The White Rook pursues a black horse on a board of \(3 \times 1969\) cells (they walk in turn according to the usual rules). How should the rook play in order to take the horse? White makes the first move.
This problem is from Ancient Rome.
A rich senator died, leaving his wife pregnant. After the senator’s death it was found out that he left a property of 210 talents (an Ancient Roman currency) in his will as follows: “In the case of the birth of a son, give the boy two thirds of my property (i.e. 140 talents) and the other third (i.e. 70 talents) to the mother. In the case of the birth of a daughter, give the girl one third of my property (i.e. 70 talents) and the other two thirds (i.e. 140 talents) to the mother.”
The senator’s widow gave birth to twins: one boy and one girl. This possibility was not foreseen by the late senator. How can the property be divided between three inheritors so that it is as close as possible to the instructions of the will?
Two play a game on a chessboard \(8 \times 8\). The player who makes the first move puts a knight on the board. Then they take turns moving it (according to the usual rules), whilst you can not put the knight on a cell which he already visited. The loser is one who has nowhere to go. Who wins with the right strategy – the first player or his partner?
In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle \(ABC\) with right angle at \(C\).
The difference of the squares of the hypothenuse and one of the arms is \(AB^2 -BC^2\). This expression can be represented in the form of a product \[AB^2 -BC^2 = (AB - BC)(AB+BC)\] or \[AB^2 -BC^2 = -(BC - AB)(AB+BC)\] Dividing the right hand sides by the product \(-(AB-BC)(AB+BC)\), we obtain the proportion \[\frac{AB+BC}{-(AB+BC)} = \frac{BC-AB}{AB-BC}.\] Since the positive quantity is greater than the negative one we have \(AB+BC>-(AB+BC)\). But then also \(BC-AB>AB-BC\), and therefore \(2BC>2AB\), or \(BC>AB\), i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!
If you are on a boat and toss a suitcase overboard, will the water level rise or fall?
This academic year Harry decided not only to attend Maths Circles, but also to join his local Chess Club. Harry’s chess set was very old and some pieces were missing so he ordered a new one. When it arrived, he found out to his surprise that the set consisted of 32 knights of different colours. He was a bit upset but he decided to spend some time on solving the problem he heard on the last Saturday’s Maths Circle session. The task was to find out if it is possible to put more than 30 knights on a chessboard in such a way that they do not attack each other. Do you think it is possible or not?
After listening to Harry’s complaints the delivery service promised him to deliver a very expensive chess set together with some books on chess strategies and puzzles. This week one of the tasks was to put 14 bishops on a chessboard so that they do not attack each other. Harry solved this problem and smiled hoping he is not getting 32 identical bishops this time. Can you solve it?
Sometimes life can make us do the craziest of things. In this problem you just need to find out how one can cut an \(8\times8\) chessboard into 20 pieces each having the same perimeter and consisting of a whole number of cells.
On the way back from his weekly maths circle Harry created the following puzzle:
Put 48 rooks on a \(10\times10\) board so that each rook attacks only 2 or 4 empty cells.
When he showed this problem to the teachers next Saturday they were very impressed and decided to include it in the next problem set. Try to find a suitable placement of rooks.
A boy is playing on a \(4\times10\) board. He is trying to put 8 bishops on the board so that each cell is attacked by one of the bishops. Finally he manages to solve this problem.
(a) Can you show a possible solution?
(b) Can you do the same thing with 7 bishops?