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Can You Roll The Perfect Bowl?

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 next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

At the recent World Indoor Bowls Championships in Great Yarmouth, England, one of the rolls by Nick Brett went viral. Here it is in all its glory:

In order for Nick’s green bowl to split the two red bowls, he needed expert precision in both the speed of the roll and its final angle of approach.

Suppose you were standing in Nick’s shoes, and you wanted to split two of your opponent’s bowls. Let’s simplify the math a little, and say that each bowl is a sphere with a radius of 1. Let’s further suppose that your opponent’s two red bowls are separated by a distance of 3 — that is, the centers of the red bowls are separated by a distance of 5. Define ÐÑ as the angle between the path your bowl is on and the line connecting your opponent’s bowls.

For example, here’s how you could split your opponent’s bowls when ÐÑ is 75 degrees:

What is the minimum value of ÐÑ that will allow your bowl to split your opponents’ bowls without hitting them?

## Riddler Classic

From Robert Berger comes a question of maximizing magnetic volume:

Robert’s daughter has a set of Magna-Tiles, which, as their name implies, are tiles with magnets on the edges that can be used to build various polygons and polyhedra. Some of the tiles are identical isosceles triangles with one 30 degree angle and two 75 degree angles. If you were to arrange 12 of these tiles with their 30 degree angles in the center, they would lay flat and form a regular dodecagon. If you were to put fewer (between three and 11) of those tiles together in a similar way, they would form a pyramid whose base is a regular polygon. Robert has graciously provided a photo of the resulting pyramids when three and 11 tiles are used:

If Robert wanted to maximize the volume contained within the resulting pyramid (presumably to store as much candy for his daughter as possible), how many tiles should he use?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Tom Lauwers ÑÑâÐ of Pittsburgh, Pennsylvania, winner of last week’s recent Riddler Express.

Last week, you analyzed the voting for the baseball Hall of Fame when there were 20 players on the ballot and 400 voters, each of whom could select up to 10 players for induction without voting for any given player more than once. To gain entry, a player must have been selected on at least 75 percent of the ballots.

Under these circumstances, what was the maximum number of players that could be inducted into the Hall of Fame?

Tom reasoned that to get as many players into the Hall of Fame as possible, each of the 400 voters should have selected the maximum allowable number of 10 players, meaning a total of 4000 votes would have been cast. Meanwhile, to gain entry, a player needed to appear on 75 percent of the ballots, meaning they needed to receive at least 300 votes. The greatest number of players who could have received 300 votes out of 4000 total votes was 4000 ÷ 300, which is slightly greater than 13. That meant the number of inductees was at most 13 — but was it possible for exactly 13 players to be voted in?

Indeed it was. Solver Ramsey Jade showed one possible way by first labeling the 13 players with the letters from A through M, and dividing the 400 voters into four groups of 100.

• The first 100 voters would select players A, B, C, D, E, F, G, H, I and J.
• The next 100 voters would select D, E, F, G, H, I, J, K, L and M.
• The third set of 100 voters would select A, B, C, G, H, I, J, K, L and M.
• The final set of 100 voters would select A, B, C, D, E, F, J, K, L and M.

Twelve of the 13 players received 300 votes, and player J apparently joined Mariano Rivera as the only unanimous inductees.

Interestingly, this answer didn’t depend on the number of eligible players or the number of voters — just the maximum number of players on a ballot and the percentage needed to get inducted.

Some solvers wondered about the minimum possible number of inductees, which was, unsurprisingly, zero. The voters weren’t obligated to list 10 players on their ballots, so they could have simply cast empty ballots. And even if they had each voted for the maximum 10 players, it would still be possible for the 20 candidates to split the votes so that no one made it into the Hall of Fame.

Solver Ravi Chandrasekaran further studied how much more crowded Cooperstown would be if voters colluded to maximize the number of inductees. According to Ravi’s analysis, there would be more than seven times as many players enshrined.

Fortunately, this is not something that Larry Walker will have to worry about anymore.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Sion Verschraege ÑÑâÐ of Ghent, Belgium, winner of the last week’s Riddler Classic.

Last week, I invited you to play a game of “Pinching Pennies,” which started off with between 20 and 30 pennies. First, I divided the pennies into two piles any way I liked. Then we alternated taking turns, with you first, until someone won the game. For each turn, a player took any number of pennies he or she liked from either pile, or instead took the same number of pennies from both piles. Each player also had to take at least 1 penny every turn. The winner of the game was the one who took the last penny.

If we both played optimally, what starting numbers of pennies (again, between 20 and 30) guaranteed that you could always win the game?

Solver Jason Ash tackled this puzzle by first looking at potential endgame strategies. For example, if it were your turn and you encountered two piles that each had 1 penny, which we can indicate with the ordered pair (1, 1), then you could win (by taking both coins). Similarly, any ordered pair of the form (n, 0), (0, n), and (n, n) would result in a one-move victory for you.

But what about the ordered pair (1, 2), meaning one pile had 1 penny and the other had 2? Here are your four possible moves and what I (your opponent) would do next:

• Remove 1 penny from the first pile, leaving (0, 2). I would then take both pennies from the second pile, and you’d lose.
• Remove 1 penny from the second pile, leaving (1, 1). I would then take one penny from each pile, and you’d lose.
• Remove 2 pennies from the second pile, leaving (1, 0). I would then take the last penny from the first pile, and you’d lose.
• Remove 1 penny from each pile, leaving (0, 1). I would then take the last penny from the second pile, and you’d lose.

As you can see, (1, 2) was a state of the game you’d prefer to avoid, as it represented a no-win scenario for you. Meanwhile, any pile arrangements of the form (1, n+2) would guarantee a victory for you, since you could remove the n pennies from the second pile and then I’d be faced with (1, 2).

This week’s winner, Sion, continued working up to larger and larger numbers, finding the game states that we’d each prefer to avoid, and an interesting pattern began to emerge:

Those red cells in Sion’s grid show the states we both want to avoid: (1, 2), (3, 5), (4, 7), (6, 10), (8, 13), (9, 15), (11, 18), etc. They appear to fall along two symmetric lines.

But I digress. The answer to the original question, which was asking for the initial numbers of pennies that guaranteed a victory for you, was 20, 22, 23, 25, 26, 27, 28 and 30. Had we instead played with the other numbers between 20 and 30, I would have won: I would have split 21 coins into piles of 8 and 13, 24 coins into piles of 9 and 15 and 29 coins into piles of 11 and 18. (Note: I also accepted answers that omitted 20 and 30 — a different interpretation of “between 20 and 30” — as well as answers of 21, 24 and 29, which were winning numbers for me.)

Now, back to the pattern in Sion’s grid. As it turns out, this game pre-dates last week’s column and is better known as Wythoff’s Game, named for the Dutch mathematician and game theorist. The game has been well studied, and those losing positions are related to the golden ratio — notice that the second number in each of the above pairs is approximately 1.6 times greater than the first, very close to the golden ratio. Another famous sequence of numbers with close ties to the golden ratio is the Fibonacci sequence, whose numbers like 5, 8 and 13 you’ll also see lurking in Sion’s grid.

For more on Wythoff’s Game, check out his original paper from 1907. (Special thanks to solver Laurent Lessard for finding this gem!)

Whether it’s 1907 or 2020, a good riddle is timeless.

## 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.

## Footnotes

1. Important small print: Please wait until Monday to publicly share your answers. In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.