Team Selection Test for IMO 2023

Answer: 1011.

Let us describe the strategy of Aslı for guaranteeing at least 1011 solutions of the equation. Aslı makes her first move arbitrarily. After that, at each move she chooses any star with already replaced neighbour and replaces this star with the sign of this neighbour. She completes the game with 1011 such moves after her first one. With this strategy, Aslı guarantees the existence of at least 1011 intervals of the form $(2^k,2^{k+1})$ such that the signs in front of both $\frac{1}{x-2^k}$ and $\frac{1}{x-2^{k+1}}$ are the same. In such an interval, by considering the values of $x$ arbitrarily close to $2^k$ and $2^{k+1}$, the equation on the board takes both arbitrarily large and arbitrarily small values. Therefore, by the Mean Value Theorem, there is a real root of this equation in this interval and hence Aslı guarantees at least 1011 solutions to this equation.

Now we will describe Zehra's strategy to keep the number of real roots less than 1012. Let us enumerate the stars as 0, 1, 2, ..., 2023. Consider the pairs (0, 2023) and (1, 2), (3, 4), (5, 6), ..., (2021, 2022). Zehra's strategy is the following: For the pair (0, 2023), whenever Aslı puts a sign in one of them, Zehra puts the same sign on the other one. For all the other pairs, whenever Aslı puts a sign in one of them, Zehra puts the opposite sign. We will show that this strategy works.

Firstly we consider the infinite intervals. Without loss of generality, assume that signs of the pair (0, 2023) is $+$. Consider the values $x\in (2^{2023},\infty )$. We will start reading all the signs from 2023 to 0. First, we will see a $+$ sign in 2023, then for each consecutive terms we read a pair of opposite signs until the last one with a $+$ sign. At any set in the range from $k$ to 2023 for $0\le k\le 2023$, we will see more positive terms than negative ones. That's because if $k$ is odd, we have $\frac{k-1}{2}$ pairs of opposite signs with a plus sign at the end, and if $k$ is even we have $\frac{k-2}{2}$ pairs of opposite signs with a plus sign at the end again, making sure that number of plus signs is always at least half of all signs. So, we can assign to each minus sign a plus one appearing later in the equation. Since the terms appearing later has larger absolute value, in these pairings each pair has a positive sum with two positive numbers remaining in hand, which means the function is always positive and there are no real roots. Same argument applies for the other infinite interval.

Now we consider the intervals between the stars numbered $(k,k+1)$ for $0\le k\le 2022$. There are two possible cases based on their signs:

† If the signs of $k,k+1$ are different, there are no real solutions in the interval $(2^k,2^{k+1})$. We will show that the equation is either always positive or always negative in this interval. We will start with the following observation.

Lemma 1: For all $x\in (2^k,2^{k+1})$ we have $$-\frac{1}{x-2^{k-2}}-\frac{1}{x-2^{k-1}}+\frac{1}{x-2^k}-\frac{1}{x-2^{k+1}}+\frac{1}{x-2^{k+2}}+\frac{1}{x-2^{k+3}}\ge 0$$ Proof: Let $x=2^{k-2}y$ then we need to show that for each $y\in (4,8)$ we have $$\frac{4}{(y-4)(8-y)}\ge \frac{14}{(y-2)(16-y)}+\frac{31}{(y-1)(32-y)}$$ We know that the product of two positive real numbers with a fixed sum is larger when the numbers are closer. Therefore, smallest value of LHS in this interval is 1 when $y=3$, while the largest value of the RHS in this interval is $\frac{14}{24}+\frac{31}{84}<1$ when $y=4$, hence we are done.

Back to the case, we work in the interval $x\in (2^k,2^{k+1})$ and W.L.O.G. assume that the sign of $k$ is $+$. Then we can swap the $-$ sign appearing in the largest element in the set of stars $\{0,1,\dots ,k-3\}$ with the positive signs in the stars numbered $k-1$ and $k-2$. If there are any $+$ signs among the two. If there are less than two signs on one of the ends, we just add new terms, and because the signs are different this will only decrease the value of the function. We perform the same operation for $k+1$ (carrying the $+$ signs to the closest two places), then we see that the new function on the blackboard takes values less than the original one for each value of $x$ in this interval. Moreover, new function is always positive for the values in this interval, since the 6 consecutive terms appearing in the lemma has positive sum and for all the other terms, just like the argument in the infinite interval claim, all positive terms can be paired up with a negative term having less absolute value (and same for the other part with opposite signs), hence the function is always positive in this interval and have no roots.

Lemma 2: For all $x\in (2^k,2^{k+1})$ we have $$\frac{1}{(x-2^k{)}^2}-\frac{1}{(x-2^{k-1}{)}^2}-\frac{1}{(x-2^{k-2}{)}^2}\ge 0$$ and $$\frac{1}{(x-2^{k+1}{)}^2}-\frac{1}{(x-2^{k+2}{)}^2}-\frac{1}{(x-2^{k+3}{)}^2}\ge 0$$ Proof: Again we will use the transformation $x=2^{k-2}y$. Then, the first inequality becomes $$\frac{1}{(y-4{)}^2}\ge \frac{1}{(y-2{)}^2}+\frac{1}{(y-1{)}^2}$$ for $y\in (4,8)$. We can even prove the stronger statement as $\frac{1}{(y-4{)}^2}\ge \frac{2}{(y-2{)}^2}$ since the linear inequality $\sqrt{2}(y-4)\le y-2$ is held in the interval $y\in (4,8)$. The second inequality becomes $$\frac{1}{(y-8{)}^2}\ge \frac{1}{(y-16{)}^2}+\frac{1}{(y-32{)}^2}$$ LHS gets its minimum value at $y=4$, RHS gets its maximum at $x=8$ and when we compare these two values we still get that LHS is greater than the RHS. Now, when we take the derivative we will see that by the lemma above, the value of the derivative is always positive for these 6 consecutive terms and by the same swapping and pairing arguments used previously, the derivative of the function on the blackboard has a value greater than 0 hence the function is strictly increasing (or decreasing) in this interval, hence it must provide exactly 1 root.

Finally, Zehra's strategy shows that there can be at most 1002 real solutions to the equation at the end. That's because there are no real solutions are coming from the infinite intervals, and 1001 among the 2023 finite intervals have opposite signs at its ends thus they provide no solutions as well. This lefts at most 1002 intervals with same signs at both ends where each such interval can provide at most one real solution, bounding the number of real solutions above by 1002. However, at the end, the numerator is a polynomial of degree 2023 since the number of signs are not equal and we will have either $2x^{2023}$ or $-2x^{2023}$ as the leading coefficient after expanding. Since the degree is odd, the number of real roots must be odd, this forces the number of real solutions to be at most 1011 since 1012 real solutions is not possible and this completes the proof.