Question:
A line in the plane is called *sunny* if it is not parallel to any of the $x$-axis, the $y$-axis, or the line $x+y=0$.

Let $n \ge 3$ be a given integer. Determine all nonnegative integers $k$ such that there exist $n$ distinct lines in the plane satisfying both of the following:

- for all positive integers $a$ and $b$ with $a+b\le n+1$, the point $(a,b)$ lies on at least one of the lines;
- exactly $k$ of the $n$ lines are sunny.

We claim that the only possible numbers of sunny lines are $k \in \{0, 1, 3\}$.

We view the set of lattice points $(a, b)$ with $a, b > 0$ and $a + b \le n + 1$ as a triangular grid of points. We define a *long line* to be one of the three boundary lines of this grid, which are the lines passing through exactly $n$ grid points. Notice that each long line is not a sunny line, as it is parallel to either the $x$-axis, the $y$-axis, or the line $x + y = 0$.

The key claim is that if $n \ge 4$, then any set of $n$ distinct lines covering all the required grid points must contain at least one long line.

Proof of the claim: Consider the $3(n - 1)$ points on the outer edge of the triangular grid. If no long line were present in our set of $n$ covering lines, then each of the $n$ lines would intersect the boundary of the triangular grid in at most two points, meaning each line could pass through at most two edge points. Since the $n$ lines must cover all $3(n - 1)$ edge points, we would obtain the inequality
\[
2n \ge 3(n - 1).
\]
Solving this inequality yields $n \le 3$, which contradicts our assumption that $n \ge 4$. Thus, the claim is proven.

Therefore, for any configuration with $n \ge 4$, at least one long line must exist among the $n$ lines. We may delete this long line and remove the $n$ grid points it covers. The remaining grid points form a smaller triangular grid corresponding to size $n - 1$. Since the deleted long line is not sunny, this reduction from size $n$ to size $n - 1$ does not change the number of sunny lines in the configuration. By repeating this process, we can continually reduce the configuration until we reach the base case of $n = 3$. Conversely, once a valid configuration of lines is known for a smaller value of $n$, we can increase $n$ by adding a long line (which covers the new boundary points and is not sunny), maintaining the exact same number of sunny lines. Consequently, the possible counts of sunny lines for any $n \ge 3$ are exactly those that arise in the base case $n = 3$.

We now classify all possible coverings of the six grid points in the $n = 3$ triangular grid by exactly three lines.

Case 1: A long line is present. This long line covers exactly three grid points. The remaining three grid points must be covered by the remaining two lines. By the Pigeonhole Principle, one of these two remaining lines must pass through at least two of the remaining three points and hence is not sunny (since any line passing through two points of the remaining grid points in this configuration is necessarily parallel to a boundary line). The third and final line may or may not be sunny. Therefore, in this case, the possible numbers of sunny lines are $0$ or $1$.

Case 2: No long line is present. Then each of the three lines can pass through at most two grid points. Since there are six grid points in total and only three lines, each line must in fact pass through exactly two points. The only possible configuration is the symmetric one in which all three lines are sunny. Hence this case gives exactly $3$ sunny lines.

Thus the only possible values of $k$ are
\[
0, \ 1, \ 3.
\]

It remains to see that all three values are achievable. The $n = 3$ case above already realizes these possibilities, and by adding long lines we obtain constructions for every larger $n$. Therefore, the answer is precisely $k \in \{0, 1, 3\}$.

(Partial) 1. Verified by explicit configurations (at least for n=3) that k=0, k=1, and/or k=3 are attainable, and explained how to extend a valid configuration from n to n+1 by adding a non-sunny boundary (“long”) line. 2. Defined the three boundary long lines of the triangular grid and observed they are exactly the non-sunny directions (parallel to x-axis, y-axis, or x+y=0), and that deleting such a long line reduces the problem from size n to size n−1 without changing k. 3. Proved the key counting inequality on boundary points for n≥4: if no long line is used then each line meets the outer edge in at most 2 grid points, giving 2n≥3(n−1) and hence a contradiction for n≥4. (Almost) 1. Proved that for n≥4 any covering must contain a long line and correctly reduced to the base case n=3, and classified the n=3 coverings to get k∈{0,1,3}, but did not explicitly justify the induction step that constructions for n=3 extend to all n (or omitted verifying attainability for all n). 2. Gave a complete reduction to n=3 and correct constructions showing all k∈{0,1,3} occur, but the boundary-point counting argument (or the claim that a non-long line hits at most two boundary points) is not fully justified or has a small logical gap. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
Alice and Bazza are playing the *inekoalaty game*, a two-player game whose rules depend on a positive real number $\lambda$ known to both players. On the $n$th turn of the game (starting with $n=1$):

- If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that
  \[
  x_1+x_2+\cdots+x_n \le \lambda n.
  \]
- If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that
  \[
  x_1^2+x_2^2+\cdots+x_n^2 \le n.
  \]

If a player cannot choose a suitable $x_n$, the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players.

Determine all values of $\lambda$ for which Alice has a winning strategy and all those for which Bazza has a winning strategy.

The threshold is $\lambda = \frac{1}{\sqrt{2}}$. Alice has a winning strategy for $\lambda > \frac{1}{\sqrt{2}}$, Bazza has a winning strategy for $\lambda < \frac{1}{\sqrt{2}}$, and neither player can force a win when $\lambda = \frac{1}{\sqrt{2}}$.

First, consider Alice when $\lambda \ge \frac{1}{\sqrt{2}}$. She plays
\[
x_{2i+1}=0
\]
on every odd turn until she decides to finish the game.
At time $n=2k+1$, we have
\[
x_1+x_2+\cdots+x_{2k} = x_2+x_4+\cdots+x_{2k}.
\]
By the Cauchy-Schwarz inequality,
\[
x_2+x_4+\cdots+x_{2k}
\le k\sqrt{\frac{x_2^2+x_4^2+\cdots+x_{2k}^2}{k}}.
\]
Because Alice plays $0$ on odd turns, we have $x_2^2+x_4^2+\cdots+x_{2k}^2 = x_1^2+x_2^2+\cdots+x_{2k}^2$. From Bazza's constraint on turn $2k$, this sum of squares is at most $2k$. Thus,
\[
x_2+x_4+\cdots+x_{2k} \le k\sqrt{\frac{2k}{k}} = \sqrt{2}k.
\]
Hence, the legal interval for Alice's next move is bounded below by $0$ and above by $\lambda(2k+1) - (x_1+\cdots+x_{2k})$, meaning she can pick any
\[
x_{2k+1} \in \bigl[0,\lambda(2k+1)-\sqrt{2}k\bigr].
\]
If $\lambda \ge \frac{1}{\sqrt{2}}$, this interval is always nonempty since $\lambda(2k+1)-\sqrt{2}k \ge \frac{1}{\sqrt{2}}(2k+1)-\sqrt{2}k = \frac{1}{\sqrt{2}} > 0$. Thus, Alice can always play $0$ to avoid losing forever.

If in fact $\lambda > \frac{1}{\sqrt{2}}$, Alice can choose $k$ large enough such that
\[
\sqrt{2}k < \lambda(2k+1)-\sqrt{2k+2}.
\]
Then on move $2k+1$, the upper bound of her legal interval is strictly greater than $\sqrt{2k+2}$, allowing Alice to choose
\[
x_{2k+1} > \sqrt{2k+2}.
\]
At the next step, Bazza would need to choose $x_{2k+2}$ such that
\[
x_1^2+\cdots+x_{2k+1}^2 + x_{2k+2}^2 \le 2k+2,
\]
but since $x_{2k+1}^2 > 2k+2$, this is impossible. Thus, Bazza has no legal move, and Alice wins for $\lambda > \frac{1}{\sqrt{2}}$.

Now, consider Bazza when $\lambda \le \frac{1}{\sqrt{2}}$. He plays on every even turn the largest possible value,
\[
x_{2i+2}=\sqrt{2-x_{2i+1}^2}.
\]
This keeps each pair $(x_{2i+1},x_{2i+2})$ contributing exactly $2$ to the quadratic budget, so $x_1^2 + \cdots + x_{2i+2}^2 = 2(i+1)$, which perfectly satisfies Bazza's condition for any even turn.
To estimate Alice's move $x_{2k+1}$, we use her constraint:
\[
\lambda(2k+1) \ge x_1+x_2+\cdots+x_{2k+1}.
\]
Grouping the terms in pairs gives
\[
\lambda(2k+1)
\ge \bigl(x_1+\sqrt{2-x_1^2}\bigr)+\cdots+\bigl(x_{2k-1}+\sqrt{2-x_{2k-1}^2}\bigr)+x_{2k+1}.
\]
For every $t\in[0,\sqrt{2}]$, we have $(t+\sqrt{2-t^2})^2 = 2 + 2t\sqrt{2-t^2} \ge 2$, which implies
\[
t+\sqrt{2-t^2} \ge \sqrt{2}.
\]
Applying this to each pair yields
\[
\lambda(2k+1) \ge \sqrt{2}k + x_{2k+1}.
\]
Therefore,
\[
x_{2k+1} \le \lambda(2k+1)-\sqrt{2}k.
\]
When $\lambda \le \frac{1}{\sqrt{2}}$, this upper bound is at most $\frac{1}{\sqrt{2}}(2k+1) - \sqrt{2}k = \frac{1}{\sqrt{2}} < \sqrt{2}$. Thus, Alice is forced to play $x_{2k+1} \le \sqrt{2}$, ensuring that Bazza can indeed play
\[
x_{2k+2}=\sqrt{2-x_{2k+1}^2}
\]
and never gets stuck.

If moreover $\lambda < \frac{1}{\sqrt{2}}$, then for all sufficiently large $k$,
\[
\lambda(2k+1)-\sqrt{2}k < 0.
\]
Since Alice must choose $x_{2k+1} \ge 0$, she eventually has no legal move at all. Hence, Bazza wins for $\lambda < \frac{1}{\sqrt{2}}$.

At the boundary value $\lambda = \frac{1}{\sqrt{2}}$, Alice's strategy of playing $0$ shows she can survive forever, while Bazza's maximal strategy also shows he can survive forever. Thus, neither player can force a win at the critical value.

