Black and White

William Shinkman published this problem in the St. Louis Globe Democrat in 1887. White is to mate in 8 moves:

shinkman chess problem - 1

It’s easier than it sounds — with the right approach, all Black’s moves are forced:

1. O-O-O Kxa7 2. Rd8 Kxa6 3. Rd7 Kxa5 4. Rd6 Kxa4 5. Rd5 Kxa3 6. Rd4 Kxa2 7. Rd3 Ka1 8. Ra3#

shinkman chess problem - 2

Remarkably, though the problem position looks contrived, it’s reachable in a legal game (discovered by Bader Al-Hajiri):

1. g4 e5 2. Nh3 Ba3 3. bxa3 h5 4. Bb2 hxg4 5. Bc3 Rh4 6. Bd4 exd4 7. Nc3 dxc3 8. dxc3 g3 9. Qd3 Rb4 10. Nf4 g5 11. h4 f5 12. h5 d5 13. h6 Bd7 14. h7 g2 15. h8B g1R!! 16. Bd4 Ba4 17. Rh4 Rg3 18. Bg2 gxf4 19. Be3 fxe3 20. Be4 fxe4 21. fxe3 exd3 22. exd3 c5 23. Rc4 dxc4 24. dxc4 b5!! 25. cxb4 Qa5 26. cxb5 Na6 27. bxa5 O-O-O!! 28. bxa6 Rd4 29. exd4 Rb3 30. cxb3 Ne7 31. bxa4 Nd5 32. dxc5 Nb6 33. cxb6 Kb8 34. bxa7+ Ka8

(Thanks, Florian.)

November 26, 2017November 25, 2017 | Puzzles

Quickie

From Martin Gardner: Each of the two equal sides of an isosceles triangle is one unit long. How long must the third side be to maximize the triangle’s area? There’s an intuitive solution that doesn’t require calculus.

	Select Click for Answer
Rotate the triangle until one of the two equal sides is horizontal, as above. Then rotate the other side through 180 degrees. The triangle’s area is half its base times its height, and this value is maximized when its height in this orientation is maximized, that is, when the rotating side is vertical, producing an isosceles right triangle whose hypotenuse is $\sqrt{2}$ . (From Wheels, Life and Other Mathematical Amusements, 1983.)

Click for Answer

November 22, 2017November 24, 2017 | Puzzles · Science & Math

Unsolved

On June 30, 1999, the body of 41-year-old Ricky McCormick was discovered near a cornfield in West Alton, Missouri. He’d last been seen alive five days earlier; now he was 15 miles from home though he owned no car. In his pockets were two handwritten notes (click to enlarge):

In the ensuing 18 years both the FBI’s Cryptanalysis and Racketeering Records Unit and the American Cryptogram Association have failed to find any meaning in these messages. In 2011 the FBI appealed to the public for their insights: If you have any you can contact them via this page.

“We are really good at what we do,” said CRRU chief Dan Olson, “but we could use some help with this one.”

November 22, 2017November 22, 2017 | Crime · Puzzles

Podcast Episode 178: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

November 20, 2017November 20, 2017 | Podcast · Puzzles

The Grapevine

A problem from the British Columbia Colleges Senior High School Mathematics Contest, 2000:

Not all of the nine members on the student council are on speaking terms. This table shows their relationships — 1 means two members are speaking to each other, and 0 means they’re not:

  A B C D E F G H I
A - 0 0 1 0 0 1 0 0
B 0 - 1 1 1 1 1 1 1
C 0 1 - 0 0 0 1 1 0
D 1 1 0 - 1 0 1 0 1
E 0 1 0 1 - 0 1 0 0
F 0 1 0 0 0 - 0 0 1
G 1 1 1 1 1 0 - 0 0
H 0 1 1 0 0 0 0 - 0
I 0 1 0 1 0 1 0 0 -

Recently councilor A started a rumor, and it was heard by each councilor once and only once. Each councilor heard it from, and passed it to, another councilor with whom she was on speaking terms. If we count councilor A as zero, then councilor E was the eighth and last councilor to hear the rumor. Who was the fourth?

