1. How many different numbers can there be among the numbers $a+2b, a+2c, b+2a, b+2c, c+2a$, and $c+2b$, if $a$, $b$, and $c$ are distinct real numbers? Find all possibilities.

(Josef Tkadlec)
SOLUTION. Since the problem is symmetric in the variables $a, b, c$, we can assume without loss of generality that $a < b < c$. The first inequality is $a+2b - (b+2a) = b-a > 0$, the second inequality is $\mathrm{z}(a+2c) - (a+2b) = 2(c-b) > 0$, and the third is $(b+2c) - (a+2c) = b-a > 0$.

Let's see which of the four numbers in (1) the remaining two numbers $c+2a$ and $c+2b$ can equal. To do this, we note that just as easily as (1), we can prove the inequalities

$$
b+2a < c+2a < a+2c \quad \text{and} \quad a+2b < c+2b < b+2c.
$$

By comparing with (1), we conclude that the only two possible equalities among the six numbers being examined (assuming $a < b < c$) are $c+2a = a+2b$ and $c+2b = a+2c$. Both of these equalities are clearly equivalent to the same equality $b = \frac{1}{2}(a+c)$, which is satisfiable under our assumption.

The result of our considerations for the case $a < b < c$ can be summarized as follows:

- If $b = \frac{1}{2}(a+c)$, there are exactly four different numbers among the examined numbers. Indeed, for them we have

$$
b+2a < a+2b = c+2a < a+2c = c+2b < b+2c.
$$

This is the case, for example, for $(a, b, c) = (1, 2, 3)$.

- If, on the other hand, $b \neq \frac{1}{2}(a+c)$, then all six examined numbers are different.

This is the case, for example, for $(a, b, c) = (1, 2, 4)$.

Conclusion. The number of different numbers in the given sextet is either 4 or 6.

Note. After writing down the inequalities (1), the question of the possible equality of the first "remaining" number $c+2a$ to any of the numbers in (1) can also be solved without using the inequalities (2), by directly testing the individual equalities

$$
c+2a = b+2a, \quad c+2a = a+2b, \quad c+2a = a+2c, \quad c+2a = b+2c.
$$

Similarly, for the second "remaining" number $c+2b$, we can test the equalities

$$
c+2b = b+2a, \quad c+2b = a+2b, \quad c+2b = a+2c, \quad c+2b = b+2c.
$$

Some of these equalities (e.g., $c+2a = b+2a$) are already excluded by the fact that the numbers $a, b, c$ are distinct. Other equalities (except those in (3)) are in conflict with the assumed ordering $a < b < c$ (for example, $c+2a = b+2c$ means that $a = \frac{1}{2}(b+c)$, so the number $a$ would lie between the numbers $b$ and $c$.)[^0]