Sections: Pure, Games, Geometry, Physics, Large.

If you make any progress on any of these puzzles, or if you'd like to contribute a new one, then please do tell me about it.

Puzzles in this section:

- 13 points on a circle --- spoiler
- Glass-wall Roman numeral determinant --- spoiler
- Pyewackett cover --- spoiler
- Sequences
- Sir Arthur Eddington's Cricket Problem --- spoiler
- Triple minor sets --- spoiler
- Two black neighbours implies black --- spoiler

Q: ??? 13 points in the plane are such that given any three
some fourth one lies on the same circle as those three. Show that all 13
are cocircular.

[Julian Gilbey, 13 Jun 1996]

Spoiler --- Top of page

Q: Find a determinant identity using Roman numerals that can be
painted on a glass wall and be valid from both sides. To be interesting
it should involve the Roman decrement convention (eg IV = V - I).

[rec.puzzles]

Spoiler --- Top of page

Q: ??? The cover of Pyewackett's "This Crazy Paradise" album includes an image something like the following:

Is this significant? If so, why?

[29 Aug 1996]

Spoiler --- Top of page

Q: ??? What is the next term in each of the following?:

- 0, 1, 10, 100, 11, 1000, 3, 20 [Neal Sloane]
- 1, 12, 66, 220, 495, 792 [rec.puzzles, 2 Feb 1996]
- 25, 7, 49, 3, 17, 6, 67, 1, 39, 4, 75 [rec.puzzles, 4 Apr 1996]

Extract from the score of a cricket match between Eastershire and Westershire:

EASTERSHIRE - Second Innings:

A. A. Atkins | 6 |

Bodkins | 8 |

D. D. Dawkins | 6 |

Hawkins | 6 |

Jenkins (J) | 5 |

Larkins | 4 |

Meakins | 7 |

Hon. P. P. Perkins | 11 |

Capt. S. S. Simkins | 6 |

Tomkins | 0 |

Wilkins | 1 |

Extras | 0 |

Total | 60 |
---|

Bowling Analysis:

Overs | Maidens | Runs | Wickets | |
---|---|---|---|---|

Pitchwell | 12.1 | 2 | 14 | 8 |

Speedwell | 6 | 0 | 15 | 1 |

Tosswell | 7 | 5 | 31 | 1 |

The score was entirely in singles and fours, with no catches, no-balls, or short runs. Speedwell and Tosswell each had only one spell of bowling. Pitchwell bowled the first over, with Mr Atkins taking the first ball. Speedwell was the other opening bowler.

Q:

- Whose wickets were taken by Speedwell and Tosswell?
- Who was not out?
- What was the score at the fall of each wicket?

[Editor's note: This is a brilliant and complex puzzle. The spoiler has two complete solutions, but each is well over 1000 words. I do urge you to have a serious crack at it before you look at the spoiler. Here are some hints to get you started if you want, hidden by rot13 so as not to spoil the fun:

- Jub objyrq rnpu bire?
- Jung fpbevat fgebxrf jrer gurer va Gbffjryy'f aba-znvqra biref? Jub pbhyq unir fpberq gurz?
- Jung qbrf guvf gryy lbh nobhg Obqxvaf?
- Jung pna lbh fnl nobhg jung unccrarq va gur svefg srj biref?]

Spoiler --- Top of page

Q: In English country dancing a "triple minor" set consists of a
line of `n` couples, numbered 1, 2, 3, 1, 2, 3, ... from the
top: each block of three couples 1, 2, 3 is called a "minor set". At
each turn of the dance the 1s move down one place and the 2s move up one
place, and the 2s and 3s then swap numbers to form new minor sets
numbered 1, 2, 3 (omitting the new top couple). During any turn only
those in *complete* minor sets dance (experienced dancers sometimes
do a fudged version in an incomplete minor set, but ignore this for
now).

The 1s move down the set: when they reach the bottom - 1 position
they immediately move to the bottom, and wait two turns to restart as
3s. The others move (more slowly) up the set, alternately dancing as 2s
and 3s: when they reach the top - 1 position they wait one turn, dance
once as 2s to get to top place, and wait two turns to restart as
1s. Normally the dance has `t` = 3`u` + 1 turns, for
some `u`.

Many people believe that couples spend a greater proportion of time
dancing if `n` is not a multiple of 3. Is this true?

[David Barnert, ECD discussion list, 19 Jun 2001]

Q: Generalise to minor sets of `m` couples, as follows:
after the first turn the 1s have moved down one place and as before
become the top of new minor sets, so the couple below them change their
number to 2, the next couple to 3, etc. End effects in this case are
left as an exercise for the reader. Let `n` = `pm` +
`q`, and while we're at it generalise the number of turns to
`t` = `um` + `v`. How does the number of
couples affect the efficiency?

Notes: There are many "duple minor" English country dances, with
`m` = 2. There are quite a few "triple minors", with
`m` = 3 as above, which were particularly popular in the 17th
and 18th centuries because they tend to give the 2s and 3s more flirting
time. Many Scottish country dances also have this type of progression,
almost always with `n` = 4, `m` = 3 and `t` =
8 (although with such a small `n` the two sets of end effects
meet...). *Hampstead Manor* is a "quadruple minor", with
`m` = 4, but I don't know details, or any other such dances. A
dance with `m` = 3 or 4 should not be confused with a duple
minor that has a triple or quadruple progression. Nor is it the same as
a double contra, such as Jim Kitch's *Double Chocolate*, which is
different again.

Spoiler --- Top of page

Q: Consider a 10 x 10 array of white squares. Colour some initial
set black. Now, if a square has two black orthogonal neighbours, colour
it black. Repeat as much as possible. Show that if the initial set has 9
squares we cannot end with the whole array black.

[Dima Fon-Der-Flaass, 24 Feb 1997]

Q: More generally, find the minimum initial # blacks required to
turn an `m` x `n` array black.

Spoiler --- Top of page

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