	Select Click for Answer
We know that A is at the start of the sequence, and E is at the end. Of the others, F and H are each speaking to only two other councilors. So F must pass the rumor between B and I (in one direction or the other, BFI or IFB), and H must pass it between B and C (BHC or CHB). Putting these fragments together, we must have either CHBFI or IFBHC. Also, this larger fragment can’t connect directly to either of the endpoints A or E, because neither C nor I is speaking to either of them. That means that the remaining councilors, D and G, must go on either end of our fragment before we can attach the endpoints. So the extended fragment is either GCHBFID or DIFBHCG, and with the endpoints A and E added we get the full sequence AGCHBFIDE or ADIFBHCGE. In either case, if A is counted as zero then the fourth person is B. (Crux Mathematicorum 27:3 [April 2001], 194.):

Click for Answer

November 16, 2017 | Puzzles

Murder Most Classificatory

In Anthony Boucher’s short story “QL 696.C9,” a librarian is found dead at her desk. She has been shot, and apparently spent her last moments typing the Library of Congress card catalog number that gives the story its title. The killer evidently saw nothing incriminating in this and so left it alone. The investigators have narrowed the list of suspects to junior librarian Stella Swift, children’s librarian Cora Jarvis, library patron James Stickney, and high school teacher Norbert Utter. Who did it?

	Select Click for Answer
In the Library of Congress system, QL refers to zoology, the 600s to vertebrates, 696 to birds, and .C9 to Cypseli — or swifts.

November 14, 2017 | Puzzles

A Mating Machine

elkies chess puzzle

A “no-brainer” by Harvard mathematician (and chess master) Noam Elkies. White has choices as to how to play, but no matter how he proceeds, he’ll wind up mating Black on the seventh move.

(From Tim Krabbé’s chess diary.)

November 10, 2017 | Puzzles

Black and White

tarrant chess problem

By Thomas Tarrant. White to mate in two moves.

	Select Click for Answer
1. Nf5! and 2. Rb5 will mate. From the Westminster Papers, 1877.

November 2, 2017November 1, 2017 | Puzzles

Coverup

A problem from the British Columbia Colleges Senior High School Contest for 2000:

If I place a 6 cm × 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the triangle on the square, I can cover up to 2/3 of the square. What is the area, in cm², of the triangle?

(a) 22 4/5
(b) 24
(c) 36
(d) 40
(e) 60

	Select Click for Answer
The answer is (d). The key is to see that these measures are one and the same: The triangle overlaps the square by the same amount that the square “underlaps” the triangle, so these areas are maximized at the same moment. If A is the area of the triangle, then 0.6A = 2/3 × 36, and A = 40 cm². From “The Skoliad Corner,” Crux Mathematicorum, December 2000.

October 18, 2017October 17, 2017 | Puzzles

Black and White

shinkman chess problem

“We now give what is acknowledged to be the finest two-move problem extant,” wrote J.H. Blackburne in the Strand in 1908. “It is by the American expert, W.A. Shinkman, and is also claimed by G.E. Carpenter, a fellow countryman of his. Here we have not only a difficult key-move, but also beauty of theme and artistic construction, the three essential qualities necessary to a perfect problem.”

White to mate in two moves.

	Select Click for Answer
“It is solved by B to R 4th [that is, 1. Ba4], moving the B from a strong position, where it defended a Kt, to, apparently, a useless square. In reply, Black has a choice of five moves”: 1. … f5 2. Qg8# 1. … f6 2. Ndc7# 1. … e4 2. Qxe4# 1. … Kxd5 2. Bb3# 1. … d6 2. Nbc7# “Truly a marvellous composition.” (“Bridge and Chess Problems,” Strand, March 1908.)

October 12, 2017October 11, 2017 | Puzzles