Alice wins for $\lambda > 1/\sqrt2$; Bazza wins for $\lambda < 1/\sqrt2$; neither can force a win at $\lambda = 1/\sqrt2$

(Partial) 1. Identified the correct threshold value \(\lambda=1/\sqrt2\) (e.g. by optimizing a Cauchy–Schwarz type inequality or by analyzing one full odd-even pair). 2. Gave a correct strategy for one player in one direction, with a real supporting estimate (e.g. Alice plays \(x_{2i+1}=0\) and shows \(x_2+x_4+\cdots+x_{2k}\le \sqrt2k\); or Bazza plays \(x_{2i+2}=\sqrt{2-x_{2i+1}^2}\) and derives \(x_{2i+1}+x_{2i+2}\ge \sqrt2\)). 3. Correctly proved at least one of the strict win statements: Alice wins for all \(\lambda>1/\sqrt2\) (by eventually choosing \(x_{2k+1}^2>2k+2\)); or Bazza wins for all \(\lambda<1/\sqrt2\) (by forcing \(\lambda(2k+1)-\sqrt2k<0\) for large \(k\)). (Almost) 1. Correctly handled both strict regimes \(\lambda>1/\sqrt2\) and \(\lambda<1/\sqrt2\) with valid strategies and inequalities, but did not correctly treat the boundary case \(\lambda=1/\sqrt2\) (e.g. claimed one player wins there, or omitted showing neither can force a win). 2. Gave essentially the full solution including the critical case, but left a localized gap in the key estimate (e.g. misapplied Cauchy–Schwarz in bounding \(x_2+\cdots+x_{2k}\), or did not justify \(t+\sqrt{2-t^2}\ge \sqrt2\) on the needed domain). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Let $a_1,a_2,a_3,\dots$ be an infinite sequence of positive integers, and let $N$ be a positive integer. Suppose that, for each $n>N$, the number $a_n$ is equal to the number of times $a_{n-1}$ appears in the list $(a_1,a_2,\dots,a_{n-1})$. Prove that at least one of the sequences $a_1,a_3,a_5,\dots$ and $a_2,a_4,a_6,\dots$ is eventually periodic.

Let $M=\max(a_1,\dots,a_N)$. We can organize the process by building towers: when the term $a_i$ appears, place a block $B_i$ on top of tower $a_i$. For $i>N$, the sequence rule states that if $B_i$ is added to tower $a_i$, then the next block $B_{i+1}$ is added to the tower whose index equals the height reached by tower $a_i$.

We first establish two basic structural facts.

1. If tower $k+1$ grows arbitrarily high, then tower $k$ also grows arbitrarily high. In fact, there is a constant $C$ such that at every time,
\[
h_k \ge h_{k+1}-C,
\]
where $h_j$ denotes the current height of tower $j$. This holds because almost every sufficiently high block in tower $k+1$ can be traced back injectively to a block in tower $k$.

2. If $a_n>M$, then $a_{n+1}\le M$. Otherwise, there would be a first moment when two consecutive terms both exceed $M$. This would force more than $M$ towers to have height greater than $M$ before that time, which is impossible by the minimality of the first such moment.

From the second fact, terms at most $M$ occur with density at least $1/2$, so some values occur infinitely often. Let $L$ be the largest index such that towers $1,2,\dots,L$ are unbounded, while towers $L+1,\dots,M$ are eventually frozen. After some time $N'>N$, we may assume that:
- towers $1$ through $L$ are already very tall;
- towers $L+1$ through $M$ will never receive another block;
- $a_{N'}\le L$.

From that point onward, the terms alternate between small and big: if $n\equiv N' \pmod 2$, then $a_n\le L$, while if $n\equiv N'+1 \pmod 2$, then $a_n>M$. Thus, one parity subsequence consists of all the small terms.

Consider only indices $n\equiv N' \pmod 2$. At such a time, define the state
\[
S(n)=(h_1,h_2,\dots,h_L; a_n).
\]
The next two steps are completely determined by this state: the intermediate large term is the height of the tower just hit, and the following small term is the number of towers among $1,\dots,L$ whose heights are at least that large. Thus, $S(n+2)$ depends only on $S(n)$.

Because only relative heights matter, it is enough to replace $S(n)$ by
\[
T(n)=(h_1-h_2, h_2-h_3, \dots, h_{L-1}-h_L; a_n).
\]
Using the first structural fact repeatedly, one obtains upper bounds for these differences. A second contradiction argument gives lower bounds as well: for every $1\le \ell<L$, there is a uniform bound
\[
h_\ell \le h_{\ell+1}+C(L-1).
\]
Hence, each coordinate of $T(n)$ is bounded, so $T(n)$ takes only finitely many values. Since the transition $T(n)\mapsto T(n+2)$ is deterministic, the sequence $T(n)$ is eventually periodic. Therefore, the subsequence $a_{N'},a_{N'+2},a_{N'+4},\dots$ is eventually periodic.

Consequently, at least one of the two parity subsequences $a_1,a_3,a_5,\dots$ and $a_2,a_4,a_6,\dots$ is eventually periodic, as required.

At least one of the two subsequences $a_1,a_3,a_5,\dots$ or $a_2,a_4,a_6,\dots$ is eventually periodic

(Partial) 1. Proved that if some value >M:=max(a1,…,aN) occurs at position n, then an adjacent term must be ≤M (e.g. established a statement of the form: there cannot be a first time when both an and a_{n+1} exceed M). 2. Showed that at least one integer occurs infinitely often (for instance, deduced from the previous item that terms ≤M occur infinitely often / with density ≥1/2, hence by pigeonhole some value repeats infinitely many times). 3. Introduced a meaningful global organizing structure (e.g. “towers/blocks” where tower k counts occurrences of k) and proved at least one nontrivial monotonicity/comparison fact between towers (such as: unboundedness of tower k+1 forces unboundedness of tower k, or a uniform bound h_k ≥ h_{k+1}−C). (Almost) 1. Reduced to a late stage N′ where the set of “unbounded small values” is {1,…,L} and values >L never reappear among ≤M, and correctly deduced the alternation pattern from that point: one parity consists entirely of terms ≤L and the other parity consists entirely of terms >M, but did not finish the periodicity-from-finite-states step. 2. Defined a correct finite state (e.g. T(n) built from bounded height differences and the current small term) and proved that T(n+2) is a deterministic function of T(n), and that the coordinates of T(n) are bounded so only finitely many states occur, but left a localized gap in bounding the differences or in justifying that only relative heights matter. 3. Completed the argument that the deterministic finite-state evolution forces eventual periodicity of one parity subsequence, but omitted/garbled one boundary case (e.g. L=1, or the justification that towers L+1,…,M are eventually frozen) or made a small but non-negligible logical slip near the end.


Question:
Turbo the snail is in the top row of a grid with $2024$ rows and $2023$ columns and wants to get to the bottom row. However, there are $2022$ hidden monsters, one in every row except the first and last, with no two monsters in the same column.

Turbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then repeatedly moves to an orthogonal neighbor. If Turbo reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move between attempts, and Turbo remembers whether or not each cell he has visited contains a monster. If he reaches any cell in the last row, his attempt ends and Turbo wins.

Find the smallest integer $n$ such that Turbo has a strategy which guarantees being able to reach the bottom row in at most $n$ attempts, regardless of how the monsters are placed.

The minimum number of attempts is $n=3$. In fact, the same argument works for any $s \times (s-1)$ grid with $s \ge 4$; here the grid is $2024 \times 2023$, so $s = 2024$.

First, we show that at least three attempts are necessary. On the first attempt, when Turbo first enters the second row, he may immediately hit a monster $M_1$. Because his attempt ends immediately, he learns nothing else about the grid. On his second attempt, he must avoid the cell containing $M_1$, so he must enter the third row in a different column. However, upon entering the third row, he may again immediately hit a monster $M_2$. In this scenario, his second attempt ends, and he still has not reached the bottom row. Thus, no strategy can guarantee success in only two attempts.

Now we prove that three attempts always suffice by providing a strategy that guarantees reaching the bottom row in at most three attempts.

On the first attempt, Turbo moves into the second row and explores it horizontally until he discovers the unique monster $M_1$ located in that row. Upon hitting $M_1$, his first attempt ends. There are two cases to consider based on the position of $M_1$.

Case 1: $M_1$ is not on an edge cell of the second row.
Since $M_1$ is not in the leftmost or rightmost column, Turbo knows that the two cells adjacent to $M_1$ in the second row (one on the left and one on the right) are safe. Starting from either of these safe cells, he can attempt to follow a path going downward on the corresponding side of $M_1$, and then continue to the bottom row. These two candidate paths lie on opposite sides of the column containing $M_1$. Because there is only one monster in each lower row and no two monsters share a column, at least one of the two downward routes avoids all monsters. On his second attempt, Turbo tries one of these routes. If he succeeds, he reaches the bottom row. If he encounters a monster, his second attempt ends, but he now knows that the other route must be completely free of monsters. Hence, on his third attempt, Turbo can take the other route and is guaranteed to succeed. Thus, he succeeds within the next two attempts.

Case 2: $M_1$ lies on an edge cell of the second row.
By symmetry, suppose $M_1$ is in the leftmost cell of the second row (column 1). On the second attempt, Turbo follows a staircase path heading down and gradually to the right. If this staircase path reaches the last row without hitting any monsters, he wins immediately on the second attempt. Otherwise, he encounters a second monster $M_2$ somewhere on the staircase path. At this point, his second attempt ends. However, because he reached $M_2$, everything strictly to the left of that encountered point on the staircase path has been verified as safe. Furthermore, Turbo also knows that the entire first column below $M_1$ is completely free of monsters, because $M_1$ already occupies the unique monster position for the first column (recall that no two monsters share a column). Using this combined information, on the third attempt Turbo can safely travel down along the verified safe region to the left of $M_2$, move horizontally into the first column (the column of $M_1$), and then continue straight down to the bottom row. This route avoids every monster.

Thus, Turbo always has a strategy that guarantees success in at most three attempts, and two attempts are impossible. Therefore, the answer is
\[
\boxed{3}
\]

(Partial) 1. Proved the lower bound n\ge 3 by exhibiting an adversarial placement causing failure in 2 attempts (e.g. first hit a monster upon first entering row 2, then forced to enter row 3 in a different column and can be hit immediately again). 2. Gave the key idea for the upper bound strategy: first locate the monster in row 2 by sweeping along that row until hitting it, and then use information from this to plan later attempts. 3. Solved Case 1 (monster in row 2 not on an edge): proposed the two “routes on opposite sides of the row-2 monster” and correctly argued that, since monsters occupy distinct columns, at least one route is monster-free so success is guaranteed within 3 attempts. (Almost) 1. Correctly proved n\ge 3 and presented a complete 3-attempt strategy for the non-edge Case 1, but handled the edge Case 2 only heuristically (missing a rigorous construction of the final safe path after the staircase hits a monster). 2. Gave the full two-case strategy (non-edge and edge) and the distinct-columns argument, but left one localized gap, such as not justifying why the “other side” route must be completely free after one failure, or not proving that the claimed safe region in the edge case can always be reached without stepping on unknown cells. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: The Bank of Oslo issues two types of coin: aluminum (denoted $A$) and bronze (denoted $B$). Marianne has $n$ aluminum coins and $n$ bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k \le 2n$, Gilberty repeatedly performs the following operation: he identifies the longest chain containing the $k$th coin from the left and moves all coins in that chain to the left end of the row. For example, if $n=4$ and $k=4$, the process starting from the ordering $AABBBABA$ would be $AABBBABA \to BBBAAABA \to AAABBBBA \to BBBBAAAA \to \dotsb$. Find all pairs $(n,k)$ with $1 \le k \le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.

