IMO Team Selection Test 2

It is clear that $n=1$ does not satisfy the property, as not every three positive real numbers $a,b$, and $c$ are the lengths of the sides of a triangle.

We first show that any $n\ge 5$ cannot satisfy the property by considering the triple $(a,b,c)=(1,2,3)$. Suppose that there exist $k,\ell ,m$ such that $n^k,2n^{\ell },3n^m$ are the lengths of the sides of a triangle. As $n\ne 2,3$, no three of these are equal. By repeatedly removing common factors $n$ if they exist, we can and do assume that one of $k,\ell ,m$ is equal to 0. Suppose that the other two of those three integers are positive, then their corresponding side lengths both are multiples of $n$. Their difference then is at least $n$, while the third side has length of at most 3. This contradicts the triangle inequality. Hence of $k,\ell ,m$, at least two must be zero. The corresponding two side lengths then sum to at most 5, so the third side must have length less than 5. As $n\ge 5$, this third side cannot have a factor $n$ either, so $k,\ell ,m$ are all equal to 0. However, then 1, 2, 3 should be the lengths of the sides of a triangle, while $3=2+1$. This is a contradiction. We deduce that $n\ge 5$ does not satisfy the property.

Now consider $n=2,3,4$. We construct $(k,\ell ,m)$ as follows. Take a triple $(a,b,c)$. If its entries already are the lengths of the sides of a triangle, we take $k=\ell =m=0$. Otherwise there exists a triangle inequality that is not satisfied. Without loss of generality, we assume that $a\ge b+c$. Multiply the lesser of $b$ and $c$ with $n$. If the right hand side still isn't larger than $a$, then take the new summands, and again multiply the lesser of the two by $n$. So: if $a\ge n^{i}b+n^{j}c$, then we multiply the lesser of $n^{i}b$ and $n^{j}c$ by $n$, and repeat if the inequality still holds. This process always stops at some point, as there exists $i$ such that $n^i>a$.

Consider the $i$ and $j$ such that $a\ge n^{i}b+n^{j}c$, such that applying the process makes the right hand side larger than $a$. We assume without loss of generality that $n^{i}b\le n^{j}c$, so we have $a<n^{i+1}b+n^{j}c$. We claim that we can take $(k,\ell ,m)$ to be equal to either $(0,i+1,j)$ or $(0,i+1,j+1)$.

By definition, we have $a<n^{i+1}b+n^{j}c$, and we have $n^{j}c<n^{i}b+n^{j}c\le a<a+n^{i+1}b$. Therefore, if $(0,i+1,j)$ doesn't satisfy the property, then we must have $n^{i+1}b\ge a+n^{j}c$. Moreover $n^{i}b\le n^{j}c$, so $n^{i+1}b\le n^{j+1}c<a+n^{j+1}c$. We also have $a<n^{i+1}b+n^{j+1}c$, so if $(0,i+1,j+1)$ does not satisfy the property, then we must have $n^{j+1}c\ge a+n^{i+1}b$. We will derive a contradiction in case both triples do not satisfy the property, i.e. in case both $n^{i+1}b\ge a+n^{j}c$ and $n^{j+1}c\ge a+n^{i+1}b$.

Adding these inequalities, and subtracting $n^{i+1}b$ on both sides, yields $n^{j+1}c\ge 2a+n^{j}c$, or equivalently $(n-1)n^{j}c\ge 2a$. Therefore $$n^{i+1}b\ge a+n^{j}c\ge a+\frac{2a}{n-1}=\left(1+\frac{2}{n-1}\right)a,$$ from which follows that $$a\ge n^{i}b+n^{j}c\ge \frac{1}{n}\left(1+\frac{2}{n-1}\right)a+\frac{2}{n-1}a=\frac{(n-1)+2+2n}{n(n-1)}a=\frac{3n+1}{n(n-1)}a.$$ So $3n+1\le n(n-1)$. For $n=2,3,4$, this inequality reads $7\le 2,10\le 6$, and $13\le 12$, all of which are false. This is a contradiction.

Hence $n=2,3,4$ are precisely the values which satisfy the property. $\square$