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You may not know this, but we have a strange penchant for optical illusions, fractals, and other mind-perplexing images. That’s why we got so impressed when we saw for the first time the “infinite-zoom” artwork by Nikolaus Baumgarten. Bearing the suitable name “The Zoomquilt”, this seemingly never-ending image instantly became our favorite screen-saver. Take a look at it by clicking the image below, and if you like it, make sure to search YouTube for other similar illustrations.
Strata is a beautiful award-winning game with mesmerizing sound and unique puzzle concept. It contains hundreds of levels with common rules and final goal. Below I present you these rules and ask you to find a universal algorithm, which will allow you to solve easily every single level of the game.
The rules are simple – you begin with an nxn board, some squares of which are colored in arbitrary colors. Then you start placing stripes of whatever color you choose over entire rows and columns of the board. Your task is after placing all available 2n stripes, the color of every (colored) square to match the color of the stripe which has been placed second over it (on top).
Can you find a simple algorithm, which results in solving any level of the game, no matter the starting position? You can watch AppSpy’s video below for better understanding of the rules.
Imagine the reverse Strata puzzle – the color of every square must match the color of the first stripe which is placed upon it. Clearly, there must be a line in the grid such that all colored squares in it have the same color. Take all such lines in the grid and place on them stripes of appropriate colors. Then erase the colors from all squares covered by the stripes and repeat the procedure until you place all 2n stripes. It is easy to see that if the reverse Strata puzzle has a solution, then we will find it using this strategy. Finally, in order to solve the original Strata puzzle, just place the stripes in reverse order.
How well can your brain focus and what happens when it does that? Watch these two awareness tests and you may get surprised by the results.
52 cards – 2 of clubs to Ace of clubs, 2 of diamonds to Ace of diamonds, 2 of hearts to Ace of hearts, and 2 of spades to Ace of spades – are arranged in a deck. We shuffle them in the following manner:
Show that this method shuffles the deck uniformly, i.e. every permutation has the same chance to appear.
Notice that at all times the cards below the King of spades are shuffled uniformly. Therefore at the end, after we put the King of spades in a random place inside the deck, the entire shuffle will be uniform as well.
The chances are you have already seen the hundreds of optical illusions we have collected for you on Puzzle Prime, but have you ever encountered any audio illusions? The YouTube channel AsapSCIENCE has created a short video in which they present and explain some of the most famous audio illusions, such as the McGurk effect and the Tritone paradox. Watch their video below and see if you can trust your ears.
Can you construct a convex polyhedron, such that no two of its faces have the same number of edges?
No, you can not construct such a polyhedron. Assume the opposite and consider its face, which has the largest number of sides, say k. Then the polyhedron contains at least k more faces with different numbers of sides, all less than m. However, this is clearly impossible.
In 1996, just a day before the election of the 40th President of US, the New York Times published a curious crossword. In the 8th row, the solver should discover a phrase – the “lead story of tomorrow’s newspaper”. More precisely – the name of the future President of the country appears there. But how could New York Times know whether it was going to be Clinton or Bob Dole?
ACROSS:
1. “___ your name” (Mamas and Papas lyric)
6. Fell behind slightly
15. Euripides tragedy
16. Free
17. Forecast
19. Be bedridden
20. Journalist Stewart
21. Rosetta ???
22. 1960s espionage series
24. ___ Perigion
25. Qulting party
26. “Drying out” program
28. Umpire’s call
30. Tease
34. Tease
36. Standard
38. “The Tell-Tale Heart” writer
39. Lead story in tomorrow’s newspaper, with 43A
43. See 39A
45. Gold: Prefix
46. ___ Lee cakes
48. Bobble the ball
49. Spanish aunts
51. Obi
53. Bravery
57. Small island
59. Daddies
61. Theda of 1917’s “Cleopatra”
62. Employee motivator
65. Otherworldly
67. Treasure hunter’s aid
68. Title for 39A next year
71. Exclusion from social events
72. Fab Four name
73. They may get tied up in knots
74. Begin, as a maze
DOWN:
1. Disable
2. Cherry-colored
3. Newspaperman Ochs
4. Easel part
5. Actress Turner
6. Ropes, as dogies
7. Place to put your feet up
8. Underskirt
9. First of three-in-a-row
10. Lower in public estimation
11. Onetime bowling alley employee
12. Threesome
13. English prince’s school
14. ’60s TV talk-show host Joe
18. Superannuated
23. Sewing shop purchase
25. TV’s Uncle Miltie
27. Short writings
29. Opponent
31. Likely
32. Actress Caldwell
33. End of the English alphabet
35. Trumpet
37. Ex-host Griffin
39. Black Halloween animal
40. French 101 word
41. Provider of support, for short
42. Much debated political inits
44. Sourpuss
47. Malign
50. “La Nausee” novelist
52. Sheiks’ cliques
54. Bemoan
55. Popsicle color
56. Bird of prey
58. 10 on a scale of 1 to 10
60. Family girl
62. Famous ___
63. Something to make on one’s birthday
64. Regarding
65. Quite a story
66. Dublin’s land
69. ___ Victor
70. Hullabaloo
The answer is simple, yet very impressive. The crossword’s author, the mathematics professor Jeremiah Farrell, created the puzzle so that it could be solved in two different ways, revealing either “Clinton Elected” or “Bob Dole Elected” in the middle row. Many of the newspaper’s readers didn’t realize the prank and assumed New York Times was displaying a bias towards one of the candidates. They started sending lots of angry letters and calling the editor, complaining about arguably the coolest crossword of all time.
Which was your favorite part of the Mathematics you
Prove that among any 9 points in (3D) space, there are three which form an obtuse angle.
Let the points be labeled A1, A2, … , A9, and P be their convex hull. If we assume that all angles among the points are not obtuse, then the interiors of the bodies P + A1, P + A2, … , P + A9 should be all disjoint. That is because, for every Ai and Aj, P must be bound between the planes passing through Ai and Aj which are orthogonal to the segment AiAj. However, all of these 9 bodies have the same volume and are contained in the body P + P, which has 8 times larger volume. This is a contradiction, and therefore our assumption is wrong.
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