The valid pairs $(n, k)$ are those satisfying $n \le k \le \left\lceil \frac{3}{2} n \right\rceil$.

Call a maximal chain a \emph{block}. Then the line of coins can be described as a sequence of blocks: it is one of
\[
\underbrace{A\dots A}_{e_1} \underbrace{B\dots B}_{e_2} \underbrace{A\dots A}_{e_3} \dots \underbrace{A\dots A}_{e_m} \quad \text{for odd } m,
\]
\[
\underbrace{A\dots A}_{e_1} \underbrace{B\dots B}_{e_2} \underbrace{A\dots A}_{e_3} \dots \underbrace{B\dots B}_{e_m} \quad \text{for even } m,
\]
or the same thing with the roles of $A$ and $B$ flipped. The main claim is the following:

\textbf{Claim.} The number $m$ of blocks will never increase after an operation. Moreover, it stays the same if and only if
\begin{itemize}
\item $k \le e_1$; or
\item $m$ is even and $e_m \ge 2n+1-k$.
\end{itemize}

\textit{Proof.} This can be seen by directly applying the operation. Moving a block to the front will strictly decrease the number of blocks unless the moved block is already the first block (which happens when $k \le e_1$) or it is the last block and does not share a type with the first block (which happens when $m$ is even and the $k$-th coin falls in the last block, meaning $e_m \ge 2n+1-k$). In these two specific cases, the number of blocks remains unchanged.

The problem asks for which values of $k$ we always reach $m=2$ eventually; we already know that $m$ is non-increasing. We consider a few cases:
\begin{itemize}
\item If $k < n$, then any configuration with $e_1 = n-1$ will never change, since $k \le e_1$.
\item If $k > \left\lceil \frac{3}{2} n \right\rceil$, then take $m=4$ and $e_1 = e_2 = \left\lfloor n/2 \right\rfloor$ and $e_3 = e_4 = \left\lceil n/2 \right\rceil$. This configuration retains $m=4$ always: the blocks simply rotate.
\item Conversely, suppose $k \ge n$ has the property that $m > 2$ stays fixed. Since $m > 2$, the first block cannot contain all $n$ coins of its type, so $e_1 < n \le k$. Thus, the condition $k \le e_1$ is never satisfied. If after the first three operations $m$ hasn't changed, then we must repeatedly fall into the second case, meaning we must have $m \ge 4$ even, and $e_m, e_{m-1}, e_{m-2} \ge 2n+1 - k$. Now, since $e_m$ and $e_{m-2}$ are blocks of the same coin type, their sum cannot exceed the total number of coins of that type, which is $n$. Therefore,
\[
n \ge e_m + e_{m-2} \ge 2(2n+1-k) \implies k \ge \frac{3}{2} n + 1,
\]
so this completes the proof.
\end{itemize}

\(n \le k \le \left\lceil \frac{3}{2} n \right\rceil\)

(Partial) 1. Introduced the block (maximal chain) decomposition and proved the monotonicity claim that the number of blocks never increases under the operation (possibly without fully characterising equality cases). 2. Gave a correct counterexample construction for one forbidden range of k (e.g. showed k<n fails by exhibiting a fixed configuration with first block length e1=n−1, or showed k>ceil(3n/2) fails by giving a 4-block “rotation” example). 3. Proved a correct necessary condition for “blocks can stay constant for several moves”, e.g. deduced that if the block count does not drop on a move with m>2 then the k-th coin must lie in the last block and m must be even (or an equivalent structural restriction). (Almost) 1. Proved that the answer set is n≤k≤ceil(3n/2), but left a small gap in justifying the exact equality cases for when the number of blocks stays the same after an operation (e.g. missed one direction of the “iff” for k≤e1 or m even with em≥2n+1−k). 2. Completed the two negative directions (k<n and k>ceil(3n/2)) and gave the main “if k is in range then we must eventually reach m=2” argument, but the final inequality step forcing k≥3n/2+1 from em+e_{m−2}≤n (or the justification that three consecutive non-decreases imply em,e_{m−1},e_{m−2}≥2n+1−k) is not fully justified. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Let $n \ge 100$ be an integer. Ivan writes the numbers $n, n+1, \dots, 2n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.

We will find three cards $a<b<c$ such that the sum of any two is a perfect square. By the Pigeonhole Principle, if these three cards are divided into two piles, at least one pile must contain two of them, yielding a pair whose sum is a perfect square.

We look for $a, b, c$ such that
\[
b+c=(2k+1)^2,\qquad
c+a=(2k)^2,\qquad
a+b=(2k-1)^2
\]
for some integer $k$. Solving the system of equations for $a$, $b$, and $c$ gives
\[
a=\frac{(2k)^2+(2k-1)^2-(2k+1)^2}{2}=2k^{2}-4k,
\]
\[
b=\frac{(2k+1)^2+(2k-1)^2-(2k)^2}{2}=2k^{2}+1,
\]
\[
c=\frac{(2k+1)^2+(2k)^2-(2k-1)^2}{2}=2k^{2}+4k.
\]

We need to show that for any given integer $n\ge 100$, one can find a suitable $k$ such that the cards $a, b, c$ are drawn from the available numbers $\{n, n+1, \dots, 2n\}$. This requires $n \le a < b < c \le 2n$. Let
\[
I_k\coloneq\{\,n\in\mathbb Z\mid n\le a<b<c\le 2n\,\}
\]
be the set of valid integers $n$ for a given $k$. Since $a < b < c$ holds for all $k \ge 2$, the condition $n \le a$ requires $n \le 2k^2 - 4k$, and the condition $c \le 2n$ requires $2k^2 + 4k \le 2n$, which simplifies to $k^2 + 2k \le n$. Thus, the interval is
\[
I_k = \{\,n\in\mathbb Z\mid k^{2}+2k\le n\le 2k^{2}-4k\,\}.
\]
When $n\in I_k$, the problem is solved for that choice of $k$. It is therefore sufficient to show that the union of these intervals $I_k$ covers all integers $n \ge 100$.

Starting with $k=9$, we have $k^2+2k = 99$ and $2k^2-4k = 126$, so
\[
I_{9}=\{\,n\in\mathbb Z \mid 99\le n\le 126\,\}.
\]
For the intervals to overlap and cover all subsequent integers, the right endpoint of $I_k$ must be greater than or equal to the left endpoint of $I_{k+1}$ minus $1$. In fact, we can show a stronger inequality for $k \ge 9$:
\[
2k^{2}-4k\ge (k+1)^{2}+2(k+1).
\]
Expanding the right side gives $k^2 + 4k + 3$. The inequality rearranges to
\[
k^2 - 8k - 3 \ge 0,
\]
which is clearly true for $k \ge 9$ since $9^2 - 8(9) - 3 = 6 \ge 0$, and the quadratic is increasing for $k \ge 4$.

This confirms that the right endpoint of $I_k$ strictly exceeds the left endpoint of $I_{k+1}$ for all $k \ge 9$. Since $I_9$ begins at $99$, the intervals $I_k$ for $k \ge 9$ collectively cover all integers $n \ge 99$. Hence, for any $n\ge 100$, a suitable $k$ exists, and the statement holds.

(Partial) 1. Reduced the problem to finding three numbers a<b<c in {n,n+1,…,2n} such that at least two of them lie in the same pile (pigeonhole), and stated that it suffices to have at least one pair among {a,b,c} with square sum. 2. Constructed (or correctly derived) a triple a,b,c with all pairwise sums being consecutive odd/even squares, e.g. a=2k^2−4k, b=2k^2+1, c=2k^2+4k, and checked that a+b, a+c, b+c are perfect squares. 3. Obtained correct inequality conditions on k ensuring a,b,c all lie in [n,2n] (typically n≤a and c≤2n, equivalent to n≤2k^2−4k and k^2+2k≤n), even if the coverage for all n≥100 is not finished. (Almost) 1. Found a valid k for each n≥100 via the interval method Ik={n: k^2+2k≤n≤2k^2−4k}, but left a small gap in proving that the intervals Ik overlap/cover all integers n≥100 (e.g. the inequality 2k^2−4k≥(k+1)^2+2(k+1) not fully justified or the starting k not checked carefully). 2. Completed the construction of a,b,c and the pigeonhole step, and correctly argued existence of k for all n≥100, but made a localized algebraic slip in computing a,b,c or in one endpoint inequality while the intended interval-covering argument is clear and otherwise correct. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Two squirrels, Bushy and Jumpy, have collected $2021$ walnuts for the winter. Jumpy numbers the walnuts from $1$ through $2021$, and digs $2021$ little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of $2021$ moves. In the $k$th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$. Prove that there exists a value of $k$ such that, on the $k$th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.

Assume for contradiction that no such $k$ exists. We will use a so-called "threshold trick". This process takes exactly $2021$ steps. Right after the $k$-th move, we consider a situation where we color walnut $k$ red, so at the $k$-th step there are $k$ red walnuts. For brevity, a non-red walnut is called black.

At each step, the $k$-th move swaps the two walnuts adjacent to $k$. By the assumption that no $a < k < b$ swap occurs, these two adjacent walnuts are either both less than $k$ (both red) or both greater than $k$ (both black). Because they share the same color, the swap does not change the color sequence. Then, walnut $k$ turns red. Thus, at each step, the walnut that becomes red is between two non-red or two red walnuts.

On the other hand, since there are $2021$ walnuts, one obtains a parity obstruction to this simplified process. We claim that after the first step, there is always a consecutive block of black walnuts of positive even length. After the first step, there is a block of $2020$ black walnuts. Thereafter, note that a length $2$ block of black walnuts can never be changed, because a black walnut turning red must be between two black walnuts, requiring a black block of length at least $3$. Meanwhile, for even lengths of at least $4$, if one places a red walnut inside it (meaning a black walnut in the block turns red), the even-length block splits into an odd-length block and an even-length block, since the remaining odd number of black walnuts is partitioned into two blocks.

(Partial) 1. Introduced the threshold/coloring idea: after the k-th move declare walnuts 1,2,…,k red (others black), and observed that if no swap with a<k<b occurs then the k-th swap exchanges two walnuts of the same color, so the cyclic red/black pattern is unchanged by the swap itself. 2. Proved the key structural consequence under the negation assumption: the walnut that turns from black to red at step k must lie between two red walnuts or between two black walnuts (equivalently, it cannot be adjacent to exactly one red and one black walnut at the moment it turns red). 3. Established a parity/block invariant for black segments on the circle (e.g. after step 1 there is a black block of length 2020, and argued that some positive even-length block of consecutive black walnuts must persist or reappear under the allowed “turn a black walnut to red between two blacks” operation). (Almost) 1. Carried out the coloring reduction and correctly analyzed how black blocks change when a black walnut between two blacks turns red (even block splits into one odd and one even block; a black block of length 2 cannot be affected), but did not finish the final contradiction (e.g. did not explicitly deduce impossibility at step 2021 when no black walnuts remain). 2. Proved the existence of a persistent positive even-length black block for all steps k≥1, but left a small gap in justifying why the process cannot eliminate all black walnuts by step 2021 (missing boundary case for a single remaining black block or an unproved monotonicity/parity claim). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n$. Each pebble is coloured in one of $n$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles such that the total weights of both piles are the same, and each pile contains two pebbles of each colour.

