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 \Gamma be the circumcircle of acute-angled triangle ABC. Points D and E lie on segments AB and AC, respectively, such that AD = AE. The perpendicular bisectors of BD and CE intersect at the minor arcs AB and AC of \Gamma at points F and G, respectively. Prove that the lines DE and FG 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 FD and GE to X and Y on \Gamma, respectively. The goal is to show that DEXY are concyclic. Then, by using transversal FX, we have

    \[ \angle XFG = \angle XYG = \angle XYE = \angle XDE, \]

which solves the problem. We do this with more angle chasing. Compute

    \[ \angle YAD = \angle YAB = \angle YGB = \angle EGB = \angle EGC - \angle BGC = \angle EGC - \angle BAC = \angle EGC - \angle DAE. \]

Also compute

    \[ \angle YED = \angle DEA - \angle AEY = \angle DEA - \angle CEG = \frac{1}{2} (180 - \angle DAE) - \frac{1}{2} (180 - \angle EGC) = \frac{\angle EGC - \angle DAE}{2}. \]

This gives us the relation \angle YED = \dfrac{1}{2} \angle YAD and similarly \angle XDE = \dfrac{1}{2} \angle XAE. This is sufficient to show that X, Y lie on the circle centered at A passing through points D, E so DEXY is cyclic and we are done.

p1 diagram

Problem 2: Find all integers n \geq 3 for which there exist real numbers a_1, a_2, \dots, a_{n + 2}, such that a_{n + 1} = a_1 and a_{n + 2} = a_2 and

    \[ a_i a_{i + 1} + 1 = a_{i + 2} \]

for i = 1, 2, \dots, n.

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 n = 3. We have the equations

    \[ \begin{aligned} a_1 a_2 + 1 &= a_3 \\ a_2 a_3 + 1 &= a_1 \\ a_3 a_1 + 1 &= a_2. \\ \end{aligned} \]

We can substitute for a_3 in the third equation to obtain

    \[ \begin{aligned} (a_1 a_2 + 1) a_1 + 1 &= a_2 \\ a_1^2 a_2 - a_2 + a_1 + 1 &= 0 \\ (a_1^2 - 1) a_2 + (a_1 + 1) &= 0 \\ (a_1 + 1)(a_1 - 1) a_2 + (a_1 + 1) &= 0 \\ (a_1 + 1)(a_1 a_2 - a_2 + 1) &= 0, \\ \end{aligned} \]

which implies either \implies a_1 = -1 or a_3 = a_2. Let’s just say for now that a_1 = -1. Then a_2 a_3 = -2 and a_2 + a_3 = 1, so a_2, a_3 are the roots of the polynomial x^2 - x - 2 = (x - 2)(x + 1). We then notice that the sequence will go on as -1, -1, 2, -1, -1, 2, \dots indefinitely, so 3 \mid n is definitely valid.

This begs the question of: are these the only solutions? This suggests looking at the indices modulo 3, and there are only a few ways to obtain some nice looking equalities from the given conditions involving i, i + 1, and i + 2. First, we know from the condition a_i a_{i + 1} + 1 = a_{i + 2} that a_i a_{i + 1} a_{i + 2} + a_{i + 2} = a_{i + 2}^2. If we substitute a_{i + 1} a_{i + 2} = a_{i + 3} - 1, we obtain

    \[ \begin{aligned} a_i a_{i + 1} a_{i + 2} + a_{i + 2} &= a_i (a_{i + 3} - 1) + a_{i + 2} \\ a_i a_{i + 1} a_{i + 2} + a_{i + 2} &= a_i a_{i + 3} - a_i + a_{i + 2} \\ a_i a_{i + 1} a_{i + 2} + a_i &= a_i a_{i + 3}, \\ \end{aligned} \]

which totally suggests that there is something going on between a_i and a_{i + 3}. 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 n):

    \[ \begin{aligned} \sum_{i = 1}^{n} a_i a_{i + 3} &= \sum_{i = 1}^{n} (a_i a_{i + 1} a_{i + 2} + a_i) \\ &= \sum_{i = 1}^{n} a_i a_{i + 1} a_{i + 2} + \sum_{i = 1}^{n} a_i \\ &= \sum_{i = 1}^{n} a_i a_{i + 1} a_{i + 2} + \sum_{i = 1}^{n} a_{i + 2} \\ &= \sum_{i = 1}^{n} (a_i a_{i + 1} a_{i + 2} + a_{i + 2}) \\ &= \sum_{i = 1}^{n} a_{i + 2}^2 = \sum_{i = 1}^{n} a_i^2. \\ \end{aligned} \]

The equality \displaystyle{\sum_{i = 1}^{n} a_i a_{i + 3} = \sum_{i = 1}^{n} a_i^2} really suggests that the sequence should be cyclic with length 3, so let’s try to prove it! Recall the AM-GM equality, where if we have non-negative numbers a, b, then \dfrac{a + b}{2} \geq \sqrt{ab}. We apply it to the LHS to obtain a_i a_{i + 3} \leq |a_i a_{i + 3}| \leq \dfrac{a_i^2 + a_{i + 3}^2}{2}. Taking the sum over the first n terms, we have

    \[ \sum_{i = 1}^{n} a_i a_{i + 3} \leq \frac{1}{2} \sum_{i = 1}^{n} (a_i^2 + a_{i + 3}^2) \leq \sum_{i = 1}^{n} a_i^2. \]

Since the two sides are actually equal by the previous computation, the equality condition of AM-GM must be met, or a_i = a_{i + 3}. Thus the cycle length divides 3, 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 a_i = c and

    \[ c^2 + 1 = c \implies c^2 - c + 1 = 0, \]

but that equation has no real roots.

The answer is then all n \geq 3 such that n \equiv 0 \pmod{3}.

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