Can You Squeeze Your Quads?

Polygons, not muscles. Never muscles.

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

Your goal is to squeeze two non-overlapping quadrilaterals within a unit circle (i.e., a circle with radius 1). The quadrilaterals can share common edges, parts of edges, or vertices, but their interiors may not overlap. Their vertices may also lie on the circle’s circumference.

What is the greatest combined area these quadrilaterals can have? The quadrilaterals must be convex or concave (or one of each), but “crossed” or “bowtie” quadrilaterals are not allowed here.

Extra Credit

From Jason Zimba comes (in my humble opinion) a trickier puzzle of polygons inscribed within circles:

For any four distinct points on a unit circle, define Z as the following sum: the area of the (convex) quadrilateral formed by all four of these points plus the area of the largest triangle (by area) formed by any three of these points.

For example, if the four points were the vertices of an inscribed square, then Z would equal 2—the area of the square—plus 1—the area of the largest triangle formed by three of the points (in this case, the area of any of them)—for a total value of 3.

Given that the four distinct points can be anywhere on the unit circle, what is the greatest possible value of Z?

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a cute logic puzzle that’s sure to give you a quick dose of “aha!”

Courtesy of Sandro Ambuehl, who cited a talk by Fiery Cushman, I’m reproducing the puzzle here:

One of the shapes below is the magic symbol. It's the middle one. Select exactly one of the three words below such that somebody who does not know which is the magic symbol will know it for sure.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Gary M. Gerken 🎻 from Littleton, Colorado. I received 48 timely submissions, of which 34 were correct—good for a 71 percent solve rate in what turned out to be a rather tricky puzzle.

Last week, you were asked to consider a large iceberg that was a perfect cube. Thanks to the difference in densities between ice and water, 10 percent of the iceberg’s mass was above sea level, while the remaining 90 percent of the mass was below.

In its “optimal” orientation, the iceberg’s center of mass was as deep in the water as possible. Suppose the highest point (or points) of the iceberg was a distance x above sea level, while the lowest point (or points) of the iceberg was a distance y below sea level. What was the ratio of x to y?

Many solvers (myself included) immediately thought of a cube with one vertex sticking straight up out of the water, while the other vertex was plunged deep into the water. Surely, this would result in the center of mass being as deep as possible. But let’s check the math on that.

Dirk Meijer made a sketch of the cube oriented this way:

That dashed equilateral triangle at the top signified the water level, above which was a tetrahedron with three perpendicular edges descending from the apex. If each of these edges had length c, then this tetrahedron had volume c³/6. And because we knew that had to be 10 percent of the cube’s volume (which we’ll set at 1 for convenience), that meant c was equal to (0.6)^1/3, or about 0.8434. (I don’t know about you, but I found this value to be surprisingly large—the three topmost edges were almost fully emerged from the water in this configuration!)

With this value of c, you next had to determine x. Because the right triangular faces of the tetrahedron were 45-45-90, you knew the side lengths of the equilateral base were c√(2). That meant the altitude of the base was c√(1.5). And that meant the distance from its centroid to one of its vertices was 2/3·c√(1.5). Applying the Pythagorean theorem to this distance, a descending edge of the tetrahedron (length c) and the tetrahedron’s altitude (length x), you could show that x was equal to c/√(3), and hence that x was equal to (1/75)^1/6, or approximately 0.487.

Since the internal diagonal of a unit cube has length √(3), that meant y was equal to √(3)−(1/75)^1/6, or about 1.245. Finally, this “optimal” ratio x/y was roughly 0.3911. Right?

Wrong!

In this orientation, the cube’s center of mass was a distance (y−x)/2 below the surface, which came out to around 0.379. Meanwhile, if the cube were orientated so that its topmost face was parallel to the surface of the water, x would have been 0.1, y would have been 0.9, and (y−x)/2 would have been 0.4. In other words, an ice cube sits deeper in the water when its topmost face is parallel to the surface than when it has a vertex sticking out.

Not only that—having a face parallel to the water turned out to have the “optimal” orientation, with a center of mass as deep in the water as possible.

In the end, no trigonometry was necessary. (But proving no trigonometry was necessary was a beast of a problem.) The ratio x/y was simply 1/9, just like the volumetric ratio.

Finally, solver Andy Quick tried this at home by dropping an ice cube into a pot of water. Sure enough, its topmost face was parallel to the surface!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jon Moon  🎻 from Sunnyvale, California. I received 47 timely submissions, of which 40 were correct—good for an 85 percent solve rate, and the second time in just three weeks that the Extra Credit had more correct solutions than the Fiddler. (Come on, Zach, do your job! The Extra Credit is supposed to have fewer solvers, right?)

For Extra Credit, the iceberg was now a perfect sphere. As before, the highest point of the iceberg was a distance x above sea level, while the lowest point of the iceberg was a distance y below sea level.

What was the ratio of x to y?

Thanks to the sphere’s symmetry, you didn’t have to worry about the iceberg’s orientation this time around. (Phew!)

A few solvers like Michael Montuori blazed through the calculus to find the volume of the “cap” of the sphere that was above the water’s surface, while Emilie Mitchell cleverly subtracted the volume of a cone from the volume of a spherical sector. Most solvers, however, simply looked up the formula for a spherical cap of thickness x in a sphere of radius r: 𝜋x²/3·(3r−x). For simplicity, let’s make this a unit sphere (i.e., r = 1), giving us the formula 𝜋x²/3·(3−x). This volume had to equal 10 percent of the sphere’s total volume, 4/3·𝜋.

Setting these volumes equal gave you a cubic equation: x²·(3−x) = 2/5. The only solution to this cubic between 0 and 2 was x ≈ 0.3916. Meanwhile, y was equal to 2−x, which meant x/y was approximately 0.24347—or, as solver Joel Lewis said, “0.24something.” So for the spherical geometry, while 90 percent of the iceberg’s mass was submerged, only about 80 percent of its length was submerged.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.