Henle’s solution:

The black king is in check, and the check must have been discovered when the white king moved from f3 to e2. But how is this possible? The white king would itself have been in check in two different ways on f3. How could Black have inflicted two checks at once?

The answer is that a black pawn must have stood on f2, blocking the rook’s check. This pawn then captured some white piece on e1, promoting to a knight and inflicting a double check. So, from this position:

1. … fxe1=N+ 2. Ke2+.

(Jim Henle, “The Entertainer,” Mathematical Intelligencer 40:2 [June 2018], 76-80.)

If one of the men is unacquainted with four of the others, then those four know one another and we’re done.

If not, then this means that each man is mutually acquainted with at least five of the others. It can’t be exactly five in every case, because the total number of acquaintances in the room must be an even number. Hence someone is acquainted with at least six men.

Among those six, three know one another, and they together with the man himself make four.

1. Kitty
2. Joan
3. Betty
4. Mary
5. Ethel

It’s possible to begin with any of the girls’ statements, but here’s one solution:

Two girls say that Mary placed fourth. If this is not true then Kitty was second (per Kitty’s statement) and Betty first (per Mary’s statement). But if Betty is first then Joan is second (per Ethel’s statement), which is a contradiction.

That means it’s true that Mary placed fourth, and it follows that Kitty is not second (per Kitty’s statement).

Hence Betty placed third (per Betty’s statement), Ethel placed fifth (per Joan’s statement), and Joan placed second (per Ethel’s statement). That leaves Kitty, who by process of elimination must have placed first.

From Problem Omnibus, 1960.

02/21/2025 UPDATE: There are at least three additional solutions. Label the columns A-D and the rows 1-4, like a miniature chessboard. Now:

A4-A1-B1-D1-D4-D2-C2-C4-C1-C3-A3-A4 (from reader Justin Hilyard)

A4–D4–B4–B2–C2–C4–C1–C3–A3–A4 (from reader Heinrich Hemme)

A4-A1-A2-D2-C2-C4-C1-C3-A3-A4 (from reader Paul Wagner)

They’re surprisingly plain once you see them — their joined lips are just to the left of Bacchus’ nose.

From “Old Puzzle Cards,” Strand, November 1899.

No. Think of the large cube as a square well with a flat bottom. If the task is possible, then this floor is tiled perfectly with cubes of different sizes. Consider the smallest of these cubes. This can’t adjoin an edge of the floor, because in that position it would be impossible to surround it with cubes larger than itself. So it lies somewhere on the expanse of the floor, surrounded by larger cubes.

But now we’re back where we started. The top of this small cube forms the floor of a square well, and all the reasoning above applies again. And so on forever: At each stage we’ll be forced to supply smaller cubes, and our task will never be finished.

Ten.

The bishops are nicely aligned with White’s king; if Black had no other option, he’d have to play … Bxg2, giving mate. But White can’t simply withdraw his bishop along the long diagonal, as then Black could shift his own bishop mildly to g2, avoiding mate.

What other options does White have? He mustn’t move his knight, as that would allow the Black king to move. 1. e6 allows Black to take the f6 pawn. Moving the f6 pawn allows Black to take the g7 pawn. Promoting the g pawn to a queen or a rook counterproductively protects the g2 bishop. Promoting it to a knight checkmates Black (!). Promoting it to a bishop almost works, but after 1. g8=B exf6 2. exf6 Bxg2+ White now has 3. Bd5, which again prevents checkmate.

The only resource that works is to promote the c7 pawn to a knight. Now Black has only two ways to avoid giving mate, and both are only temporary:

1. … exf6 2. exf6 Bxg2#
2. … e6 2. g8=B Bxg2# (as now the promoted bishop can’t reach d5)

Note that the c7 pawn must promote to a knight — any other piece could block the final check.