Knights and Coins

Bob and Jane are taking turns, placing knights and coins respectively on a chessboard. If Bob is allowed to place a knight only on an empty square which is not attacked by another knight, how many pieces at most can he place before running out of moves? Assume that Jane starts second and plays optimally, trying to prevent Bob from placing knights on the board.

Bob can place at most 16 knights. One way to do this is to keep placing knights only on the 32 white squares. In order to see that Jane can prevent Bob from placing more than 16 knights, split the board in four 4×4 grids. Then, group the squares in each grid in pairs, as shown on the image below. If Bob places a knight on any square, then Jane will place a coin on its paired square. This way Bob can place at most one knight on each of the four red squares, one knight on each of the four green squares, one knight on each of the four brown squares, and one knight on each of the four blue squares. Therefore, he can not place more than 64/4 = 16 knights on the board.

Light Bulbs in the Attic

There are three light bulbs in your attic. All of them are turned off and their switches are installed downstairs. You can play with the switches as much as you want and after that, you can visit the attic above just once. How can you find out which switch to which bulb corresponds?

You turn the first switch on, then wait for 30 minutes and turn the second switch on. After that go upstairs and examine the bulbs. The one which is turned off corresponds to the third switch. The one which is turned on and is still cold corresponds to the second switch. The one which is turned on and is hot corresponds to the first switch.

Cover the Table

100 coins are placed on a rectangular table, such that no more coins can be added without overlapping. Show that you can cover the entire table with 400 coins (overlapping allowed).

Since we can not place any more coins on the table, each point of it is at distance at most 2r from the center of some coin, where r is the radius of the coin. Now shrink the entire table twice in width and length, then replace every shrunk coin with a full sized one. This way the small table will be completely covered because every point of it will be at distance at most r from the center of some coin. Add three more of these smaller tables, covered with coins, to create a covering of the big table.

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.