“What Does The Mirror Say?”

The above line of figures does not appear very interesting at first sight, but if one asks some charming member of the fair sex to turn it upside down and hold it to a mirror to read it, a hidden meaning becomes apparent.

— Strand, December 1908

July 14, 2010August 26, 2010 | Puzzles

Side Business

Here’s proof that one leg of a triangle always equals the sum of the other two.

ABC is our triangle. Extend it make a parallelogram, as shown, and divide the parallelogram into a grid. Obviously,

AB + BC = (AG + HJ + KL + MN) + (GH + JK + LM + NC).

Now let the grid grow increasingly fine: Instead of dividing the parallelogram into a 4×4 grid, make it 5×5, then 6×6, and so on. With each iteration, the stairstep figure described above will approximate AC more closely, and yet its total length will always equal AB + BC. Thus, at the limit, AB + BC = AC. Where is the error?

(From Henry Dudeney’s Canterbury Puzzles, via W.W. Rouse Ball’s Mathematical Recreations and Essays, 1892.)

July 3, 2010August 26, 2010 | Puzzles

The 65th Square

65th square

By Eric Angelini, Europe Echecs, 1990.

White adds one square at the edge of the board and then mates in two.

	Select Click for Answer
Add the square at e9. Then 1. Ne9 Kxe9 (forced) 2. Qc7#.

July 3, 2010January 24, 2013 | Puzzles

A Remade Man

A tangram paradox by Sam Loyd. Each of these gentlemen is assembled from the same seven pieces. Yet one has a foot and the other doesn’t. How is this possible?

June 26, 2010August 29, 2010 | Puzzles

Car Hop

A classic railroad shunting puzzle. The segment at the top can accommodate either freight car, but not the locomotive. The freight cars can be joined if desired, and the locomotive can push or pull either or both cars from either direction. The task is to use the locomotive to swap the positions of the two cars.

	Select Click for Answer
Push A into the common area at the top. Return to the main line, back up, push B into A, and couple them. Draw the coupled cars back onto the main track and center them there. Now uncouple the cars, pull B to the right, then push it up the right-hand siding and into the common area. Return to the main track, pull A to the right, then push it into B’s former position. Then return to the main track, move to the left, back up the left-hand siding, and pull B into A’s former position.

Click for Answer

June 23, 2010January 24, 2013 | Puzzles

Track Meet

From Henry Dudeney:

A better class of puzzle is the well-known one of the Railway. If New York and San Francisco are just seven days’ journey apart, and if trains start from both ends every day at noon, how many trains coming in an opposite direction will a train leaving New York meet before it arrives at its destination at San Francisco?

	Select Click for Answer
Solution to Track Meet: This puzzle, which many people find so perplexing, is quite easy if you refer to the diagram I have given. It will be seen that at the moment the train starts from New York there are six trains already on the way in the positions indicated, since it is a seven days’ journey. The train numbered 7 also starts at the same time. Therefore the train has to meet all these seven. But when the New York train is at point 1, an eighth train will be leaving San Francisco, when it is at point 2 a ninth train will be leaving, at point 3 a tenth, at point 4 an eleventh, at point 5 a twelfth, and at point 6 a thirteenth train will be leaving San Francisco. Therefore the train will meet thirteen others on the journey. Of course, at the moment of its arrival a train will be just starting, but it cannot be said to ‘meet’ this one on the journey. UPDATE: An insightful reader points out that New York and San Francisco are in different time zones. This means that at the instant we start, the intermediate trains won’t precisely occupy the numbered positions in Dudeney’s diagram — they’ll be offset a bit. And that changes the answer to 14! In effect, the “train that will be just starting” will be fairly out of the station when we arrive (if only just) and thus counts as an extra encounter. That’s my best interpretation, anyway, in viewing Dudeney’s puzzle in this new light.

Click for Answer

June 19, 2010January 24, 2013 | Puzzles

Immaterialism

George Berkeley proposed that material things exist only because they are perceived to exist. In 1993, L.C. Rodó offered a suitably Berkeleyan puzzle in El Acertijo. Any piece, including a king, that is not regarded (attacked or guarded) by another piece disappears from the board. In the diagram above, every man is either attacked or guarded. But a single move (it may be by either White or Black) will set in motion a chain reaction in which all the pieces vanish from the board. What is the move?

	Select Click for Answer
Play 1. … d6 and you’ll clear the squares in roughly this order: c6, e6, d5, b5, f5, e4, c4, d3, g3, g2, e1, d2, d1, b3, g4, h6, a3, e3, f4, b2, h7, f6, c3, d8, h8, b4, a8, d6, f8, b7, b6, f7.

June 15, 2010January 24, 2013 | Puzzles

Assisted Suicide

From this position, how can White force Black to checkmate him in two moves?

	Select Click for Answer
1. Ba2 c2 (Black’s only legal move) 2. Rf8 Kxf8#. From Theodore Morris Brown, Book of Chess Problems, 1874.

June 2, 2010January 24, 2013 | Puzzles

“The Caged Bird”

Very little is known about Theophilus Thompson, a chess problemist who was born a slave in Maryland in 1855. That’s a pity, because his work is often beautiful. Here’s a sample:

thompson chess puzzle

White to mate in two.

	Select Click for Answer
White plays the surprising 1. Ne4!, threatening 2. Qa8#. If the king takes the knight then 2. Bc6 is mate, and if 1. … d3 then 2. Qxd3#. Sadly, Thompson disappeared shortly after publishing his only book in 1873. It’s feared he was lynched. Here’s another problem, to show the depth of his conception: “White to play and capture the Black Bishop with the King: In how few moves can this be effected? And in what manner?” The answer is to bully Black into locking his own bishop in a corner so that the white king can approach. All Black’s moves are forced: 1. Bg3+ hxg3 2. Rxe3+ Bxe3 3. Rxg1+ Bxg1 4. Qc1+ Kf2 5. Kd7 Bh2 6. Ke6 Bg1 7. Kf5 Bh2 8. Kg4 Bg1 9. Nf5 Bh2 10. Qd2+ Kg1 11. Nxg3 Bxg3 12. Kxg3 “Quod erat demonstrandum.”

Click for Answer

May 23, 2010January 24, 2013 | Puzzles

Undefined

smullyan undefined chess puzzle

Another retrograde chess puzzle by Raymond Smullyan. The pawn on g3 is of unknown color. During the game no piece has moved from a light square to a dark one or vice versa. Is the unknown pawn black or white?

	Select Click for Answer
The d2 and f2 pawns have not moved; how did the white king escape onto the board? White must have castled kingside. Now, if the pawn on g3 were white then it must have come from h2, and the king could not have left his castle. Therefore the pawn must be black.

May 16, 2010January 24, 2013 | Puzzles