A tangram paradox from Sam Loyd’s Eighth Book of Tan (1903). Each of these cups was composed using the same seven geometric shapes. But the first cup is whole, and the others contain vacancies of different sizes.
“Of course it is a fallacy, a paradox, or an optical illusion, for you will say the feat is impossible!” But how is it done?
nikhedonia
n. the pleasure of anticipating victory or success
On June 30, 1999, sheriff’s officers discovered the body of 41-year-old Ricky McCormick in a field in St. Louis. In his pockets were the two hand-printed documents above. Both the FBI and the American Cryptogram Association have failed to decipher the notes, so they’ve issued an appeal for help from the public.
Investigators believe the notes were written up to three days before McCormick’s death; his family says he’d used such encrypted notes since he was a boy. “Breaking the code could reveal the victim’s whereabouts before his death and could lead to the solution of a homicide,” said Dan Olson, chief of the FBI’s Cryptanalysis and Racketeering Records Unit. “Not every cipher we get arrives at our door under those circumstances.”
This is not an April Fools’ joke — the FBI’s appeal, including larger versions of the images, is here.
(Thanks, Bunk.)
Grant Wood’s American Gothic is a bit ersatz — the artist recruited a Cedar Rapids dentist, B.H. McKeeby, to pose as the farmer, and his sister Nan plays the woman (conceived as the farmer’s spinster daughter, not his wife).
But the setting was inspired by a real cottage in Wood’s native Iowa, and by his admiration for “the kind of people I fancied should live in that house.”
“I tried to characterize them honestly, to make them more like themselves than they are in actual life,” he said. “To me they are basically good and solid people.”
The lowly 3×3 magic square has modest pretensions — each row, column, and diagonal produces the same sum.
But perhaps it’s magicker than we suppose:
6182 + 7532 + 2942 = 8162 + 3572 + 4922 (rows)
6722 + 1592 + 8342 = 2762 + 9512 + 4382 (columns)
6542 + 1322 + 8792 = 4562 + 2312 + 9782 (diagonals)
6392 + 1742 + 8522 = 9362 + 4712 + 2582 (counter-diagonals)
6542 + 7982 + 2132 = 4562 + 8972 + 3122 (diagonals)
6932 + 7142 + 2582 = 3962 + 4172 + 8522 (counter-diagonals)
(R. Holmes, “The Magic Magic Square,” The Mathematical Gazette, December 1970)
More: Any of the equations above will still hold if you remove the middle digit or any two corresponding digits in each of the six addends.
Yet more: (6 × 1 × 8) + (7 × 5 × 3) + (2 × 9 × 4) = (6 × 7 × 2) + (1 × 5 × 9) + (8 × 3 × 4)
I think everything above will work for any rotation or reflection of the square (that is, for any normal 3×3 magic square). I haven’t checked, though.
If you have ever, like me,
Missed the “r” and hit the “t,”
Addressing some fat blister
As “Mt.” instead of “Mr.,”
I trust you left it unamended?
Splendid.
— J.B. Boothroyd, Punch, 1948
In a 1772 letter to Joseph Priestley, Ben Franklin described a method “for arriving at decisions in doubtful cases.” He would divide a sheet of paper into two columns, labeled Pro and Con, and during the course of three or four days record all the motives for and against the idea. Then he’d assign a weight to each consideration. Where he could find arguments, sometimes in combination, that counterbalanced one another, he would strike them out:
Should I enter into business with Mr. Smith?
(This example is from Paul C. Pasles, Benjamin Franklin’s Numbers, 2008.) This exercise would show him where the balance lay, and if after a day or two of further reflection no additional considerations occurred to him, he would come to a decision.
“Though the weight of reasons cannot be taken with the precision of algebraic quantities, yet, when each is thus considered separately and comparatively, and the whole lies before me, I think I can judge better, and am less liable to make a rash step; and in fact I have found great advantage from this kind of equation, in what may be called moral or prudential algebra.”
A perpetual-motion scheme from Henry Dircks’ Perpetuum Mobile (1861). Each bellows is fitted with a weight and filled with quicksilver, and a canal connects each opposing pair of bellows. Thus the weights will continually compress the bellows on the left and expand those on the right, forcing the quicksilver always into the rightmost bellows and ensuring that the wheel turns forever. Won’t they?
Imagine yourself to be a surgeon, a truly great surgeon. Among other things you do, you transplant organs, and you are such a great surgeon that the organs you transplant always take. At the moment, you have five patients who need organs. Two need one lung each, two need a kidney each, and the fifth needs a heart. If they do not get those organs today, they will all die; if you find organs for them today, you can transplant the organs and they will all live. But where to find the lungs, the kidneys, and the heart? The time is almost up when a report is brought to you that a young man who has just come into your clinic for his yearly check-up has exactly the right blood-type and is in excellent health. Lo, you have a possible donor. All you need to do is cut him up and distribute his parts among the five who need them. You ask, but he says, ‘Sorry. I deeply sympathize, but no.’ Would it be morally permissible for you to operate anyway?
— Judith Jarvis Thompson, “The Trolley Problem,” Yale Law Journal, 1985
Is a legal chess game possible in which all the pawns promote and each player has nine queens?
Yes — Freidrich Burchard of Germany and Friedrich Hariuc of Romania reached nearly identical solutions in 1980:
1. e4 f5 2. e5 Nf6 3. exf6 e5 4. g4 e4 5. Ne2 e3 6. Ng3 e2 7. h4 f4 8. h5 fxg3 9. h6 g5 10. Rh4 gxh4 11. g5 g2 12. g6 Bg7 13. hxg7 g1=Q 14. f4 h3 15. f5 h2 16. b4 a5 17. b5 a4 18. b6 a3 19. Bb2 Ra7 20. bxa7 axb2 21. a4 b5 22. a5 b4 23. a6 b3 24. c4 h1=Q 25. c5 h5 26. c6 Bb7 27. cxb7 c5 28. d4 c4 29. d5 Nc6 30. dxc6 c3 31. c7 c2 32. c8=Q c1=Q 33. b8=Q Qc7 34. a8=Q d5 35. a7 d4 36. Nc3 dxc3 37. Qa6 c2 38. Qa8b7 c1=Q 39. a8=Q Qd5 40. gxh8=Q+ Kd7 41. g7 bxa1=Q 42. g8=Q b2 43. f7 b1=Q 44. f8=Q h4 45. f6 h3 46. f7 h2 47. Qfa3 h1=Q 48. f8=Q exf1=Q+
This may be the shortest possible such game.