We begin by pairing the weights such that the sum of each pair is constant:
\[ 1+4n = 2+(4n-1) = 3+(4n-2) = \dots = 2n+(2n+1) = 4n+1. \]
Place all four pebbles of the same colour into a single box, yielding $n$ boxes in total. For each $k=1,2,\dots,2n$, we connect pebble $k$ and pebble $4n+1-k$ with a piece of string. This creates $2n$ strings, each connecting a pair of pebbles whose weights sum to $4n+1$.

To solve the problem, it suffices to paint each string either blue or green such that each box has exactly two blue strings and two green strings attached to its pebbles (where a string connecting two pebbles within the same box counts twice for that box). If we can achieve this, we can assign the pairs of pebbles connected by blue strings to the first pile and those connected by green strings to the second pile. Since each box has two string ends of each colour, each pile will contain exactly two pebbles of each colour. Furthermore, since there are exactly $n$ blue strings and $n$ green strings, each pile will consist of $n$ pairs of pebbles, giving both piles an equal total weight of $n(4n+1)$.

We can rephrase the problem in terms of graph theory by viewing the $n$ boxes as vertices and the $2n$ strings as edges. This forms a $4$-regular multigraph on $n$ vertices, where self-loops are allowed and contribute $2$ to the degree of a vertex. We wish to colour the edges blue and green such that each vertex is incident to exactly two blue edges and two green edges.

Since every vertex in this multigraph has an even degree of $4$, each connected component of the graph admits an Eulerian circuit. Consider a connected component with $k$ vertices. The sum of the degrees of the vertices in this component is $4k$, so it contains exactly $2k$ edges. Because the Eulerian circuit for this component has an even length ($2k$ edges), we may colour the edges along the circuit in alternating colours: green and blue.

This alternating colouring ensures that every time the circuit passes through a vertex (entering via one edge and leaving via the next), the two edges involved have different colours. Since each vertex has a degree of $4$, the Eulerian circuit passes through each vertex exactly twice. If the vertex has no self-loops, it is visited on two separate occasions, with each visit contributing one blue and one green edge, yielding exactly two blue and two green edges in total. If a vertex has a self-loop, the self-loop is traversed as a single edge of one colour, contributing two ends of that colour to the vertex; the alternating nature of the circuit ensures that the edges immediately preceding and following the self-loop are of the opposite colour, providing two ends of the other colour. A vertex with two self-loops simply receives one blue self-loop and one green self-loop. Thus, every vertex receives exactly two blue and two green ends. This yields the desired arrangement, completing the proof.

(Partial) 1. Used the pairing 1+4n=2+(4n-1)=...=4n+1, and reduced the task to selecting exactly n of these 2n pairs so that each colour appears exactly twice among the selected pebbles. 2. Reformulated the problem as a 4-regular multigraph on n vertices (colours as vertices; each complementary-weight pair as an edge, allowing loops), and stated the goal as a 2-edge-colouring with exactly two edges of each colour incident to every vertex. 3. Proved (or correctly invoked) that each connected component of an even-degree graph has an Euler tour, with the intention to alternate colours along such a tour. (Almost) 1. Carried out the Euler-circuit alternating-colouring argument to obtain two edges of each colour at every vertex, but did not correctly justify the handling of self-loops (counting two incidences) or multiple edges in the incidence count. 2. Completed the graph-colouring construction, but made an arithmetic/logic slip in translating back to pebbles (e.g. did not explicitly conclude both piles have weight n(4n+1), or miscounted the number of pairs/pebbles per pile while the colouring condition is correct). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points in the plane such that the distance between any two different points in $\mathcal S$ is at least $1$. Prove there is a line $\ell$ separating $\mathcal S$ such that the distance from any point of $\mathcal S$ to $\ell$ is at least $\Omega\!\bigl(n^{-1/3}\bigr)$. (A line $\ell$ separates a set of points $S$ if some segment joining two points in $\mathcal S$ crosses $\ell$.)

Let $\delta$ be the maximum possible distance between two consecutive projections of the points of $\mathcal S$ onto any line $m$. (This maximum is achieved by a standard compactness argument.) We will prove that $\delta \ge \Omega(n^{-1/3})$. This solves the problem, because we can choose a line $m$ achieving this maximum gap $\delta$ and place a line $\ell$ perpendicular to $m$ in the exact middle of the gap. The line $\ell$ separates $\mathcal S$, and the distance from any point of $\mathcal S$ to $\ell$ is at least $\frac{\delta}{2} = \Omega(n^{-1/3})$.

Let $A$ and $B$ be two points in $\mathcal S$ that are farthest apart. Let $R = AB$; since all pairwise distances between points in $\mathcal S$ are at least $1$, we have $R \ge 1$. All points of $\mathcal S$ lie in the intersection of the disks centered at $A$ and $B$ with radius $R$. Consider the circle $\odot(B)$ centered at $B$ with radius $R$. Let $\overline{XY}$ be the chord of $\odot(B)$ such that $\overline{XY} \perp \overline{AB}$ and the distance from $A$ to $\overline{XY}$ is exactly $\frac12$. (Such a chord exists since $R \ge 1 > \frac12$.) Let $\mathcal T$ denote the smaller region bounded by $\odot(B)$ and the chord $\overline{XY}$ (this is the cap containing $A$).

First, note that $R = AB \le (n-1)\delta < n\delta$, since the $n$ projections of the points in $\mathcal S$ onto the line $AB$ are spaced at most $\delta$ apart. The distance from $B$ to the chord $\overline{XY}$ is $R - \frac12$. The Pythagorean theorem gives the length of the chord as
\[
XY = 2\sqrt{R^{2} - \left(R-\frac12\right)^{2}} = 2\sqrt{R - \frac14} < 2\sqrt{R} < 2\sqrt{n\delta}.
\]

By definition, the region $\mathcal T$ corresponds to an interval of length $\frac12$ along the line $AB$, starting from $A$. Since all points of $\mathcal S$ lie in the intersection of the disks $D(A, R)$ and $D(B, R)$, their projections onto the line $AB$ all lie in the segment $AB$. Let these projections be $x_1 \le x_2 \le \dots \le x_n$, where $x_1 = 0$ corresponds to $A$ and $x_n = R$ corresponds to $B$. The points in $\mathcal T$ are exactly those whose projections satisfy $x_i \le \frac12$. Let $k = |\mathcal T|$ be the number of such points. Then $x_k \le \frac12 < x_{k+1}$. Because the maximum gap between consecutive projections is at most $\delta$, we have
\[
\frac12 < x_{k+1} \le x_1 + k\delta = k\delta.
\]
Thus, the number of points in $\mathcal T$ satisfies
\[
\lvert\mathcal T\rvert = k > \frac{1}{2\delta}.
\]

Next, consider the projections of the points in $\mathcal T$ onto the line $XY$. Any two points $P, Q \in \mathcal T$ are at distance at least $1$. The distance between their projections onto the line $AB$ is at most the width of $\mathcal T$ in the direction of $AB$, which is $\frac12$. Therefore, the distance between their projections onto the line $XY$ is at least
\[
\sqrt{PQ^2 - \left(\frac12\right)^2} \ge \sqrt{1^2 - \frac14} = \frac{\sqrt3}{2}.
\]
The region $\mathcal T$ is bounded by the chord $\overline{XY}$ and the arc of $\odot(B)$. Thus, the projections of all points in $\mathcal T$ onto the line $XY$ lie on the segment $\overline{XY}$. The distance between the extreme projections among these $|\mathcal T|$ points is at most the length of $\overline{XY}$. On the other hand, since the projections are spaced at least $\frac{\sqrt3}{2}$ apart, this distance is at least $\frac{\sqrt3}{2}(\lvert\mathcal T\rvert - 1)$. This implies
\[
XY \ge \frac{\sqrt3}{2}\left(\lvert\mathcal T\rvert - 1\right).
\]

Substituting the lower bound for $|\mathcal T|$, we obtain
\[
XY > \frac{\sqrt3}{2}\left(\frac{1}{2\delta} - 1\right).
\]
Combining this with the upper bound $XY < 2\sqrt{n\delta}$, we have
\[
2\sqrt{n\delta} > \frac{\sqrt3}{4\delta} - \frac{\sqrt3}{2}.
\]
If $\delta \ge \frac{1}{4}$, then $\delta \ge \frac{1}{4} n^{-1/3}$ trivially holds since $n \ge 2$. Otherwise, if $\delta < \frac{1}{4}$, we have $\frac{1}{2\delta} > 2$, so $\frac{\sqrt3}{4\delta} - \frac{\sqrt3}{2} > \frac{\sqrt3}{8\delta}$. This gives
\[
2\sqrt{n\delta} > \frac{\sqrt3}{8\delta} \implies \delta^{3/2} > \frac{\sqrt3}{16\sqrt{n}} \implies \delta > \left(\frac{3}{256}\right)^{1/3} n^{-1/3}.
\]
In either case, we have $\delta \ge \Omega(n^{-1/3})$. This completes the proof.

(Partial) 1. Reformulated the task via projections: defined $\delta$ as the maximum gap between consecutive projections of the $n$ points onto some line, and explained how a gap of size $\delta$ yields a separating line at distance $\gtrsim \delta$ (e.g. by taking a line parallel to the projection line placed in such a gap). 2. Introduced the diameter endpoints $A,B$ (farthest pair) and deduced a global bound $R=AB \le (n-1)\delta$ (or $R < n\delta$) from spacing of projections onto $AB$. 3. Constructed the chord $XY$ of the circle centered at $B$ with $XY \perp AB$ and $\operatorname{dist}(A,XY)=1/2$ (or an equivalent thin cap region $T$), and correctly computed/estimated $XY = 2\sqrt{R^2-(R-1/2)^2}=2\sqrt{R-1/4}$ to relate $XY$ to $R$ and hence to $\delta$. (Almost) 1. Completed the cap-counting strategy (using the thin region $T$ cut off by $XY$) and derived $\delta \ge c n^{-1/3}$, but left a small gap in justifying one of the key spacing/count estimates inside $T$ (e.g. why projections onto $XY$ are at least $\sqrt{3}/2$ apart, or why $|T| \gtrsim (1/2)\delta^{-1}$). 2. Obtained the final inequality $\delta \ge \Omega(n^{-1/3})$ with correct dependence on $n$, but made a localized computational slip in combining bounds (e.g. lost a constant/power when passing from $XY < 2\sqrt{n\delta}$ and $XY > c(\delta^{-1}-1)$ to $\delta \ge c'n^{-1/3}$), while the main argument is otherwise complete. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
A social network has $2019$ users, some pairs of which are friends (friendship is symmetric). If $A$, $B$, $C$ are three users such that $AB$ are friends and $AC$ are friends but $BC$ is not, then the administrator may perform the following operation: change the friendships such that $BC$ are friends, but $AB$ and $AC$ are no longer friends.

