## Task 6B - 181246B

a) Let $M$ be the set of all triples $(x, y, z)$ of real numbers for which the following inequalities (1) to (5) are satisfied:

$$
\begin{aligned}
55 x + z & \leq 54 \\
55 y + z & \leq 54 \\
55 x - 4 z & \geq 4 \\
55 y - 4 z & \geq 4 \\
z & \geq -1
\end{aligned}
$$

Investigate whether there exists a triple $\left(x_{0}, y_{0}, z_{0}\right) \in M$ with the property that for all triples $(x, y, z) \in M$ the inequality

$$
f\left(x_{0}, y_{0}, z_{0}\right) \geq f(x, y, z)
$$

holds. If this is the case, determine $f\left(x_{0}, y_{0}, z_{0}\right)$.

b) Let $M^{\prime}$ be the set of all triples $(x, y, z)$ of integers for which the inequalities (1) to (5) are satisfied.

Investigate whether there exists a triple $\left(x_{1}, y_{1}, z_{1}\right) \in M^{\prime}$ with the property that for all triples $(x, y, z) \in M^{\prime}$ the inequality

$$
f\left(x_{1}, y_{1}, z_{1}\right) \geq f(x, y, z)
$$

holds. If this is the case, determine $f\left(x_{1}, y_{1}, z_{1}\right)$.
a) Let $x, y, z$ be real numbers that satisfy inequalities (1) to (5). Then, due to (3) and (1),

$$
4 z - 55 x \leq -4 ; \quad z + 55 x \leq 54 \quad \text{thus} \quad 5 z \leq 40 \rightarrow z \leq 10
$$

Due to (5), it follows that $-1 \leq z \leq 10$ (7). Now, from (1) and (3),

$$
\frac{4 + 4 z}{55} \leq x \leq \frac{54 - z}{55}
$$

and from (2) and (4),

$$
\frac{4 + 4 z}{55} \leq y \leq \frac{54 - z}{55}
$$

Due to (7), $\frac{4 \cdot 4 z}{55} \geq 0.00 ; \quad g(-1) = 3 ; \quad g(10) = 101.28
\end{aligned}
$$

Thus, this function attains its maximum in its domain for $z = 10$. Therefore, $f(x, y, z) \leq 101.28$ (12). For $z_0 = 10, x_0 = y_0 = \frac{54 - z_0}{55} = 0.8 \quad$ (13), we now have $f(x_0, y_0, z_0) = 101.28 \quad$ (14), meaning the triplet $(x_0, y_0, z_0) = (0.8; 0.8; 10)$ has the required property.

b) Now let $x, y, z$ be integers that satisfy inequalities (1) to (5). Then, inequalities (7), (8), (9), and (10) still hold, so $x$ can only be either 0 or 1, and $y$ can only be either 0 or 1.

If $x = 0$, then from (3) $z \leq -1$, and hence, due to $z \geq -1: z = -1$.

If $x = 1$, then from (1) $z \leq -1$, so again $z = -1$.

Therefore, there are at most the following integer triplets that satisfy inequalities (1) to (5):

$$
(0; 0; -1), \quad (0; 1; -1), \quad (1; 0; -1), \quad (1; 1; -1)
$$

These triplets indeed satisfy these inequalities, and furthermore,

$$
f(0; 0; -1) = 1, \quad f(0; 1; -1) = f(1; 0; -1) = 2, \quad f(1; 1; -1) = 3
$$

Thus, the triplet $(x_1, y_1, z_1) = (1; 1; -1)$ has the required property with $f(x_1, y_1, z_1) = 3$. Adapted from [3]

### 9.21 XIX. Olympiad 1979

### 9.21.1 First Round 1979, Class 12