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,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.
Riddler Express
From Ivor Traber comes a probability triathlon:
You have three fair coins, three fair dice and a full deck of cards in your possession. First, you flip all three coins and note the number of heads. Next, you toss all three dice and note the number of ones or sixes. Finally, you draw three random cards from the deck of 52 and note the number of hearts.
What is the probability that all three numbers are the same?
The solution to this Riddler Express can be found in the following column.
Riddler Classic
Two ants named Geo and Desik are racing along the surface of a cone. The circular base of the cone has a radius of 2 meters and a slant height of 4 meters. Geo and Desik both start the race on the base, a distance of 1 meter away from its center.
The race’s finish is halfway up the cone, 90 degrees around the cone’s central axis from the start, as shown in the following diagram:
Geo and Desik both want your help in strategizing for the race. What is the length of the shortest path from the start to the finish?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Justin Roberts 👏 of Austin, Texas, winner of last week’s Riddler Express.
Last week, I had a mystery function that passed through the points (0, 1), (1, 2), (2, 4), (3, 8) and (4, 16). I also knew the function was continuous and smooth. In other words, I could draw it in a single stroke without any sharp corners or cusps.
Moreover, this function was not f(x) = 2x. In fact, it was decreasing at x = 2. Could you find an expression for such a function?
There were countless solutions here. Many solvers, like Amy Teegarden of Salem, Oregon, created such a function by starting with f(x) = 2x and then adding or subtracting an oscillatory term. This term had to be zero at integer values of x so it would still pass through the five required points. But between those points it moved up and down more freely. One such function was f(x) = 2x − 3sin(𝜋x), which, sure enough, was decreasing at x = 2:
Sinusoids weren’t the only functions that could equal zero at five places — polynomials worked, too! Solver Mark Goodrich of Oxford, Pennsylvania, found the function f(x) = 2x − x(x−1)(x−2)(x−3)(x−4). Whenever x was 0, 1, 2, 3 or 4, the function was equal to 2x. However, the additional polynomial term ensured it was decreasing at x = 2:
Of course, you could have forgotten about the exponential term entirely and made the entire function a sixth-degree polynomial that passed through the five given points, which is exactly what the students in The Hewitt School Problem Solving & Problem Posing class did.
For extra credit, you were told that the function was always increasing. Could you find a function that was concave down at x = 2?
Once again, one approach was to include a sinusoidal term. But to ensure that the function was always increasing, its amplitude had to scale with the exponential function itself. Solver Christopher Clark found such a function, f(x) = 2x(1−cos(2𝜋x)/16):
Sure enough, this function was always increasing, but you can see how its upward trajectory slowed around x = 2, which indicated that it was indeed concave down.
And kudos to solver Izumihara Ryoma of Toyooka, Japan, who found a polynomial solution that checked all the boxes, f(x) = 1/24·(x9 − 13x8 + 62x7 − 123x6 + 51x5 + 129x4 − 104x3 − 41x2 + 62x + 24):
With infinitely many solutions, there was plenty of room for creativity. Well done, everyone! This all goes to show that knowing a few discrete points of a function is a far cry from precisely knowing the function, or even what it does between those points.
Solution to last week’s Riddler Classic
Congratulations to 👏 Justin Ahmann 👏 of Bloomington, Indiana, winner of last week’s Riddler Classic.
Last week’s Classic took you back to high school algebra.
Suppose you had two distinct points on the x-axis of the coordinate plane. If I told you a parabola (with a vertical line of symmetry) passed through those two points, then the parabola’s vertex had to be on the perpendicular bisector of those two points. (Neat!)
Now, suppose the two distinct points were anywhere on the coordinate plane. If I told you that a parabola (again, with a vertical line of symmetry) passed through those two points, where on the plane could that parabola’s vertex have been?
For (relative) simplicity, you could have assumed that the two points were symmetric about the origin. (In other words, you could redefine the origin as the midpoint of the two points. Then, once you had the locus of possible vertices, you could undo that initial translation.) That meant the two points had coordinates (a, b) and (-a, –b) for some values of a and b.
Now, the vertical distance between a parabola’s vertex and any other point on the parabola is proportional to the square of the horizontal distance. So if the vertex had coordinates (x, y), that meant y−b = k(x−a)2 and y+b = k(x+a)2, where both equations had the same constant of proportionality k. Subtracting the first equation from the second allowed you to solve for that constant: k = b/(2ax).
Next, adding the equations and dividing by 2 gave you y = k(x2+a2). From there, you could plug in that newly-found expression for k to get y = b(x2+a2)/(2ax). As x approached zero, this function went off to positive of negative infinity. And as x got very large, it approached the line y = b/(2a)·x. In other words, it had two asymptotes. Now if that realization didn’t bring you back to high school algebra, I don’t know what could have.
Here’s a plot with a = 2 and b = 1:
The different parabolas are shown in blue, while the locus of their vertices are shown in red. The answer was a hyperbola! What’s more, the two given points were the local maximum and minimum of the two halves of the hyperbola.
Interestingly, as the two points get closer and closer to being horizontally aligned, the hyperbola deforms, approaching the union of a vertical line (the aforementioned perpendicular bisector) and a horizontal line that passes through both points (a deformed parabola).
Anyway, I hope you enjoyed this foray into the world of high school algebra!
Want more riddles?
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.