Initially, $1009$ users have $1010$ friends and $1010$ users have $1009$ friends. Prove that the administrator can make a sequence of operations such that all users have at most $1$ friend.

We reformulate the problem in graph-theoretic terms. Let the users be vertices and the friendships be edges of a graph $G$. The given operation, which we call a \emph{toggle}, consists of taking three vertices $a, b, c$ such that $ab$ and $ac$ are edges but $bc$ is not, and replacing the edges $ab$ and $ac$ with the edge $bc$.

\begin{claim*}
  Let $G$ be a connected graph. Then one can toggle $G$ without disconnecting the graph, unless $G$ is a clique, a cycle, or a tree.
\end{claim*}
\begin{proof}
  Assume $G$ is connected and is not a clique, a cycle, or a tree. Since $G$ is not a tree, it contains a cycle. Let $C$ be a cycle of minimal length in $G$; by hypothesis, $C \neq G$.

  If $C$ is not a triangle (equivalently, $G$ is triangle-free), then since $G$ is connected, there exists a vertex $b \notin C$ adjacent to some vertex $a \in C$. Let $c \in C$ be a neighbor of $a$ on the cycle. Since $G$ is triangle-free, $b$ is not adjacent to $c$. Thus, we can toggle $a, b, c$, replacing edges $ab$ and $ac$ with $bc$. This operation does not disconnect the graph because $a$ and $c$ remain connected through the remaining path in the cycle $C$, and $b$ remains connected to the rest of the graph via $c$.

  Now assume that $G$ contains a triangle. Let $K$ be a maximal clique of size at least $3$ (which exists since $G$ contains a triangle). By hypothesis, $K \neq G$. Since $G$ is connected, there exists an edge $e = ab$ with $a \in K$ and $b \notin K$. Since $K$ is a maximal clique, $b$ cannot be adjacent to all vertices in $K$. Choose a vertex $c \in K$ that is not adjacent to $b$. We can then toggle $a, b, c$, replacing edges $ab$ and $ac$ with $bc$. This does not disconnect the graph because $K$ has size at least $3$, meaning there is another vertex $d \in K$ through which $a$ and $c$ remain connected (since $ad$ and $cd$ are edges in $K$), and $b$ remains connected to the rest of the graph via $c$.
\end{proof}

Back to the original problem; let $G_{\text{imo}}$ be the given initial graph. We can apply toggles (by the claim) repeatedly, without disconnecting the graph, until we reach a tree. This is guaranteed because:
\begin{itemize}
  \item $G_{\text{imo}}$ is initially connected. Any two non-adjacent vertices have a common neighbor by the Pigeonhole Principle, since the sum of their degrees is at least $1009 + 1009 = 2018$, and there are only $2017$ other vertices in the graph ($1009 + 1009 + 2 > 2019$).
  \item $G_{\text{imo}}$ cannot become a cycle. Initially, $1010$ vertices have an odd degree of $1009$. A toggle at vertex $a$ involving neighbors $b$ and $c$ decreases the degree of $a$ by $2$ and leaves the degrees of $b$ and $c$ unchanged, so toggles preserve the parity of the degree of every vertex. Since $G_{\text{imo}}$ always contains vertices of odd degree, it can never become a cycle (which requires all vertices to have an even degree).
  \item $G_{\text{imo}}$ is not a clique initially, since its maximum degree is $1010$, which is strictly less than $2018$. It cannot become a clique later because each toggle strictly decreases the total number of edges.
\end{itemize}
Thus, repeatedly applying these connectivity-preserving toggles must eventually reduce $G_{\text{imo}}$ to a tree.

Once $G_{\text{imo}}$ is a tree, we repeatedly apply toggles arbitrarily until no more are possible. In a forest, any toggle replaces edges $ab$ and $ac$ with $bc$. This operation cannot create a cycle because, after removing $ab$ and $ac$, $b$ and $c$ belong to different connected components, so adding $bc$ simply merges these components. Thus, the graph remains acyclic (a forest). Since each toggle strictly decreases the total number of edges, the process must terminate. The process can only terminate when no vertex has degree $2$ or more; otherwise, in a forest, any two neighbors of a vertex are non-adjacent, allowing for another toggle. Therefore, the terminal graph has a maximum degree of at most $1$, as desired.

(Partial) 1. Reformulated the operation as a graph toggle on a triple (a,b,c): delete edges ab,ac and add bc, and correctly observed that each toggle strictly decreases the number of edges. 2. Proved the parity invariant: in a toggle, only deg(a) changes (by −2), so the parity of the degree of every vertex is preserved throughout. 3. Proved the initial graph is connected, e.g. by showing any two non-adjacent vertices have a common neighbor using the degree condition (so there is a path of length 2 between any such pair). (Almost) 1. Gave the full reduction strategy “apply toggles while keeping the graph connected until reaching a tree; then keep toggling in the forest until no vertex has degree ≥2”, but left a small gap in justifying that one can always choose a toggle that does not disconnect the graph unless the graph is a clique/cycle/tree. 2. Proved termination and the final step in a forest (if a vertex has degree ≥2 then a toggle is possible; edge count decreases so the process ends with max degree ≤1), but did not fully justify why the process cannot get stuck earlier in a connected non-tree state (e.g. missed ruling out the cycle/clique possibilities via invariants). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, which can show either heads or tails. He does the following operation: if there are $k$ coins which show heads and $k > 0$, then he flips the $k$th coin over; otherwise he stops the process. (For example, the process starting with $THT$ would be $THT \to HHT \to HTT \to TTT$, which takes three steps.)

Prove the process will always terminate, and determine the average number of steps this takes over all $2^{n}$ configurations.

We will prove that the process always terminates and that the expected number of steps over all $2^n$ initial configurations is
\[
E_n = \frac{1}{2} (1 + 2 + \dots + n) = \frac{1}{4} n(n+1),
\]
which is finite.

We represent the operation by a directed graph $G_n$ whose vertices are the strings in $\{0,1\}^n$, where $1$ corresponds to heads and $0$ corresponds to tails. Each string points to its successor under the operation. For a bit $b \in \{0,1\}$, we let $\overline{b} = 1-b$, and we denote binary strings as a sequence of $n$ symbols.

The main claim is that the graph $G_n$ can be described explicitly in terms of $G_{n-1}$. Specifically, the vertex set of $G_n$ can be partitioned into two halves, which we will call $X$ and $Y$, such that each half is isomorphic to $G_{n-1}$ in its edge structure, with a single connecting edge from $Y$ to $X$. The construction is as follows:
\begin{itemize}
    \item We take two copies $X$ and $Y$ of the graph $G_{n-1}$.
    \item To form the vertices of $X$, we take each string of length $n-1$ from $G_{n-1}$ and append a $0$ to it. In symbols, we apply the mapping $s_1 s_2 \dots s_{n-1} \mapsto s_1 s_2 \dots s_{n-1}0$.
    \item To form the vertices of $Y$, we take each string of length $n-1$ from $G_{n-1}$, toggle every bit, reverse the order of the bits, and then append a $1$ to the end. In symbols, we apply the mapping $s_1 s_2 \dots s_{n-1} \mapsto \overline{s}_{n-1} \overline{s}_{n-2} \dots \overline{s}_1 1$.
    \item Finally, we add one new edge from $Y$ to $X$, specifically from the state $11\dots1$ to the state $11\dots10$.
\end{itemize}

To prove this claim, we need only show that the arrows in this recursively defined directed graph match the actual operation rules.

First, consider the subgraph $X$. For any string in $X$, the $n$-th bit is $0$. Appending a $0$ to a string $s_1 \dots s_{n-1}$ does not change the total number of $1$s. Thus, if the original string had $k$ ones, the new string also has $k$ ones, and the operation will flip the $k$-th bit. This perfectly matches the transitions in $G_{n-1}$, meaning $X$ is a valid subgraph of $G_n$ and the process operates exactly as it does in $G_{n-1}$.

Next, consider the subgraph $Y$. Suppose a string $s = s_1 \dots s_{n-1}$ in $G_{n-1}$ has exactly $k$ ones. The modified string in $Y$ is given by $\overline{s}_{n-1} \dots \overline{s}_1 1$. The number of ones in the first $n-1$ bits of this new string is $(n-1) - k$, because every bit was toggled. Including the final $1$, the total number of ones in the new string is $(n-1-k) + 1 = n-k$.
Therefore, the operation on this new string will flip its $(n-k)$-th bit. The $(n-k)$-th bit of the string $\overline{s}_{n-1} \dots \overline{s}_1 1$ corresponds to the bit $\overline{s}_{n - (n-k)} = \overline{s}_k$. Flipping this bit changes it to $s_k$. Thus, the transition in $G_n$ maps the string to:
\[
\overline{s}_{n-1} \dots \overline{s}_{k+1} s_k \overline{s}_{k-1} \dots \overline{s}_1 1.
\]
This is exactly the image we would get by applying the transformation for $Y$ to the successor of $s$ in $G_{n-1}$ (since the successor of $s$ in $G_{n-1}$ is obtained by flipping $s_k$). This confirms that the internal transitions within $Y$ are isomorphic to those in $G_{n-1}$.

The only exception in $Y$ is the string corresponding to the terminal state of $G_{n-1}$. The terminal state of $G_{n-1}$ is $00\dots0$ (which has zero $1$s, so the process stops). Under the mapping for $Y$, the state $00\dots0$ becomes $11\dots1$. In $G_n$, the string $11\dots1$ has exactly $n$ ones, so the operation flips its $n$-th bit, changing it to $11\dots10$. This exactly matches the single new edge from $Y$ to $X$ defined in our construction.

Since $G_{n-1}$ is assumed by induction to terminate at $00\dots0$, all paths in $Y$ eventually reach $11\dots1$, which then transitions to $11\dots10$ in $X$. From there, the process follows the rules of $X$ (which is isomorphic to $G_{n-1}$) and must eventually terminate at $00\dots0$. This proves that the process always terminates for any initial configuration in $G_n$.

To finish the problem, let $E_n$ denote the expected number of steps for a randomly chosen initial configuration in $G_n$.
The vertices of $G_n$ are split equally between $X$ and $Y$.
For a string starting in $X$, the expected number of steps to reach the terminal state $00\dots0$ is exactly $E_{n-1}$.
For a string starting in $Y$, the process first traverses the internal edges of $Y$. Since $Y$ is isomorphic to $G_{n-1}$, the expected number of steps to reach the local sink $11\dots1$ is $E_{n-1}$. Once at $11\dots1$, the process transitions to $11\dots10$, and from there it takes exactly $n-1$ more steps to reach $00\dots0$ (since $11\dots10 \to 11\dots00 \to \dots \to 00\dots0$). Thus, from $11\dots1$, it takes exactly $n$ steps to finish. The expected number of steps for a string in $Y$ is therefore $E_{n-1} + n$.

