Hard. Let \(\mathcal{S}\) be a finite set of at least two points in the plane. Assume that no three points of \(\mathcal S\) are collinear. A windmill is a process that starts with a line \(\ell\) going through a single point \(P \in \mathcal S\). The line rotates clockwise about the pivot \(P\) until the first time that the line meets some other point belonging to \(\mathcal S\). This point, \(Q\), takes over as the new pivot, and the line now rotates clockwise about \(Q\), until it next meets a point of \(\mathcal S\). This process continues indefinitely. Show that we can choose a point \(P\) in \(\mathcal S\) and a line \(\ell\) going through \(P\) such that the resulting windmill uses each point of \(\mathcal S\) as a pivot infinitely many times.
In the Land of Linguists live \(m\) people, who have opportunity to speak \(n\) languages. Each person knows exactly three languages, and the sets of known languages may be different for different people. It is known that \(k\) is the maximum number of people, any two of whom can talk without interpreters. It turned out that \(11n \leq k \leq m/2\). Prove that then there are at least \(mn\) pairs of people in the country who will not be able to talk without interpreters.
The king decided to reward a group of \(n\) wise men. They will be placed in a row one after the other (so that everyone is looking in the same direction), and each is going to wear a black or a white hat. Everyone will see the hats of everyone in front, but not those behind them. The wise men will take turns (from the last to the first) to name the color (white or black) and the natural number of their choice.
At the end, the number of sages who have named the color of their hat correctly is counted: that is exactly how many days the whole group will be paid a salary raise. The wise men were allowed to agree in advance on how to respond. At the same time, the wise men know that exactly \(k\) of them are insane (they do not know who exactly). Any insane man names the color white or black, regardless of the agreement. What is the maximum number of days with a pay supplement that the wise men can guarantee to a group, regardless of the location of the insane in the queue?
In a graph \(G\), we call a matching any choice of edges in \(G\) in such a way that all vertices have only one edge among chosen connected to them. A perfect matching is a matching which is arranged on all vertices of the graph.
Let \(G\) be a graph with \(2n\) vertices and all the vertices have degree at least \(n\) (the number of edges exiting the vertex). Prove that one can choose a perfect matching in \(G\).
On a \(10\times 10\) board, a bacterium sits in one of the cells. In one move, the bacterium shifts to a cell adjacent to the side (i.e. not diagonal) and divides into two bacteria (both remain in the same new cell). Then, again, one of the bacteria sitting on the board shifts to a new adjacent cell, either horizontally or vertically, and divides into two, and so on. Is it possible for there to be an equal number of bacteria in all cells after several such moves?
Let \(a\), \(b\) and \(c\) be the three side lengths of a triangle. Does there exist a triangle with side lengths \(a+1\), \(b+1\) and \(c+1\)? Does it depend on what \(a\), \(b\) and \(c\) are?
There is a triangle with side lengths \(a\), \(b\) and \(c\). Can you form a triangle with side lengths \(\frac{a}{b}\), \(\frac{b}{c}\) and \(\frac{c}{a}\)? Does it depend on what \(a\), \(b\) and \(c\) are? Give a proof if it is always possible or never possible. Otherwise, construct examples to show the dependence on \(a\), \(b\) and \(c\).
Recall that a triangle can be drawn with side lengths \(x\), \(y\) and \(z\) if and only if \(x+y>z\), \(y+z>x\) and \(z+x>y\).
There is a triangle with side lengths \(a\), \(b\) and \(c\). Does there exist a triangle with side lengths \(|a-b|\), \(|b-c|\) and \(|c-a|\)? Does it depend on what \(a\), \(b\) and \(c\) are?
Recall that a triangle can be formed with side lengths \(x\), \(y\) and \(z\) if and only if all the inequalities \(x+y>z\), \(y+z>x\) and \(z+x>y\) hold.
There is a triangle with side lenghts \(a\), \(b\) and \(c\). Does there exist a triangle with sides of lengths \(a^2+bc\), \(b^2+ca\) and \(c^2+ab\)? Does it depend on the values of \(a\), \(b\) and \(c\)?
Suppose you meet a person inhabiting this planet and they ask you “Am I a Goop?" What would you conclude?