# IMO 2018 Day 1

I have solved problems 1 and 2. I have not yet attempted problem 3 but I believe that it is beyond me, especially within a time frame of four and a half hours.

Problem 1: Let be the circumcircle of acute-angled triangle . Points and lie on segments and , respectively, such that . The perpendicular bisectors of and intersect at the minor arcs and of at points and , respectively. Prove that the lines and are parallel (or are the same line).

Solution 1: This was more of an exercise in angle chasing than an actual problem and felt really out of place, even for a problem 1.

Extend and to and on , respectively. The goal is to show that are concyclic. Then, by using transversal , we have which solves the problem. We do this with more angle chasing. Compute Also compute This gives us the relation and similarly . This is sufficient to show that lie on the circle centered at passing through points so is cyclic and we are done.

Problem 2: Find all integers for which there exist real numbers , such that and and for .

Solution 2: This problem seemed kind of easy for a problem 2, or maybe I’m just getting better at math despite not doing any contest math for several years now.

We start by examining the case . We have the equations We can substitute for in the third equation to obtain which implies either or . Let’s just say for now that . Then and , so are the roots of the polynomial . We then notice that the sequence will go on as indefinitely, so is definitely valid.

This begs the question of: are these the only solutions? This suggests looking at the indices modulo , and there are only a few ways to obtain some nice looking equalities from the given conditions involving , , and . First, we know from the condition that . If we substitute , we obtain which totally suggests that there is something going on between and . When you have a general equality and you have no idea what to do with it, taking the sum is usually not a bad choice so we do that here to obtain (we also extend the sequence so that is infinite with cycle of length ): The equality really suggests that the sequence should be cyclic with length , so let’s try to prove it! Recall the AM-GM equality, where if we have non-negative numbers , then . We apply it to the LHS to obtain . Taking the sum over the first terms, we have Since the two sides are actually equal by the previous computation, the equality condition of AM-GM must be met, or . Thus the cycle length divides , so it must either be three or one. We have seen previously that three works, so we focus our efforts on one. A cycle length of one means that the sequence is constant, or and but that equation has no real roots.

The answer is then all such that .

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