Averaging over all $2^n$ configurations (half in $X$ and half in $Y$), we obtain the recurrence relation:
\[
E_n = \frac{1}{2} \left[ E_{n-1} + (E_{n-1} + n) \right] = E_{n-1} + \frac{n}{2}.
\]
With the base case $E_0 = 0$, we can solve this recurrence by induction:
\[
E_n = \sum_{i=1}^n \frac{i}{2} = \frac{1}{2} (1 + 2 + \dots + n) = \frac{1}{4} n(n+1).
\]
This completes the proof.

(Partial) 1. Proved termination by exhibiting a strictly decreasing invariant (e.g. interpreting the configuration as a binary number / a sum of distinct powers of 2) and checking that one move always decreases it. 2. Set up a correct recursion/partition of the state space into two parts of size 2^{n-1} each which evolve like the (n−1)-coin process (e.g. identifying the two copies corresponding to the last coin being T/H after a suitable transformation). 3. Derived a correct recurrence for the expectation E_n of the form E_n = E_{n-1} + n/2 (or an equivalent averaging argument), even if not fully justified or not fully solved. (Almost) 1. Gave the full decomposition into two copies of G_{n-1} and obtained E_n = E_{n-1} + n/2, but made a localized mistake in verifying the unique connecting transition (typically the behaviour at the all-heads state) or in proving the isomorphism on one half. 2. Solved the recurrence to get E_n = n(n+1)/4 and proved termination, but left a minor gap in the step-count for the special path after reaching 11…1 (e.g. claiming it takes n or n−1 steps without a complete justification of the exact number). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
An \emph{anti-Pascal triangle} is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following array is an anti-Pascal triangle with four rows which contains every integer from $1$ to $10$.
\begin{center}
\begin{tikzpicture}[scale = 0.8]
\node at (1.5,2.58) {$4$};
\node at (1,1.72) {$2$};
\node at (2,1.72) {$6$};
\node at (0.5,0.86) {$5$};
\node at (1.5,0.86) {$7$};
\node at (2.5,0.86) {$1$};
\node at (0,0) {$8$};
\node at (1,0) {$3$};
\node at (2,0) {$10$};
\node at (3,0) {$9$};
\end{tikzpicture}
\end{center}
Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1+2+\dots +2018$?

The answer is no; there is no anti-Pascal triangle with the required properties.

Let $n = 2018$ and $N = 1+2+\dots+n$. For every number $d$ not in the bottom row, draw an arrow from $d$ to the larger of the two numbers immediately below it (i.e., if $d=|a-b|$ and $a>b$, draw $d \to a$). This creates an *oriented forest* of directed paths (which look like lightning strikes). Notice that along any directed path, the values strictly increase, with the increment at each step being the smaller of the two numbers below the current one.

Consider the directed path starting from the top vertex $A$. The final value on the bottom row is the sum of the starting number at $A$ and the increments at each step along the path. Since the starting number and the increments at each step are distinct positive integers, and there are $n$ such numbers in total ($1$ starting number and $n-1$ increments), their sum is at least $1+2+\dots+n = N$. Since no number in the triangle can exceed $N$, equality must hold. Thus, the path from the top ends exactly at $N$, and the starting number together with the increments must exactly comprise the set $\{1,2,\dots,n\}$. Let $B$ be the position of $N$ on the bottom row.

Consider the two left and right neighbors $X$ and $Y$ of the endpoint $B$ on the bottom row. Assume without loss of generality that $B$ is to the right of the midpoint of the bottom side. Complete the equilateral sub-triangle whose base is the segment of the bottom row to the left of $B$ (ending at $X$), to an apex $C$. Consider the lightning strike (directed path) starting from $C$ and hitting the bottom row at $D$. It travels at least $\lfloor n/2-1 \rfloor$ steps by construction, since the base of this sub-triangle has length at least $\lfloor n/2 \rfloor$.

Because all numbers in the triangle are distinct, and the integers $1,2,\dots,n$ are already used on or adjacent to the $A \to B$ lightning path, the increases along the path from $C$ must be at least $n+1$, $n+2$, \dots. Then the number at $D$ is at least the sum of these increments:
\[
 (n+1) + (n+2) + \dots + \left( n+\left( \lfloor n/2-1 \rfloor \right) \right) > 1 + 2 + \dots + n
\]
for $n \ge 2018$, which is a contradiction since no number in the triangle can exceed $N$.

(Partial) 1. Introduced the directed-arrow ("lightning") construction: for each non-bottom entry d=|a-b|, oriented an edge from d to the larger of {a,b}, and proved that values strictly increase along each directed path with increment equal to the smaller child. 2. From the top vertex A, proved a valid lower bound on the bottom endpoint on its lightning path by expressing it as (top value) + (sum of positive increments), and used distinctness to obtain a nontrivial bound such as endpoint ≥ 1+2+…+n. 3. Identified that if the endpoint of the top lightning path attains the maximal possible value N=1+2+…+n, then the multiset consisting of the top value and all increments on that path must equal {1,2,…,n} (or an equivalent rigidity statement). (Almost) 1. Carried out the lightning-path argument to force the top path to end at N and to use exactly the labels {1,2,…,n} as (start + increments), then selected an appropriate second apex C in a large subtriangle and set up the second lightning path to have length at least ⌊n/2−1⌋, but did not fully justify this step-count/geometric subtriangle construction. 2. Completed the contradiction scheme by arguing that the increments on the second path must avoid {1,2,…,n} and hence are ≥ n+1,n+2,…, and summed them to exceed N, but left a localized gap (e.g., why the second path’s increments are all distinct and disjoint from the first path’s used labels, or an inequality/off-by-one in the final sum). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: A \emph{site} is any point $(x,y)$ in the plane for which $x,y \in \{1, \dots, 20\}$. Initially, each of the $400$ sites is unoccupied. Amy and Ben take turns placing stones on unoccupied sites, with Amy going first; Amy has the additional restriction that no two of her stones may be at a distance equal to $\sqrt5$. They stop once either player cannot move. Find the greatest $K$ such that Amy can ensure that she places at least $K$ stones.

The answer is $K = 100$.

First, we show Amy can always place at least $100$ stones. Indeed, treat the problem as a grid with a checkerboard coloring. Since a distance of $\sqrt{5}$ corresponds to a knight jump, it always connects squares of different colors. Thus, Amy can choose to always play on one of the $200$ black squares without violating the distance condition. In this way, she can guarantee half the black squares, i.e., she can get $\frac{1}{2}\cdot 200 = 100$ stones.

Second, we show Ben can prevent Amy from placing more than $100$ stones. Divide the grid into several $4\times4$ squares and then further partition each $4\times4$ square according to the following labeling:
\[
\begin{array}{cccc}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2 \\
2 & 1 & 4 & 3 \\
4 & 3 & 2 & 1
\end{array}
\]
The squares with each label form $4$-cycles by knight jumps. For each such cycle, whenever Amy plays in the cycle, Ben plays in the opposite point of the cycle, preventing Amy from playing any more stones in that original cycle. Hence Amy can play on at most $\frac{1}{4}$ of the sites, meaning she can place at most $100$ stones, as desired.

(Partial) 1. Observed that the forbidden distance √5 is exactly a knight move on the grid, and in particular connects opposite colors in a checkerboard coloring. 2. Gave a valid strategy for Amy guaranteeing at least 100 moves, e.g. “always play on black squares”, with a correct argument that this never violates the √5 condition and yields at least 100 stones. 3. Set up a correct partition/mirroring idea for Ben to limit Amy (e.g. decomposition into 4-cycles under knight moves, or into 4×4 blocks with a repeated 1–4 pattern), even if the cycle structure or the final counting is not fully finished. (Almost) 1. Proved both bounds K≥100 and K≤100 using the intended ideas, but left a minor gap in the game-theoretic justification of Ben’s reply strategy (e.g. why the “opposite point in the 4-cycle” is always unoccupied and always legal when needed). 2. Completed Ben’s blocking strategy up to showing Amy gets at most one stone per 4-cycle, but made a localized counting/tiling slip (e.g. not fully justifying that the 20×20 grid is partitioned into exactly 100 disjoint 4-cycles). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
A hunter and an invisible rabbit play a game in the plane.
The rabbit and the hunter start at the same point $A_0 = B_0$.
In the $n$th round of the game ($n \ge 1$), three events occur in order:

(i) The rabbit moves invisibly from $A_{n-1}$ to a point $A_n$ such that $A_{n-1}A_n = 1$.

(ii) The hunter has a tracking device (e.g., a dog) which reports an approximate location $P_n$ of the rabbit, satisfying $P_nA_n \le 1$.

(iii) The hunter moves visibly from $B_{n-1}$ to a point $B_n$ such that $B_{n-1}B_n = 1$.

Let $N = 10^9$. Can the hunter guarantee that $A_NB_N < 100$?

No.

We will show how to increase the distance in the following way:

\begin{claim*}
  Suppose the rabbit is at a distance $d \ge 1$ from the hunter
  at some point in time.
  Then it can increase its distance to at least
  $\sqrt{d^2+\frac12}$ in $4d$ steps
  regardless of what the hunter already knows about the rabbit.
\end{claim*}
\begin{proof}
  Consider a positive integer $n > d$, to be chosen later.
  Let the hunter start at $B$ and the rabbit at $A$.
  Let $\ell$ denote line $AB$.

  Now, we may assume the rabbit reveals its location $A$,
  so that all previous information becomes irrelevant.

  The rabbit chooses two points $X$ and $Y$ symmetric about $\ell$
  such that $XY = 2$ and $AX = AY = n$. Let $M$ be the midpoint of segment $XY$; by symmetry, $M$ lies on $\ell$, and $MX = 1$.
  The rabbit can then hop to either $X$ or $Y$,
  pinging a point on $\ell$ each time.
  This takes $n$ hops.

  Now among all points $H$ the hunter can go to,
  $\min \max \{HX, HY\}$ is clearly minimized with $H \in \ell$ by symmetry.
  So the hunter moves to a point $H$ on the ray $BA$ such that $BH = n$ as well.
  In that case the new distance is $HX = HY$.

  Since $H$ and $M$ lie on $\ell$, we have $AH = BH - AB = n - d$. Furthermore, in the right triangle $AMX$, $AM = \sqrt{AX^2 - MX^2} = \sqrt{n^2 - 1}$. Thus, $HM = AM - AH = \sqrt{n^2-1} - (n-d)$.

