote-content="
These dice are more commonly known as d4, d6, d8, d12 and d20.
“>2
When I roll two of the cubes, there is a single most likely sum: seven. But when I roll one cube and two tetrahedra, there is no single most likely sum — eight and nine are both equally likely.
Which whole numbers are never the single most likely sum, no matter which combinations of dice I pick?
Solution to last week’s Riddler Express
Congratulations to Andrew Busch of London, winner of last week’s Riddler Express.
Last week, in an effort to break open the gates of the city Tinas Mirith, an army of orcs first tried using a battering ram, but to no avail. They next erected a 100-foot pole with a very massive weight at the top (i.e., the weight was much, much heavier than the rest of the pole). The pole was also anchored at the bottom, so that as the weight fell the entire pole rotated around its bottom without slipping.
How far away should the orcs have positioned the vertical pole from the gates so that when the weight came crashing down, its horizontal speed was as great as possible?
First, let’s examine an animation of the falling weight:
As the weight fell, it moved faster. At the same time, due to the rotation, its horizontal motion turns into vertical motion. Now at some point, its horizontal speed was at a maximum. One way to find this maximum was to do what many students do time and again in their first physics course: Write an equation describing conservation of energy.
As the weight came down, its gravitational potential energy became kinetic energy. If we call the pole length L (100 feet in the puzzle), then by the time the pole made an angle ? with the ground, the weight was at a height Lsin?. That meant the change in the height was L−Lsin?, or L(1−sin?). Therefore, the change in gravitational potential energy was mgL(1−sin?), where m was the mass of the weight and g was the acceleration due to gravity — roughly 32 feet/sec2.
This difference in potential energy was what became kinetic energy, which could be expressed as 1/2mv2, where v was the speed of the weight. Setting these equal gave you v = √(2gL(1−sin?)). But you didn’t want the velocity, you wanted the horizontal velocity, which was equal to vsin(?). And so the function you were trying to maximize was sin?√(2gL(1−sin?)). Pulling out a few constants that didn’t matter, this became sin?√(1−sin?). To maximize this function, you had to take the derivative with respect to ? and set that equal to zero. After some friendly cancelation, this left you with sin? = 2/3.
That was a nice result, but this wasn’t what the riddle was asking for. What was the right distance between the pole and the gates? That was L times the cosine of this optimal angle. If sin? was 2/3, then cos? was (√5)/3, which meant the distance was 100/3·√5, or about 74.5 feet.
A few folks, like Ricky Reusser, opted for a more challenging, “bad approach.” As Ricky stated: “This problem is very easily solved with a simple energy argument, but let’s brute force it from the equations of motion instead!” Ricky proceeded to solve second-order differential equations with Jacobi elliptic functions. In the end, the answer was the same.
After placing their pole in just the right spot, the orcs successfully knocked down the gates of Tinas Mirith, securing a victory for Sord Lauron. And that was the end of all things.
Solution to last week’s Riddler Classic
Congratulations to David Cohen of Silver Spring, Maryland, winner of last week’s Riddler Classic.
Last week, I was slicing a square peanut butter and jelly sandwich. But rather than making a standard horizontal or diagonal cut, I instead picked two random points along the perimeter of the sandwich and made a straight cut from one point to the other. (These points could have been on the same side.)
My slice was “reasonable” if I cut the square into two pieces and the smaller resulting piece had an area that was at least one-quarter of the whole area. What was the probability that my slice was reasonable?
To see what’s happening here, we can start by looking at special cases, like when the first point I picked was one of the square’s corners or at the midpoint of one of its sides, as shown below.
When the first point was in a corner, the cut was reasonable whenever the other point was closer to the opposite corner than any other corner, a region that made up a quarter of the square’s perimeter. And when the first point was a midpoint, the cut was reasonable whenever the other point was somewhere on the opposite side, another region that made up a quarter of the square’s perimeter. And so, for both of these special cases, my probability of making a reasonable cut was 1/4. Could the answer then simply have been 1/4?
This being a Classic, that was not the answer. As it turned out, whenever the first point was anywhere else, the probability of making a reasonable cut was greater than 1/4. Solvers Trey Goesh and Starvind made animated graphs suggesting this was the case.
To determine the exact probability, suppose the first point was a distance x from the nearest corner of the square, and let’s assume that the square had side length 1, as shown below. Note here that x was equally likely to be anywhere from 0 to 0.5.
If the second point was on the opposite side, the cut was reasonable if it was within 1/2+x of the opposite corner, creating a trapezoid whose area was at least 1/4. But if the second point was on the other side touching that opposite corner, then the cut was reasonable when that second point was within (1−2x)/(2−2x) of that corner, creating a triangle whose area was at least 1/4.
To find the probability that the cut was reasonable, you had to add these two distances together, divide by 4 (the total perimeter of the square), integrate over x from 0 to 0.5, and finally divide by 0.5 (i.e., multiply by 2), since we wanted the average probability over this range of x. The result of this integral was relatively concise: (7−ln(16))/16, or about 26.4 percent.
In the end, 25 percent was not a bad guess. It seems I am only slightly more reasonable than that.
Want more puzzles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com.
