Source: IMO-2018-P3
Category: Existence and Construction

Problem:
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 $1+2+\dots +2018$?

Solution:
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 $N$.

Answer: No

Grading guidelines:
(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.