# Problem 5.

In-1 If the real numbers $a, b, c$ are ordered in non-decreasing order, resulting in the triplet $x_{1} \leqslant x_{2} \leqslant x_{3}$, then the number $x_{2}$ will be called the middle number of $a, b, c$. Find all values of $t$ for which the middle number of the three numbers

$$
a=t^{3}-100 t ; \quad b=2^{t}-16 ; \quad c=\sin t-\frac{1}{2}
$$

is positive.
Answer: $\left(-10 ;-4 \pi+\frac{5 \pi}{6}\right) \cup\left(-2 \pi+\frac{\pi}{6} ;-2 \pi+\frac{5 \pi}{6}\right) \cup\left(2 \pi+\frac{\pi}{6} ; 2 \pi+\frac{5 \pi}{6}\right) \cup(10 ;+\infty)$

Solution. Directly comparing the values of $a, b, c$ is difficult. However, for the average of three numbers to be positive, it is necessary and sufficient that at least two of the numbers in the triplet are positive.

$$
a=t^{3}-100 t=t(t-10)(t+10)>0 \text { when } t \in(-10 ; 0) \cup(10 ;+\infty) \text {. }
$$

$b=2^{t}-16>0$ when $t>4$.

$c=\sin t-\frac{1}{2}>0$ when $t \in\left(2 \pi n+\frac{\pi}{6} ; 2 \pi n+\frac{5 \pi}{6}\right)$, where $n \in \mathbb{Z}$.

We need at least two of the three numbers to be positive.

$a>0$ and $b>0$ when $t>10$, the interval $(10,+\infty)$ is part of the answer.

$a \leqslant 0$ and $b \leqslant 0$ when $t \in(-\infty,-10] \cup[0,4]$, this interval cannot be part of the answer.

On the remaining interval $t \in(-10,0) \cup(4,10]$, only one of the numbers $a, b$ is positive. Therefore, the parts of this interval where $c>0$ will be part of the answer. Let's see how $t \in(-10,0) \cup(4,10]$ intersects with $t \in\left(2 \pi n+\frac{\pi}{6} ; 2 \pi n+\frac{5 \pi}{6}\right)$, where $n \in \mathbb{Z}$.

For $n=0$, we get the interval $t \in\left(\frac{\pi}{6} ; \frac{5 \pi}{6}\right)$. It does not intersect with the interval $t \in(-10,0) \cup(4,10]$, since $0<\frac{\pi}{6}<\frac{5 \pi}{6}<4$.

For $n=1$, we get the interval $t \in\left(2 \pi+\frac{\pi}{6} ; 2 \pi+\frac{5 \pi}{6}\right)$. It lies entirely within the interval, since $4<2 \pi+\frac{\pi}{6}<2 \pi+\frac{5 \pi}{6}<10$. The interval is part of the answer.

For $n=-1$, we get the interval $t \in\left(-2 \pi+\frac{\pi}{6} ;-2 \pi+\frac{5 \pi}{6}\right)$. It also lies entirely within the interval, since $-10<-2 \pi+\frac{\pi}{6}<-2 \pi+\frac{5 \pi}{6}<0$. The interval is part of the answer.

For $n=-2$, we get the interval $t \in\left(-4 \pi+\frac{\pi}{6} ;-4 \pi+\frac{5 \pi}{6}\right)$. Here we get the inequality $-4 \pi+\frac{\pi}{6}<-10<-4 \pi+\frac{5 \pi}{6}$, the interval intersects with the interval $t \in(-10,0) \cup(4,10]$, the intersection is the set $\left(-10,-4 \pi+\frac{5 \pi}{6}\right)$, which is part of the answer.

For other values of $n$, the intervals are either far to the left of -10 or to the right of 10, and do not affect the answer.

In the end, the answer is the union of the sets $(10,+\infty),\left(2 \pi+\frac{\pi}{6} ; 2 \pi+\frac{5 \pi}{6}\right),\left(-2 \pi+\frac{\pi}{6} ;-2 \pi+\right.$ $\left.\frac{5 \pi}{6}\right),\left(-10,-4 \pi+\frac{5 \pi}{6}\right)$.

V-2 If the real numbers $a, b, c$ are ordered in non-decreasing order, resulting in the triplet $x_{1} \leqslant x_{2} \leqslant x_{3}$, then the number $x_{2}$ will be called the median of the numbers $a, b, c$. Find all values of $t$, for each of which the median of the three numbers

$$
a=t^{3}-81 t ; \quad b=11^{t}-121 ; \quad c=\sin t-\frac{1}{2}
$$

is positive.

Answer: $\left(-2 \pi+\frac{\pi}{6} ;-2 \pi+\frac{5 \pi}{6}\right) \cup\left(2 ; \frac{5 \pi}{6}\right) \cup\left(2 \pi+\frac{\pi}{6} ; 2 \pi+\frac{5 \pi}{6}\right) \cup(9 ;+\infty)$

V-3 If the real numbers $a, b, c$ are ordered in non-decreasing order, resulting in the triplet $x_{1} \leqslant x_{2} \leqslant x_{3}$, then the number $x_{2}$ will be called the median of the numbers $a, b, c$. Find all values of $t$, for each of which the median of the three numbers

$$
a=t^{3}-121 t ; \quad b=2^{t}-32 ; \quad c=\sin t-\frac{\sqrt{3}}{2}
$$

is positive.

Answer: $t \in\left(-11 ;-4 \pi+\frac{2 \pi}{3}\right) \cup\left(-2 \pi+\frac{\pi}{3} ;-2 \pi+\frac{2 \pi}{3}\right) \cup\left(2 \pi+\frac{\pi}{3} ; 2 \pi+\frac{2 \pi}{3}\right) \cup(11 ;+\infty)$

V-4 If the real numbers $a, b, c$ are ordered in non-decreasing order, resulting in the triplet $x_{1} \leqslant x_{2} \leqslant x_{3}$, then the number $x_{2}$ will be called the median of the numbers $a, b, c$. Find all values of $t$, for each of which the median of the three numbers

$$
a=t^{3}-144 t ; \quad b=2^{t}-256 ; \quad c=\sin t-\frac{\sqrt{3}}{2}
$$

is positive.

Answer: $\left(-4 \pi+\frac{\pi}{3} ;-4 \pi+\frac{2 \pi}{3}\right) \cup\left(-2 \pi+\frac{\pi}{3} ;-2 \pi+\frac{2 \pi}{3}\right) \cup\left(8 ; 2 \pi+\frac{2 \pi}{3}\right) \cup(12 ;+\infty)$

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final stage $2021 / 22$ academic year for 11th grade