Yes — here’s one possibility:

(From E.G. Kozlova, Skazki i podskazki: Zadachi dlya matematicheskogo kruzhka [Fairy Tales and Hints: Problems for a Mathematical Circle], 2004 [in Russian], via Jennifer Beineke and Jason Rosenhouse, The Mathematics of Various Entertaining Subjects, 2017.)

1. e4 e6 2. Bb5 Ke7! 3. Bxd7 c6 4. Be8 Kxe8.

(Thanks, Jacques.)

No, it’s not possible. Imagine that the corner sub-cubes are black and that any two sub-cubes that share a face are of different colors:

Because the mouse always travels orthogonally, it will visit black and white squares alternately. And because the large cube contains an odd number of sub-cubes, any such route that starts in a black sub-cube and tours the whole cube must end in a black sub-cube. But the center sub-cube is white. So the task is impossible.

(From “Mathematical Mayhem,” Crux Mathematicorum 28:1 [February 2002], 34.)

For special arc PQ, construct the smallest isosceles right triangle ABC with hypotenuse AB on line PQ that contains the arc. Let arc PQ touch the legs of this triangle at R and S. Now reflect arc PR in AC and arc SQ in BC, obtaining arcs RP’ and SQ’. Now arc PRSQ equals arc P’RSQ’ in length and lies between lines AP’ and BQ’, which are distance AB apart. The arc length is 1, and it’s not less than AB, which proves the theorem. “Wetzel concludes, ‘Pause and reflect!'”

(From Clayton W. Dodge, “Reflections of a Problems Editor,” in Joby Milo Anthony and Howard Whitley Eves, In Eves’ Circles, 1994.)

Place triangle ABC into a square and draw the grid shown here, in which each line is parallel to EB or AC and contains a midpoint or quarter point on the side of the square. With CE and BE having lengths 3 and 4, we can determine that hypotenuse CB is of length 5.

“We have therefore shown the nature of this right triangle using a cleverly constructed diagram, and we have avoided lots of tedious computation,” write Alfred S. Posamentier and Ingmar Lehmann in Mathematical Curiosities. “This is an example where we can see how an elegant solution underscores the beauty of mathematics.”

1. Qa5! leads to four different mates:

1. … Kxb3 2. Bd1#
1. … Kb1 2. Bg6#
1. … b1=Q/R/B 2. Qc3#
1. … b1=N 2. Qa2#

From American Chess Bulletin, September-October 1940. In How to Solve Chess Problems (1945), Kenneth Howard writes, “The black pawn promotion to queen or to knight, with two different mates, is the most interesting feature of the problem. … While the solution presents little difficulty, a position like this delights the solver because of its daintiness.”

Any two of the tetrahedron’s skew perpendicular opposite edges define a series of parallel planes. The plane that passes midway between these edges produces a square cross section.

Say a player is in position H if H is the number she’s named.

Any player in position 90-99 is lost, because her opponent can advance immediately to 100.

That means that a player in position 89 will win, since her opponent will be forced to enter this losing territory.

But then position 78 is also winning, since the opponent must open a road to 89.

And so on backward: 67, 56, 45, 34, 23, 12, and 1 are all winning positions.

With correct play, then, the first player always wins: She chooses 1 and then aims for each of these milestones.

The key is to notice that the 1 × 1 × 1 blocks must fall along a diagonal of the finished cube, to ensure that each 3 × 3 × 1 “slice” of the cube (in every orientation) can contain an odd number of blocks: