Butler University mathematician Jerry Farrell has telekinesis. Here’s a demonstration. Toss a coin and enter the result (HEAD or TAIL) as 1 Across in the grid below. Then solve the rest of the puzzle:
Across Down
1 Your coin shows a ______ 1 Half a laugh
5 Wagner's earth goddess 2 Station terminus?
6 Word with one or green 3 Dec follower?
4 Certain male

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Here’s the solution without your coin result:
The fact that your result fits so neatly into the grid shows that Farrell controlled the coin flip. And how can he have done that but by telekinesis?
(From Alan Connor, The Crossword Century, 2014. Related: a clairvoyant penny.)

November 18, 2015  Puzzles
On the floor of a room of area 5, you place 9 rugs. Each is an arbitrary shape but has area 1. Prove that there are two rugs that overlap by at least 1/9.

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Suppose that this isn’t true — that every pair of rugs overlaps by less than 1/9. Place the rugs on the floor one by one and note how much of the bare floor each succeeding rug must cover. The first rug must cover area 1, or 9/9. The second must cover an area greater than 8/9 (because less than 1/9 can overlap the first rug). The third rug can overlap the first two somewhat, but must still cover an area greater than 7/9 of the bare floor. And so on: The fourth, fifth, … ninth rug will cover an area greater than 6/9, 5/9, … 1/9 on the bare floor. Since 9/9 + … + 1/9 = 5, the nine rugs together must cover an area greater than 5. That’s a contradiction, so our supposition can’t be true — some pair of rugs must overlap by at least 1/9.
From Arthur Engel, ProblemSolving Strategies, 1998.

November 12, 2015  Puzzles
By W. Speckmann, 1973. White to mate in two moves.
November 9, 2015  Puzzles
In his 1943 book The Life of Johnny Reb, Emory University historian Bell Wiley collects misspellings found in the letters of Confederate soldiers. Can you decipher these words?
 agetent
 bregad
 cerce
 crawsed
 furteege
 orpital
 perperce
 porchun
 regislatury
 ridgement
Bonus: What does A brim ham lillkern mean?
Click for answers …
October 31, 2015  History, Language, Puzzles
By O. Wurzburg, 1919. If Black does not move at all, in how few moves can the white king reach f4? White can move only his king; as in regular play, it can capture enemy pieces but cannot enter check.

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Seventy! In order to fulfill the task, the white king has to capture every black piece but the king while stepping carefully around the guarded squares.
Ka4 2. 2. Ka5 3. Ka6 4. Kxa7 5. Kb6 6. Kc7 7. Kd7 8. Ke7 9. Kf7 10. Kg6 11. Kf5 12. Kg4 13. Kg3 14. Kf2 15. Ke1 16. Kd1 17. Kxc1 18. Kd1 19. Ke1 20. Kf2 21. Kg3 22. Kg4 23. Kf5 24. Kg6 25. Kf7 26. Ke7 27. Kd7 28. Kc7 29. Kb6 30. Ka5 31. Ka4 32. Ka3 33. Ka2 34. Kxa1 35. Ka2 36. Ka3 37. Ka4 38. Ka5 39. Kb6 40. Kc7 41. Kd7 42. Ke7 43. Kf7 44. Kg7 45. Kxh8 46. Kg8 47. Kf7 48. Ke8 49. Kd8 50. Kc8 51. Kb8 52. Kxa8 53. Kb7 54. Kc6 55. Kd5 56. Ke5 57. Kf5 58. Kg4 59. Kg3 60. Kg2 61. Kxh1 62. Kh2 63. Kh3 64. Kg4 65. Kh5 66. Kh6 67. Kxh7 68. Kg6 69. Kxg5 70. Kf4
From Stephen Addison, The Book of Extraordinary Chess Problems, 1989.

October 27, 2015  Puzzles
Some “ridiculous questions” from Martin Gardner:
1. A convex regular polyhedron can stand stably on any face, because its center of gravity is at the center. It’s easy to construct an irregular polyhedron that’s unstable on certain faces, so that it topples over. Is it possible to make a model of an irregular polyhedron that’s unstable on every face?
2. The center of a regular tetrahedron lies in the same plane with any two of its corner points. Is this also true of all irregular tetrahedrons?
3. An equilateral triangle and a regular hexagon have perimeters of the same length. If the area of the triangle is 2 square units, what is the area of the hexagon?

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1. No. If this were possible then a perpetual motion machine could be built — on a horizontal plane the object would be constantly moving.
2. Yes. Any three points lie in the same plane.
3. Three square units:
(Martin Gardner, “Ridiculous Questions,” Math Horizons 4:2 [November 1996], 2425.)

October 24, 2015  Puzzles, Science & Math
AB = XY. Find z°.

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Construct equilateral triangle ABW. Now:
∠WAY = 40° (because each interior angle of ABW is 60° and the interior angles of triangle XAY must total 180°)
p = p (because triangle ATY is isosceles)
q = q (because AB = AW = XY = p + q)
∠AWY = 40° (because triangles ATX and YTW are similar)
r = r (because triangle AYW is isosceles)
This shows that ABWY is a kite, a quadrilateral with two pairs of equallength sides, each pair adjacent. The diagonals of a kite are perpendicular, so the triangle that contains angle z also contains interior angles measuring 90° and 80°. Angle z is 10°.
(Thanks, Derek.)

October 22, 2015  Puzzles, Science & Math
If the proportion of blonds among blueeyed people is greater than among the population as a whole, is it also true that the proportion of blueeyed people among blonds is greater than among the population as a whole?

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Yes. Let a be the number of blonds with blue eyes, b be the number of all blonds, c be the number of all blueeyed people, and n be the whole population. Then by the information we’re given, a/c > b/n, so an > bc and a/b > c/n.
(By A. Savin)

October 19, 2015  Puzzles
My employer has nine workers. The nine of us want to determine what our average salary is, but none of us wants to divulge his own salary. Can we find the average without doing so?

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Yes. I privately pick some large random number — say, 645,743 — add my own salary to it, and whisper the total to Worker #2. He adds his own salary to that sum and whispers the new total to Worker #3, and so on. When the last worker has added his own salary, he whispers the final amount back to me. I subtract the random number that I’d chosen and divide the remainder by 9. Now we know the average salary of the nine workers, but none of us knows anything about what the others make.
11/03/2015 UPDATE: The solution above is not quite optimal because the workers can still make some inferences about their coworkers’ maximum salaries. For example, if I give Worker #2 the number 701,229, he knows that my salary cannot be higher than that. This can be improved by allowing the random number to be negative — in that case no inferences are possible, and the math still works. (Thanks, Daniel and Jose.)

October 17, 2015  Puzzles
A poser by F.H. von Meyenfeldt, 1967. What move must Black play to enable a forced mate in two by White?

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… Bh3 enables 1. Rh1, and Black can’t stop 2. Bg7#.
From Stephen Addison, The Book of Extraordinary Chess Problems, 1989.
(10/12/2015 This seems to be cooked. After … Bg4, … Bd7, or Bc8, White can withdraw his bishop along c1h6 and mate on the long diagonal. Thanks, Malcolm.)

October 11, 2015  Puzzles