## Problem III - 6

We say that three distinct natural numbers form an additive triplet if the sum of the first two of them is equal to the third. Find, with reasoning, the maximum number of additive triplets that can exist in a given set of 20 natural numbers.
Solution.

Let twenty distinct natural numbers be arbitrarily chosen:

$$
x_{1}<x_{2}<\cdots<x_{20}
$$

We will obtain an upper bound for the number of additive triples that can be found among them, i.e., triples $\left\{x_{i}, x_{j}, x_{k}\right\}$ that satisfy $1 \leq i<j<k \leq 20$, and $x_{i}+x_{j}=x_{k}$. First, we will do this for a fixed index $k \in\{3,4, \cdots, 20\}$: let these triples be

$$
\left.\left\{x_{i_{1}}, x_{j_{1}}, x_{k}\right\}\right),\left\{x_{i_{2}}, x_{j_{2}}, x_{k}\right\}, \cdots\left\{x_{i_{p}}, x_{j_{p}}, x_{k}\right\}
$$

Then the numbers

$$
x_{i_{1}}, x_{j_{1}}, x_{i_{2}}, x_{j_{2}}, \cdots, x_{i_{p}}, x_{j_{p}}
$$

are mutually distinct and all belong to the set

$$
\left(x_{1}, x_{2}, \cdots, x_{k-1}\right)
$$

Therefore, their number $2 p$ must be subject to the constraint

$$
2 p \leq k-1
$$

i.e., $p$ must be less than or equal to the integer part of $\frac{1}{2}(k-1)$:

$$
p \leq\left[\frac{k-1}{2}\right]
$$

Consequently, the total number of additive triples cannot exceed the sum

$$
\sum_{k=3}^{20}\left[\frac{k-1}{2}\right]=1+1+2+2+3+3+\cdots+9+9=90
$$

The example of the set

$$
M=\{1,2,3, \cdots, 20)
$$

demonstrates that the upper bound 90 can indeed be achieved, since for any

$$
k \in(3,4, \cdots, 20)
$$

we can choose for $i$ any number from the set

$$
\left\{1,2, \cdots,\left[\frac{k-1}{2}\right]\right\}
$$

the corresponding $j=k-i$ indeed satisfies the inequalities $i<j<k$, so $\{i, j, k\}$ is a triple contained in $M$.