Hungry Lion

A hungry lion runs inside a circus arena which is a circle of radius 10 meters. Running in broken lines (i.e. along a piecewise linear trajectory), the lion covers 30 kilometers. Prove that the sum of all turning angles is at least 2998 radians.

Source: The Puzzle Toad

Imagine the lion is static and instead the center of the arena moves around. Then, the problem translates to a point inside the arena alternating between traveling straight South and then moving north along arcs around the center of the arena. Since the total distance traveled straight South is 30KM and the distance between the starting and the ending points is at most 20M, the total distance traveled North must be at least 30KM – 20M = 29980M. The radius of the arena is 10M, so this corresponds to at least 2998 radians.

Houses on a Farm

Is it possible to connect each of the houses with the well, the barn, and the mill, so that no two connections intersect each other?

No, it is impossible. Here is a convincing, albeit a informal proof.

Imagine the problem is solvable. Then you can connect House A to the Well, then the Well to House B, then House B to the Barn, then the Barn to House C, then House C to the Mill, and finally the Mill to House A. Thus, you will create one loop with 6 points on it, such that houses and non-houses are alternating along the loop. Now, you must connect Point 1 with Point 4, Point 2 with Point 5, Point 3 with Point 6, such that the three curves do not intersect each other. However, you can see that you can draw no more than one such curve neither on the inside, nor the outside of the loop. Therefore, the task is indeed impossible.

More rigorous, mathematical proof can be made using Euler’s formula for planar graphs. We have that F + V – E = 2, where F is the number of faces, V is the number of vertices, and E is the number of edges in the planar graph. We have V = 6 and E = 9, and therefore F = 5. Since no 2 houses or 2 non-houses can be connected with each other, every face in this graph must have at least 4 sides (edges). Therefore, the total number of sides of all faces must be at least 20. However, this is impossible, since every edge is counted twice as a side and 20/2 > 9.

Game of Coins

Kuku and Pipi decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then alternating, each player takes on his turn one of the two coins at the end of the line and keeps it. Kuku and Pipi continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as his opponent.

Let’s assume Kuku starts first. In the beginning, he calculates the total value of the coins placed on odd positions in the line and compares it with the total value of the coins placed on even positions in the line. If the former has a bigger total value, then on every turn he takes the end coin which was placed on odd position initially. If the latter has bigger value, then on every turn he takes the end coin which was placed on even position initially. It is easy to see that he can always do this because after each of Pipi’s turns there will be one “odd” coin and one “even” coin at the ends of the line.

10 Dots, 10 Coins

If you have 10 dots on the ground, can you always cover them with 10 pennies without the coins overlapping?

Assume the dots lie in a plane and the radius of a penny is 1. Make an infinite grid of circles with radii 1, as shown on the picture, and place it randomly in the plane.

If we choose any point in the plane, the probability that it will end up inside some circle of the grid is equal to S(C)/S(H), where S(C) is the area of a coin and S(H) is the area of a regular hexagon circumscribed around it. A simple calculation shows that this ratio is larger than 90%. Therefore, the probability that some chosen point in the plane will not end up inside any circle is less than 10%. If we have 10 points, the probability that neither of them will end up inside a circle is less than 100%. Therefore, we can place the grid in the plane in such a way that every dot ends up in some circle. Now, just place the given coins where these circles are.

Coins on a Chessboard

There is a room with a chessboard inside. On each of its 64 squares, there is placed a coin, either heads up or heads down. You enter the room and a person inside points towards one special square on the chessboard and gives you the chance to flip one of the coins (whichever you choose). Then you leave the room, your friend enters and has to guess which was the special square on the chessboard. If you two could devise a plan before entering the room, how would you make sure your friend always guesses correctly which is the special square?

First, you must enumerate the coins with numbers from 1 to 64, locate the mystery coin, and calculate the binary representation of its number, padded with zeros on the left to 6 digits length. For example, if the mystery coin is the 5th one on the 4th row, its number would be 29 and will have a binary representation 011101. Then, consider the following sets of coins:

A1 = {row 1, row 2, row 3, row 4}
A2 = {row 1, row 2, row 5, row 6}
A3 = {row 1, row 3, row 5, row 7}
A4 = {column 1, column 2, column 3, column 4}
A5 = {column 1, column 2, column 5, column 6}
A6 = {column 1, column 3, column 5, column 7}

Now, the strategy is to flip the coin which makes the parity of heads in set Ai odd if and only if the i-th digit in the binary representation of the mystery coin is 1. It is easy to check that this is always a possible thing to do.