Skip to main content
Menu
Can You Crack The Case Of The Crystal Key?

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 Nilay Shroff comes a puzzle that’s perfect for baseball season:

As of this week, there have already been six no-hitters this MLB season, well on pace to break the record for no-hitters for a season in the modern era, which stands at nine in 1990.

To achieve a no-hitter, a pitcher must pitch a complete game (recording 27 outs over nine innings) without allowing a hit (i.e., walks and other means of reaching base are allowed). However, to achieve a perfect game, a pitcher must record 27 consecutive outs without allowing anyone from the opposing team on base. There have been only 23 perfect games in MLB history. Two were thrown in 2010, and three were thrown in 2012.

In 2009, the leaguewide on-base percentage was 0.333. That figure has fallen over the past decade, and this year, it’s all the way down to 0.313, which helps explain the surge in no-hitters.

How low would a batter’s chances of reaching base have to be for you to expect one perfect game per season? (You can make the following simplifying assumptions: All batters have the same chances of reaching base; at-bats are independent from each other; there are 30 MLB teams, and each club plays 162 games; and no games go into extra innings.)

The solution to this Riddler Express can be found in the following column.

Riddler Classic

Dakota Jones is back in action! To gain access to a hidden temple deep in the Riddlerian Jungle, she needs a crystal key.

She already knows the crystal is a polyhedron. And according to an ancient text, it has exactly six edges, five of which are 1 inch long. Cryptically, the text does not specify the length of the sixth edge. Instead, it says that the key is the largest such polyhedron (i.e., with six edges, five of which have length 1) by volume.

Once again, Dakota Jones needs your help. What is the volume of the crystal key?

The solution to this Riddler Classic can be found in the following column.

Solution to last week’s Riddler Express

Congratulations to 👏 Seth Hollar 👏 of Raleigh, North Carolina, winner of last week’s Riddler Express.

Last week, the Riddler Cheese Company was producing “craft triples” — triangular slices of cheese whose side lengths were Pythagorean triples when measured in inches.

However, the company’s slicing machine had recently malfunctioned, producing a stock of square slices with side lengths of 5 inches. To salvage this situation, what was the greatest number of whole Pythagorean slices that could be made from each 5-inch square? (Note: You could only cut pieces out of the square. No melting or gluing pieces together was allowed!)

With a little work, you could immediately find an upper bound for the solution. The 5-inch square had an area of 25 square inches. Meanwhile, the smallest right triangle with whole number side lengths was 3-4-5, which had an area of 6 square inches. That meant you couldn’t have more than 25/6 (i.e., 4, because partial triangles were not allowed).

But was it possible to fit four craft tiples in a 5-inch square? Indeed it was! John from Washington, D.C., accomplished this using actual cheddar by lining up the four hypotenuses along the four sides of the square:

A square slice of cheese that has been sliced into four congruent 3-4-5 right triangles. Each hypotenuse is a side of the larger square. There is a small square left over in the middle.

Because the non-right angles in each right triangle were complementary, you could prove that none of the four triangles overlapped. And, as expected, there was 1 square inch of cheese left over in the middle.

For extra credit, you had to find the smallest square of cheese such that 100 percent of it could be partitioned into craft triples.

If you were using just 3-4-5 triangles — again, with an area of 6 square inches — then the square’s area had to be a multiple of 6. The smallest such square had side lengths of 6 inches and an area of 36 square inches. However, there was no way to squeeze six 3-4-5 triangles into this space.

The next-smallest square had side lengths of 12 inches and an area of 144 square inches. Sure enough, it was possible to carve 24 3-4-5 triangles out of this square. You could also use two 5-12-13 triangles and another 14 3-4-5 triangles, as demonstrated by solver Pierre Bierre below. It turned out that these were the smallest such squares.