  We now compute
  \begin{align*}
    HX^2 &= 1 + HM^2 = 1 + \left( \sqrt{AX^2-1}-AH \right)^2 \\
    &= 1 + \left( \sqrt{n^2-1}-(n-d) \right)^2 \\
    &\ge 1 + \left( \left( n-\frac1n \right) - (n-d) \right)^2 \\
    &= 1 + \left(d-\frac{1}{n}\right)^2
  \end{align*}
  which exceeds $d^2 + \frac12$ whenever $n \ge 4d$.
\end{proof}

In particular we can always take $n = 400$ even very crudely;
applying the lemma $2 \cdot 100^2$ times,
this gives a bound of $400 \cdot 2 \cdot 100^2 < 10^9$, as desired.

(Partial) 1. Gave a correct rabbit strategy idea to increase distance: choose two symmetric candidate positions X,Y about the line AB, hop among them while reporting points on AB, with the aim that the hunter’s best response is forced onto AB by symmetry. 2. Established a quantitative distance-growth inequality of the form: starting from distance d, after some number of steps the rabbit can force the distance to be at least sqrt(d^2+c) for a fixed constant c>0 (even with a weaker constant or with an unspecified-but-correct dependence on d for the number of steps). 3. Correctly argued that if the rabbit can repeatedly increase distance by a fixed positive amount in squared distance (or by a multiplicative factor >1) within O(d) steps, then after sufficiently many iterations (and hence within N=10^9 steps) it can exceed 100, concluding the hunter cannot guarantee AB<100. (Almost) 1. Presented the full symmetric two-point construction (X,Y with XY=2, AX=AY=n, midpoint M on AB, rabbit pings on AB each hop) and the hunter optimality-by-symmetry reduction to a position H on AB, but made a minor computational slip in the final estimate for HX^2 (e.g., the bound sqrt(n^2-1) >= n-1/n or the concluding choice n>=4d). 2. Proved the key claim ‘from distance d the rabbit can force distance at least sqrt(d^2+1/2) in at most 4d steps’ but did not correctly finish the iteration to reach >100 within 10^9 steps (e.g., chose the number of iterations or total-step bound incorrectly, or omitted checking d>=1/base case). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Fix $N \ge 1$. A collection of $N(N+1)$ soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove $N(N-1)$ players from this row to obtain a new row of $2N$ players in which the following $N$ conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, \dots, no one stands between the two shortest players. Prove that this is possible.

We begin by partitioning the $N(N+1)$ players into $N$ groups based on their heights. Order the players by height and define the groups as follows:
Let $G_1$ consist of the $N+1$ shortest players,
$G_2$ consist of the next $N+1$ shortest players,
and so on, up to $G_N$, which consists of the $N+1$ tallest players.

If we select exactly two players from each group $G_k$, then in the final row of $2N$ players, the two players from $G_N$ will be the 1st and 2nd tallest, the two from $G_{N-1}$ will be the 3rd and 4th tallest, and so on, down to the two from $G_1$ being the two shortest. Thus, to satisfy the problem's conditions, it suffices to select two players from each group $G_k$ such that for each $k$, the two selected players from $G_k$ end up adjacent in the final row of $2N$ players.

We prove that this is always possible by proving a more general claim by induction on $M$:
**Claim:** *For any integer $M \ge 1$, given a row of players and $M$ disjoint groups of players such that each group has at least $M+1$ players present in the row, we can select two players from each group such that, after removing all unselected players, the two selected players from each group are adjacent in the remaining row.*

**Proof of Claim:**
We proceed by induction on $M$.

*Base case:* For $M=1$, we have $1$ group with at least $2$ players. We simply select the first two players from this group in the row. Since all other players are removed, these two players will trivially be adjacent.

*Inductive step:* Assume the claim holds for some $M-1 \ge 1$. We are given $M$ groups, each with at least $M+1$ players.
Scan the row of players from left to right, and stop as soon as we encounter two players belonging to the same group. Let this group be $G_k$, and let the two players be $A$ and $B$, where $B$ is the player we just scanned (so $B$ is to the right of $A$).

By the definition of our stopping condition, the scanned prefix of the row contains exactly two players from $G_k$, and at most one player from each of the other $M-1$ groups. We select $A$ and $B$ as the two players for group $G_k$.

Next, we delete the entire scanned prefix of the row (up to and including $B$), and we also delete all remaining players of $G_k$ from the rest of the row. For each of the remaining $M-1$ groups, we have deleted at most $1$ player (since at most one player from each of these groups appeared in the scanned prefix). Therefore, each of the remaining $M-1$ groups still has at least $(M+1) - 1 = M = (M-1) + 1$ players present in the remaining row.

By the induction hypothesis, we can select two players from each of these $M-1$ groups from the remaining row such that they are adjacent in the final selection. Because $A$ and $B$ are the only players selected from the deleted prefix, they will appear at the very left of the final row of $2M$ selected players. Thus, no other selected player can stand between $A$ and $B$, meaning $A$ and $B$ are adjacent. The pairs selected from the remaining groups are also adjacent by the induction hypothesis. This completes the inductive step.

**Conclusion:**
Applying the claim for $M = N$ to our initial partition of $N$ groups (each of size $N+1$), we can successfully choose two players from each group such that they are adjacent in the final row. As established, this selection satisfies all the conditions of the problem.

(Partial) 1. Made the key reduction by partitioning the players into height groups (e.g. the k-th group consists of the (N+1) players with ranks (k-1)(N+1)+1 through k(N+1)), and observed that it suffices to choose two players from each group so that each chosen pair is adjacent in the remaining row. 2. Stated/proved the inductive auxiliary claim (or an equivalent Hall-type/greedy statement): for M disjoint groups, each appearing at least M+1 times in the row, one can choose two from each group so that within the chosen subsequence the two from each group are consecutive. 3. In the induction/greedy step, correctly identified the “scan left to right until the first repeated group” idea and proved that in the scanned prefix there is at most one element from each other group, hence deleting this prefix reduces the required lower bound for each remaining group by at most 1. (Almost) 1. Gave the full inductive construction selecting a first pair by scanning to the first repeated group, then deleting the scanned prefix and all remaining members of that group, and applying induction to the remaining M−1 groups, but did not fully justify why the selected first pair ends up adjacent in the final row (e.g. did not argue that all other selected players come from the suffix). 2. Correctly completed the auxiliary claim, and correctly applied it to the N groups of size N+1 to finish, but left a small gap in the counting after deletions (e.g. did not justify that each remaining group still has at least (M−1)+1 elements in the reduced row). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.

\begin{enumerate}[(a)]
\item Prove that Geoff can always fulfill his wish if $n$ is odd.
\item Prove that Geoff can never fulfill his wish if $n$ is even.
\end{enumerate}

Geoff can fulfill his wish if and only if $n$ is odd.

The preliminary geometric framework operates as established context. A large circle $\omega$ encases all $\binom{n}{2}$ intersection points. The boundary points $P_1, P_2, \dots, P_{2n}$ are ordered clockwise. Each line segment connects $P_i$ and $P_{i+n}$.

(a) For odd $n$, place the frogs on $P_1, P_3, \dots, P_{2n-1}$. Parity dictates that no two frogs intersect simultaneously.

(b) For even $n$, frogs cannot occupy consecutive $P_i$, requiring placement on alternating points. Frogs cannot occupy diametrically opposite points, yielding an immediate contradiction.

(Partial) 1. Set up (or correctly used) the standard circular model: all intersections inside a circle, the 2n boundary points P1,…,P2n in order, and each segment is the chord PiPi+n; and interpreted a frog’s positions after k claps as the k-th intersection along that chord. 2. For odd n, gave the correct placement pattern (e.g. frogs start at P1,P3,…,P2n−1 or an equivalent alternating set) and proved a nontrivial invariant/observation implying two frogs cannot land on the same intersection at the same time (e.g. a parity argument on indices of endpoints/intersections). 3. For even n, proved a necessary restriction on any collision-free placement, such as: starting endpoints cannot be consecutive among the P_i (otherwise an immediate collision occurs), hence frogs must be placed on alternating boundary points. (Almost) 1. Solved both (a) and (b) in the circle/chord model with the correct constructions and contradictions, but left a small gap in justifying the key translation between “same intersection at same time” and the derived parity/index condition on the corresponding boundary labels. 2. Gave the odd-n construction and essentially complete no-collision proof, and for even n obtained the forced “alternating endpoints” condition, but did not finish the final contradiction (e.g. did not fully justify why some diametrically opposite pair must be chosen / why that forces a collision). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
We say that a finite set $\mathcal{S}$ of points in the plane is *balanced* if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is *center‑free* if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there are no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

\begin{enumerate}
\item[(a)] Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.
\item[(b)] Determine all integers $n\ge 3$ for which there exists a balanced center‑free set consisting of $n$ points.
\end{enumerate}

**Part (a)**

We show that a balanced set exists for all integers $n \ge 3$. Let $O$ be a point and $C$ be a circle centered at $O$. We construct the set $\mathcal{S}$ by choosing $O$ and $n-1$ additional points on $C$. We can group these $n-1$ points into disjoint pairs and triples as follows:
- A *pair* consists of two points $A, B$ on $C$ separated by an arc of $60^\circ$.
- A *triple* consists of three points $A, B, C$ on $C$ such that $A$ and $B$ are separated by $60^\circ$, and $B$ and $C$ are separated by $60^\circ$ (so $A, B, C$ are consecutive vertices of a regular hexagon).

Since $n \ge 3$, we have $n-1 \ge 2$. Any integer $k \ge 2$ can be expressed as $2x + 3y$ for some non-negative integers $x, y$. Thus, we can place $x$ pairs and $y$ triples on $C$ such that all $n-1$ points are distinct.

We now verify that $\mathcal{S}$ is balanced. For any two distinct points $X, Y \in \mathcal{S} \setminus \{O\}$, they lie on $C$, so $OX = OY$. Thus, $O$ serves as the required third point. For $O$ and a point $X \in \mathcal{S} \setminus \{O\}$, we need a point $Y \in \mathcal{S}$ such that $YX = YO$. Since $Y$ must lie on $C$, $YO$ is the radius of $C$, which means $\triangle OXY$ must be equilateral, so $X$ and $Y$ must be separated by $60^\circ$.
- If $X$ belongs to a pair $\{A, B\}$, then $A$ and $B$ are separated by $60^\circ$, so $A$ works for $O$ and $B$, and $B$ works for $O$ and $A$.
- If $X$ belongs to a triple $\{A, B, C\}$, then $B$ works for $O$ and $A$, $B$ works for $O$ and $C$, and both $A$ and $C$ work for $O$ and $B$.

Thus, the set $\mathcal{S}$ is balanced.

**Part (b)**

The answer is all odd integers $n \ge 3$.

First, we show that for any odd $n \ge 3$, the set of vertices of a regular $n$-gon is balanced and center-free.
- **Balanced:** For any two distinct vertices $A$ and $B$, their perpendicular bisector passes through the center of the $n$-gon and exactly one vertex $C$ of the $n$-gon (since $n$ is odd). Thus, $CA = CB$, so the set is balanced.
- **Center-free:** Any three distinct vertices $A, B, C$ uniquely determine a circumcircle, which is the circumcircle of the regular $n$-gon. The only point equidistant from $A, B,$ and $C$ is the center of this circumcircle. However, the center is not a vertex of the $n$-gon, so it is not in the set. Thus, the set is center-free.

