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Let be a sequence of real numbers. Compute:
First and foremost we set and it is obvious that . We are making use of probabilistic methods. Suppose than an infinite number of coins are flipped. Let be the probability that the -th coin toss lands heads and let us consider the first time heads comes up. Then is the probability that the first head appears in the – th flip and is the probability that all flips come up tails. Thus,
Prove that in any triangle it holds that
where denotes the circumradius and the inradius.
Using the law of sines we have that
and if we denote the area of the triangle then
Let be three positive real numbers such that . Prove that
By AM – GM we have,
Hence and the exercise is complete.
Let be positive integers such that divides . Show that
is the square of an integer.
This is a very well known problem. It first appeared in IMO held in Canberra, Australia. It even has its own Wikipedia page.
One of the organisers had sent this exercise to all professors of Number Theory to check if the exercise is original and try to solve it within hours. No one was able to do so. However, one Bulgarian student managed to solve the problem in less than hours ( his solution was fallen from the sky ) and for that he was awarded a special reward beyond the medal.
Choose integers such that Now, for fixed , out of all pairs choose the one with the lowest value of . Label . Thus, is a quadratic in . Should there be another root, , the root would satisfy: Thus, isn’t a positive integer (if it were, it would contradict the minimality condition). But , so is an integer; hence, . In addition, so that . We conclude that so that .
This construction works whenever there exists a solution for a fixed , hence is always a perfect square.
Let be vectors of – dimensional Euclidean space such that . Prove that there exists a permutation of the integers such that
for each .
We define inductively. Set .Assume is defined for and also
Note that is true for . We choose in a way that is fulfilled for instead of . Set and . Assume that for all . Then and in view of ones gets which is impossible. Hence , there is such that
Put . Then using and we have
which verifies for . Thus we define for every . Finally from we get