Consider the following sum: \[\frac1{1 \times 2} + \frac1{2 \times 3} + \frac1{3 \times 4} + \dots\] Show that no matter how many terms it has, the sum will never be larger than \(1\).
Draw how Robinson Crusoe should put pegs and ropes to tie his goat in order for the goat to graze grass in the shape of a square, or slightly harder in a shape of a given rectangle.
Draw how Robinson Crusoe should put pegs and ropes to tie his goat in order for the goat to graze grass in the shape of a shape like this
Draw how Robinson Crusoe should put pegs and ropes to tie his goat in order for the goat to graze grass in the shape of a given triangle.
Draw a picture how Robinson used to tie the goat and the wolf in order for the goat to graze the grass in the shape of half a circle.
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 \(G\).
Prove that the set of all real numbers is not countable.
Today we will solve some geometric problems using the triangle inequality. This is an inequality between the lengths of the sides of any triangle, or between the distances of any three points.
The shortest path between any two points \(A\) and \(B\) is a straight segment - every other path is longer. In particular, a path through another point, \(C\), is equal or longer. \[AC + BC \ge AB\] The triangle inequality says that the sum of lengths of any two sides of a triangle is always larger than the length of the third side. The inequality only becomes an equality if \(ABC\) is not actually a triangle and the point \(C\) lies on the segment from \(A\) to \(B\).
Even though it is a simple idea, it can be a really helpful tool in problem solving.
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?
Split the numbers from \(1\) to \(9\) into three triplets such that the sum of the three numbers in each triplet is prime. For example, if you split them into \(124\), \(356\) and \(789\), then the triplet \(124\) is correct, since \(1+2+4=7\) is prime. But the other two triples are incorrect, since \(3+5+6=14\) and \(7+8+9=24\) are not prime.