## Task B-4.5.

In a basketball tournament, the teams "Wolves" and "Bears" played the first quarter to a draw. The points scored by the Wolves in each of the 4 quarters form an increasing geometric sequence, while the points scored by the Bears in each quarter form an increasing arithmetic sequence. In the end, the Wolves won by a single point. Neither team scored more than 100 points. Determine the total number of points both teams scored together at the end of the first half.
## Solution.

Let $a$ be the number of points both teams scored in the first quarter. Then the sequence of numbers, or points, that the Wolves scored in each quarter is $a, a q, a q^{2}, a q^{3}$, where $q$ is the common ratio of the geometric sequence and $q>1$ because the sequence is increasing.

The sequence of points that the Bears scored in each quarter is the following increasing arithmetic sequence: $a, a+d, a+2 d, a+3 d, d>0$.

The sum of the geometric sequence is the total number of points the Wolves scored at the end, and the sum of the arithmetic sequence is the total number of points the Bears scored at the end. Therefore,

$$
\begin{aligned}
& a+a q+a q^{2}+a q^{3} \leqslant 100 \\
& a+a+d+a+2 d+a+3 d \leqslant 100
\end{aligned}
$$

or

$$
\begin{aligned}
& a\left(1+q+q^{2}+q^{3}\right) \leqslant 100 \\
& 4 a+6 d \leqslant 100 \\
& a\left(1+q+q^{2}+q^{3}\right)=4 a+6 d+1(*)
\end{aligned}
$$

From the last equation, considering the parity of the left and right sides, we conclude that $a$ is an odd number, and $q$ is an even number. From the first inequality, we conclude that $q \leqslant 4$.

We have the following possibilities: $q=4$ or $q=2$.

If $q=4$, then $85 a \leqslant 100$, which means $a=1$. From $(*)$ for $q=4$ and $a=1$, we get $d \notin \mathbb{N}$.

If $q=2$, then $15 a \leqslant 100$, which means $a \leqslant 6, a \in\{1,3,5\}$. Now we use $(*)$ to check which of the numbers $a \in\{1,3,5\}$ give an integer solution for $d$.

$$
\begin{aligned}
& 15 a=4 a+6 d+1 \\
& 6 d=11 a-1
\end{aligned}
$$

The last equation has an integer solution only for $a=5, d=9$.

For these values, the Wolves had the following points: 5, 10, 20, 40, and the Bears had 5, 14, 23, 32.

We need to calculate the total sum of points that both teams scored at halftime, i.e., $a+a q+a+a+d=3 a+a q+d$.

The total sum of points is $5+10+5+14=34$.