Borromean Rings

Borromean rings are rings in the 3-dimensional space, linked in such a way that if you cut any of the three rings, all of them will be unlinked (see the image below). Show that rigid circular Borromean rings cannot exist.

Assume the opposite. Imagine the rings have zero thickness and reposition them in such a way, that two of them, say ring 1 and ring 2, touch each other in two points. These two rings lie either on a sphere or a plane which ring 3 must intersect in four points. However, this is impossible.

Saavedra Position

White to play. Is this game a win for White, Black, or a draw?

This game is a win for White.

1. c7 Rd6+
2. Kb5 Rd5+
3. Kb4 Rd4+
4. Kb3 Rd3+
5. Kc2! Rd4!
6. c8=R! Ra4
7. Kb3

Now Black will either lose the rook, or get mated in one. If White promoted a Queen instead of a Rook, then 6… Rc4+ would lead to 7. Qxc4, which is a stalemate.

Securing the Box

There are 5 people who possess a box. You are allowed to secure the box with as many different locks as you like and distribute any combination of keys for these locks to any people among the 5. Find the least number of locks needed, so that no 2 people can open the box, but any cannot people can open it.

For every subset of 2 people you pick among the 5, there should be a lock which none of the 2 can unlock, and each of the remaining 3 people can unlock. Clearly, the lock in question cannot be the same for any two different subsets of 2 people you choose. Therefore the number of locks you need is at least the number of different 2-element subsets of a 5-element set, which is 5!/(2!3!)=10. This number is sufficient as well – just give keys to a different group of 3 people for every lock.

Relabeling Dices

Can you relabel two 6-sided dices, so that every face has a positive number of dots, and also their sum has the same probability distribution?

Yes, you can do this. The easiest way is to use generating functions. Using simple polynomial algebra, you can see that

(x + x2 + x3 + x4 + x5 + x6)2 = (x + 2x2 + 2x3 + x4)(x + x3 + x4 + x5 + x6 + x8).

Therefore, if you take a dice with spots {1, 2, 2, 3, 3, 4}, and a dice with spots {1, 3, 4, 5, 6, 8}, their sum will have the same probability distribution.

The Lion and the Zebras

The lion plays a deadly game against a group of 100 zebras that takes place in the steppe (an infinite plane). The lion starts in the origin with coordinates (0,0), while the 100 zebras may arbitrarily pick their 100 starting positions. The lion and the group of zebras move alternately:

  • In a lion move, the lion moves from its current position to a position at most 100 meters away.
  • In a zebra move, one of the 100 zebras moves from its current position to a position at most 100 meters away.
  • The lion wins the game as soon as he manages to catch one of the zebras.

Will the lion always win the game after a finite number of moves? Or is there a strategy for the zebras that lets them survive forever?

The zebras can survive forever. They choose 100 parallel strips with width 300m each, then start on points on their mid-lines. If the lion lands on some zebra’s strip, the zebra simply jumps 100m away from the lion, along its mid-line.

Source:

Puzzling StackExchange

Wobbling Table

A perfectly symmetrical square 4-legged table is standing in a room with a continuous but uneven floor. Is it always possible to position the table in such a way that it doesn’t wobble, i.e. all four legs are touching the floor?

The answer is yes. Let the feet of the table clockwise are labeled with 1, 2, 3, 4 clockwise. Place the table in the room such that 3 of its feet – say 1, 2, 3, touch the ground. If foot 4 is on the ground, then the problem is solved. Otherwise, it is easy to see that we can not put it there if we keep legs 2 and 3 in the same places. Now start rotating the table clockwise, keeping feet 1, 2 and 3 on the ground at all times. If at some point foot 4 touches the ground as well, the problem is solved. Otherwise, continue rotating until foot 1 goes to the place where foot 2 was and foot 2 goes to the place where foot 3 was. Foot 3 will be on the ground, but this contradicts the observation that initially we couldn’t place legs 2, 3 and 4 on the ground without replacing feet 2 and 3.

Thank You!

A cowboy walks into a bar and asks the barman for a glass of water. The barman pulls out a gun instead and points it at the man. The man genuinely says “Thank you” and walks out.

What happened?

The cowboy had hiccups and needed water. The barman shocked him with his gun instead and that cured the hiccups.

11×11 Grid

All integer numbers between 1 and 121 are written in the cells of a square grid with size 11 by 11. Then the product of the numbers in every row and the product of the numbers in every column are calculated. Is it possible that the set of all 11 column products coincides with the set of all 11 row-products?

No, it is not possible. There are 13 prime number between 61 and 121. Since there are only 11 rows, two of them, X and Y, appear in the same row. Now that row is divisible by XY, but clearly, no column is divisible by that number.