## IMO 1995 Day 2

**Problem 4** Find the maximum value of for which there exists a sequence of positive reals with , such that

for all .

**Solution:** First note that for positive real numbers the equality

is equivalent to

and so we can conclude that

Define the following functions by

Then we have . So where is a composition of functions equal to or . We can give an inductive formula for such a composition of functions in the following way:

where means the composition of with itself of times and we take into account the possibility where or in which case is the identity. Note that in fact counts the number of times appears in the composition. Also, we don’t take into account compositions of with itself of more than one time, since is an involution, moreover, all we need in our further considerations is to know the parity of , which is an invariant even if we put compositions of an odd number of in the formula.

Note that when is even we would have and so we would need to have the equality

and we would end up with an even number of compositions. This contradicts the fact that we have compositions. Therefore the only valid case is when is odd, and then implies

which is maximal when is minimal while is maximal. This gives us two informations: first we need to minimize the number of ‘s on even position, which can be achieved if we keep only the position with ; secondly to maximize the sum of we need to have the least number of ‘s in the composition and this can be achieved if we consider only one .

Regarding our discussion above, the function which maximizes is so the maximal value for is

**Problem 5** Let be a convex hexagon with and , such that . Suppose and are points in the interior of the hexagon such that . Prove that .

**Solution:** This problem cannot be more simple if we make a good drawing. Construct outside the hexagon the equilateral triangles and . Due to the symmetry of the new figure with respect

to we have . Because are on the circumscribed circles of triangles and , respectively on the arcs which do not contain and , respectively, we have

so the inequality to be proven becomes the trivial inequality

**Problem 6** Let be an odd prime number. How many -element subsets of are there, the sum of whose elements is divisible by ?

**Solution:** Denote . For a set we define a translation by a number to be

which is a fancy way to write that we translate modulo inside . We define the same thing for .

Now note that if and then , or . After these considerations consider and define

Note that since the sum of elements of takes every residue modulo out of which only one is good for our problem. We can define the equivalence relation on sets with elements, as follows:

This equivalence relation determines a partition of the family of -elements subsets of different from . Each class of the partition has elements, and from each class exactly one element satisfies the conditions of the problem. Therefore the answer of the problem is