1.- Consider sets A of one hundred distinct natural numbers, which have the property that if a, b, and c are any elements of A (equal or distinct), there exists a non-obtuse triangle whose sides measure a, b, and c units.

The sum of the perimeters considered in the definition of A is denoted by $S(A)$. Calculate the minimum value of $S(A)$.
Solution:

If n is the smallest of the elements of A and m the largest, since A has a hundred distinct elements, it will be $\mathrm{m} \geq \mathrm{n}+99$. For the isosceles triangle with sides $\mathrm{n}, \mathrm{n}, \mathrm{m}$ to be non-obtuse, it must be that $\mathrm{m}^{2} \leq 2 \mathrm{n}^{2}$; if $\mathrm{m}$ is as small as possible, $\mathrm{m}=\mathrm{n}+99$ should satisfy $(\mathrm{n}+99)^{2} \leq 2 \mathrm{n}^{2}$, that is:

$$
\mathrm{n}^{2}-198 \mathrm{n}-99^{2} \geq 0 \Leftrightarrow \mathrm{n} \geq 99+\sqrt{99^{2}+99^{2}} \Leftrightarrow \mathrm{n} \geq 99(1+\sqrt{2}) \Leftrightarrow n \geq 240
$$

If n $<240$, it is certain that the set does not meet the condition in the statement because

$$
m^{2} \geq(n+99)^{2} \geq 2 n^{2}
$$

and the triangle with sides n, n, m cannot be non-obtuse. In particular, the condition is met for the set:

$$
A=\{240,241,242, \ldots ., 339\}
$$

Any other set that meets the condition will have its elements respectively equal to or greater than those of this set. This is, therefore, the one that gives rise to the minimum S(A).

The number of triangles to be considered is the number of ternary variations with repetition of the elements of $\mathrm{A}$, which is $100^{3}=1000000$, so the total number of sides will be 3000000; of these, there will be 30000 of length 240, the same number of length 241, etc. Thus,

$$
S(A)=30000 \cdot(240+241+242+\ldots+339)=30000 \cdot \frac{100 \cdot(240+339)}{2}=868500000 \text { units }
$$

This is the minimum value sought.