Next, we use a standard double-counting argument to show that no balanced center-free set exists for even $n$. Assume for the sake of contradiction that $\mathcal{S}$ is a balanced center-free set of $n$ points, where $n$ is even.

By the balanced property, for each of the $\binom{n}{2}$ pairs of distinct points $\{A, B\}$ in $\mathcal{S}$, there exists at least one point $C \in \mathcal{S}$ such that $CA = CB$. We assign one such point $C$ to each pair. By the Pigeonhole Principle, since there are $n$ points available to be assigned, some point $P \in \mathcal{S}$ must be assigned to at least
$$ \left\lceil \frac{1}{n}\binom{n}{2} \right\rceil = \left\lceil \frac{n-1}{2} \right\rceil $$
pairs of points. Because $n$ is even, $\left\lceil \frac{n-1}{2} \right\rceil = \frac{n}{2}$.

Thus, $P$ is on the perpendicular bisector of at least $\frac{n}{2}$ pairs of points in $\mathcal{S} \setminus \{P\}$. This means $P$ is equidistant from the two points in each of these $\frac{n}{2}$ pairs.

However, because $\mathcal{S}$ is center-free, $P$ cannot be equidistant from any three points in $\mathcal{S}$. This implies that the points in $\mathcal{S} \setminus \{P\}$ can be partitioned into equivalence classes based on their distance to $P$, and each class can contain at most $2$ points. The pairs assigned to $P$ correspond to these classes of size exactly $2$.

Since there are exactly $n-1$ points in $\mathcal{S} \setminus \{P\}$, the maximum number of disjoint pairs of points equidistant from $P$ is
$$ \left\lfloor \frac{n-1}{2} \right\rfloor = \frac{n-2}{2}. $$
This contradicts the fact that $P$ is assigned to $\frac{n}{2}$ pairs, because $\frac{n}{2} > \frac{n-2}{2}$. Therefore, no balanced center-free set exists for even $n$.

Part (a): Exists for all integers $n \ge 3$. Part (b): All odd integers $n \ge 3$.

(Partial) 1. For (a), gave a valid construction idea on a circle with a distinguished point O (e.g. O plus points on a circle) and correctly verified the balanced condition for pairs of points lying on the circle (using O as the equidistant point). 2. For (a), correctly reduced the remaining balanced requirement to ensuring that for each point X on the circle there is another chosen point Y with ∠XOY=60° (equivalently, an equilateral triangle with OX=OY=XY). 3. For (b), proved that for odd n the vertices of a regular n-gon form a balanced set (e.g. argued that for any two vertices A,B there is a vertex C on the perpendicular bisector of AB, or an equivalent symmetry argument). 4. For (b), established a key obstruction for even n, e.g. started the assignment/double-counting argument by assigning to each unordered pair {A,B} a point C with CA=CB and deduced that some P is assigned to at least n/2 pairs when n is even. (Almost) 1. Solved (a) completely and proved existence for all odd n in (b), and for even n carried out the assignment/pigeonhole step, but did not finish the final contradiction from center-free (e.g. did not justify why P cannot be assigned to more than ⌊(n-1)/2⌋ pairs). 2. Solved (b) completely (odd n works and even n impossible), but in (a) gave the circle construction without fully justifying that any n−1≥2 can be decomposed into disjoint 2- and 3-blocks (or an equivalent bookkeeping step ensuring all points are distinct). 3. Provided the full intended strategy for both parts, but left one localized gap such as: uniqueness/existence of the vertex C on a perpendicular bisector in the odd n regular n-gon, or the claim that center-free implies no distance-to-P class has size ≥3.


Question: A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position it is possible to colour at least $\sqrt{n}$ lines blue in such a way that none of its finite regions has a completely blue boundary.

Suppose we have colored $k$ of the lines blue, and that it is not possible to color any additional lines without creating a finite region with a completely blue boundary. This means that any of the $n-k$ non-blue lines is a side of some finite region with an otherwise entirely blue perimeter.

For each such non-blue line $\ell$, select one such finite region, and take the next counterclockwise vertex along the region's perimeter; this vertex $v$ is the intersection of two blue lines. We say that $\ell$ is the *eyelid* of $v$.

Every intersection of two blue lines has at most two eyelids. Since there are exactly $\binom{k}{2}$ such intersections of blue lines, we obtain the inequality
\[
n-k \le 2 \binom{k}{2} = k^{2} - k.
\]
This simplifies to $n \le k^{2}$, which implies $k \ge \sqrt{n}$. Thus, it is always possible to color at least $\sqrt{n}$ lines blue such that none of the finite regions has a completely blue boundary.

(Partial) 1. Introduced the maximality setup: start with a blue set of size k such that no further line can be colored blue without creating a finite region whose boundary is all blue, and correctly deduced that every non-blue line lies on the boundary of some finite region whose other sides are all blue. 2. Defined a way to associate (injectively up to bounded multiplicity) each non-blue line to an intersection point of two blue lines (e.g. by choosing a witness finite region and taking a specific adjacent vertex on its boundary), and justified that the chosen point is indeed an intersection of two blue lines. 3. Proved the local bound that a fixed intersection point of two blue lines can be associated to at most two non-blue lines (e.g. by analyzing the cyclic order of lines through that vertex and the two possible incident finite regions). (Almost) 1. Obtained the inequality n-k ≤ 2·C(k,2) (or an equivalent bound n ≤ k^2+O(k)) from the association argument, but did not fully justify why each non-blue line can be assigned to a finite region with otherwise all-blue boundary. 2. Carried out the counting argument correctly to conclude k ≥ √n, but left a small gap in the “at most two eyelids per blue-blue intersection” claim (e.g. did not exclude a third candidate region or did not use general position correctly). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
A configuration of $4027$ points in the plane is called *Colombian* if it consists of $2013$ red points and $2014$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is *good* for a Colombian configuration if the following two conditions are satisfied:
\begin{enumerate}
\ii[(i)] No line passes through any point of the configuration.
\ii[(ii)] No region contains points of both colors.
\end{enumerate}
Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.

The least value of $k$ is $2013$.

**Lower Bound:**
We first show that $k \ge 2013$ lines are necessary. Consider a configuration where $4026$ of the points form the vertices of a regular $4026$-gon, and the colors of these vertices alternate between red and blue. The $4027$th point is the remaining blue point, which can be placed anywhere such that no three points in the entire configuration are collinear.

The perimeter of the regular $4026$-gon consists of $4026$ line segments, each connecting a red point and a blue point. In any good arrangement of lines, every such segment must be intersected by at least one line, because the endpoints of the segment are of different colors and must therefore belong to different regions. Since the $4026$-gon is strictly convex, any straight line can intersect its perimeter in at most two points. Therefore, a single line can intersect at most two of the $4026$ segments. To cut all $4026$ segments, we require at least $4026 / 2 = 2013$ lines. Thus, $k \ge 2013$.

**Upper Bound:**
We now prove that $k = 2013$ lines are always sufficient. Consider the convex hull of the $4027$ points. Since the points are not all collinear, the convex hull is a convex polygon with at least three vertices. We distinguish two cases based on the vertices of this convex hull:

*Case 1: The convex hull contains at least one red point.*
Let $R$ be a red point on the convex hull. Since $R$ is a vertex of the convex hull, we can draw a single line that separates $R$ from the other $4026$ points.

There are $2012$ red points remaining. We group these $2012$ red points into $1006$ pairs arbitrarily. For each pair $\{A, B\}$, we draw two lines parallel to the segment $AB$ and sufficiently close to it such that the narrow strip between these two lines contains $A$ and $B$ but no other points of the configuration. (This is possible because no three points are collinear, so no other point lies on the line $AB$.)

We have drawn $1 + 2 \times 1006 = 2013$ lines. These lines partition the plane into several regions. We claim this is a good arrangement. Any point in the plane not on the drawn lines falls into one of the following categories:
- It lies in the half-plane containing $R$. Since this half-plane contains no other points of the configuration, any region here contains at most the red point $R$, and no blue points.
- It lies within one of the $1006$ narrow strips. By construction, the entire strip contains only the red points $A$ and $B$ and no other points of the configuration. Thus, any region within this strip contains no blue points.
- It lies outside the half-plane and outside all $1006$ strips. Since every red point is either $R$ or belongs to one of the pairs, all red points have been accounted for. Thus, any region here contains no red points (it may contain blue points or be empty).

In all cases, no region contains points of both colors. Thus, the arrangement is good.

*Case 2: The convex hull contains no red points.*
If there are no red points on the convex hull, then all vertices of the convex hull must be blue points. Since the convex hull has at least three vertices, there must exist two consecutive blue points on the perimeter of the convex hull, say $U$ and $V$.

Since the segment $UV$ is an edge of the convex hull, we can draw a single line that separates $U$ and $V$ from the other $4025$ points.

There are $2012$ blue points remaining. We group these $2012$ blue points into $1006$ pairs arbitrarily. For each pair, we draw two lines parallel to the segment connecting them, forming a narrow strip that contains exactly these two blue points and no other points of the configuration.

Again, we have drawn $1 + 2 \times 1006 = 2013$ lines. By the exact same reasoning as in Case 1, any point not on the lines either lies in the half-plane containing $U$ and $V$ (which contains no red points), lies in one of the narrow strips (which contain only blue points), or lies outside all of these (which contains no blue points, and thus only red points or empty). Hence, no region contains points of both colors.

In all cases, a good arrangement of $2013$ lines can be found. Therefore, the least value of $k$ is $2013$.

(Partial) 1. Established the lower bound idea by giving (or clearly describing) a configuration with alternating colors on the vertices of a convex 4026-gon and arguing that every red–blue edge must be crossed by the drawn lines, and that each line crosses the boundary of a strictly convex polygon at most twice, hence k≥2013. 2. Proved a correct geometric reduction for the upper bound: isolated one point (or two adjacent hull vertices) by a single line using a supporting line to the convex hull, ensuring that half-plane contains only those chosen point(s) of one color. 3. Gave the key construction step for the upper bound: for many same-colored points, paired them and for each pair drew two parallel lines close to the segment so that the strip contains exactly those points and no others (with a correct justification using ‘no three collinear’/small perturbation). (Almost) 1. Presented the full 2013-line construction (one separating line plus 1006 strips) but did not completely justify that the chosen strips can be made disjoint/placed so that each contains no other configuration points, leaving only a local existence/perturbation gap. 2. Handled one of the two convex-hull cases (a red hull vertex exists, or all hull vertices are blue and isolate two consecutive blue vertices) correctly and finished the region-color argument for that case, but omitted or inadequately treated the other case. 3. Proved both bounds k≥2013 and k≤2013 with correct overall structure, but made a small counting/edge-case slip (e.g. miscounting how many points remain to be paired, or not checking that no line passes through a configuration point), while the main idea is correct.