A 12x12 square that is divided up into Pythagorean triples. The top row is 3 units tall, consisting of six 3-4-5 triangles. The middle row is 4 units tall, consisting of eight 3-4-5 triangles. The bottom row is 5 units tall, consisting of two 5-12-13 triangles.

With this puzzle, I sincerely hope you had fun cutting the cheese!

Solution to last week’s Riddler Classic

Congratulations to 👏 Lowell Vaughn 👏 of Bellevue, Washington, winner of last week’s Riddler Classic.

Last week, Jordan Ellenberg (author of the new book, “Shape”) was asking about what he called anti-isosceles sets.

If you considered an N×N grid of points, what was the greatest subset of these N2 points such that no three of them formed an isosceles triangle? (Note: Degenerate triangles, or triangles with zero area, counted as triangles here, as long as their three vertices are distinct.)

As shown below, the largest anti-isosceles set in a 2×2 grid had two points; for a 3×3 grid, the largest anti-isosceles set had four points. (Note: For both grids, there were multiple sets with these maximum numbers of points.)

Two grids. On the left is a 2x2 grid with opposite points highlighted. On the right is a 3x3 grid with the two leftmost points in the top row and the two rightmost points in the bottom row selected. These represent anti-isosceles sets.

How many points were in the largest anti-isosceles set for a 4×4 grid?

As is often the case, there was a way to computationally brute force your way through this problem. With 16 total points in the grid, that meant there were 216 — or 65,536 — ways to group these points into sets, since each point was either in or out of a given set. Several solvers wrote programs to check all of these sets by choosing each grouping of three points within the set. If the distances between any two of these were the same, the entire set could not be anti-isosceles.

It’s worth noting that many of these sets were rotations or reflections of each other, meaning if one of them was anti-isosceles then others were as well. But since the total number of sets was not prohibitively large (for a computer) large for a computer, you didn’t have to worry about being extra efficient with your code.

If your code worked correctly, it revealed that there were exactly four anti-isosceles sets, each with the maximal six points. They were all rotations of each other, and here’s one of them:

Anti-isosceles set for a 4x4 grid, consisting of six points. In the top row, the first and third points are selected. In the second row, the first and last points are selected. In the bottom row, the last two points are selected.

You can check for yourself: With six total points, there were 20 ways to choose three of them. And none of these 20 triangles was isosceles.

For extra credit, you had to find the largest anti-isosceles sets in a 5×5 grid and a 6×6 grid (and even larger grids if you dared). A 5×5 grid had 225 — or 33,554,432 — total subsets, just at the edge of what was possible for a desktop computer to sift through in a few minutes. For larger grids, solvers like Eric Dallal of Boston, Massachusetts, had to use “every optimization I could think of.”

Meanwhile, Goh Pi Han of Petaling Jaya, Malaysia, went a different route, running simulations that added one point at a time in positions that didn’t result in any isosceles triangles. This was a clever way to identify lower bounds on the sizes of anti-isosceles grids with too many points to exhaustively handle.

Here are examples of the largest non-isosceles sets for the 5×5 grid and the 6×6 grid, which had seven and nine points, respectively.

Anti-isosceles sets in a 5x5 grid (with seven points) and a 6x6 grid (with nine points).

Friend-of-The-Riddler Dean Ballard explored non-isosceles sets in three-dimensional grids. In a 5×5×5 grid, the largest such set he found had 16 points!

Three-dimensional 5x5x5 grid with 16 selected points that represent an anti-isosceles set.

Finally, returning to two-dimensional grids, Jordan wanted to know if anyone could show that the size of the largest anti-isosceles set for an N×N grid was of the order Nk, with 1 < k < 2. Some hypothesized that the largest anti-isosceles set had approximately 3/2·N points, in which case k would be 1. Ravi Chandrasekaran looked to the number of unique distances on the grid for inspiration.

However, to my knowledge, no one has (yet) proven upper or lower bounds on k between 1 and 2. Jordan’s challenge remains an open one!

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: 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.

Comments