7. Variant 1.

In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cinema hall is the one located at the center of the hall, i.e., at the midpoint of the height drawn from the vertex of the triangle corresponding to seat number 1. In the figure, an example of such a triangular hall with 45 seats is shown, where the best seat is number 13.

CINEMA "TRIANGLE"

![](https://cdn.mathpix.com/cropped/2024_05_06_1ac95f298a5c1be791f9g-08.jpg?height=729&width=974&top_left_y=1760&top_left_x=558)

How many seats are in the cinema hall where the best seat has the number $265?$
Answer: 1035

Solution.

1st method.

Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats in each part will be the same, see the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_1ac95f298a5c1be791f9g-09.jpg?height=571&width=1030&top_left_y=508&top_left_x=570)

Thus, the problem reduces to finding the number $n$. We need to find the largest integer $n$ that satisfies the inequality: $1+2+\ldots+n<265$ or $n^{2}+n-530<0$, i.e., $n=22$. Then, in each of the four marked triangles, there will be $1+2+\ldots+22=23 \cdot 11=253$ seats, and in total, in the hall, there will be $253 \cdot 4+23=1035$ seats.

2nd method.

Note that the "height" will consist of the numbers located in the center of the rows with odd numbers.

Let $a_{k}$ be the number of the seat located in the center of the row with number $k$ (where $k$ is odd). To the right and left of it in the given row, there are $(k-1)/2$ seats. Therefore, when transitioning from $a_{k}$ to $a_{k+2}$, $(k-1)/2$ last seats from the row with number $k$, the $(k+1)$-th seat from the row with number $k+1$, and $(k+1)/2$ first seats from the row with number $k+2$ are added.

Thus, $a_{k+2}=a_{k}+(k-1)/2+(k+1)+(k+1)/2+1=a_{k}+2(k+1)$.

$a_{1}=1$

$a_{3}=a_{1}+2 \cdot 2=1+4$

$a_{5}=a_{3}+2 \cdot 4=1+4+8$

$a_{7}=a_{3}+2 \cdot 6=1+4+8+12$

$\cdots$

$a_{k}=1+4+8+12+\ldots+2(k-1)=1+4(1+2+\ldots+(k-1)/2)=1+(k+1)(k-1)/2$,

Using this formula, we will find the row number where the best seat is located.

$$
\begin{gathered}
1+(k+1)(k-1) / 2=265 \\
\left(k^{2}-1\right) / 2=264 \\
k^{2}=264 \cdot 2+1=529 \\
k=\sqrt{529}=23 .
\end{gathered}
$$

The total number of rows is $2 k-1$ (before and after the row with the best seat, there are $k-1$ rows each). Therefore, the total number of seats in the hall is given by the formula
$(1+2 k-1) \cdot(2 k-1) / 2=k \cdot(2 k-1)$. For $k=23$, we get $23 \cdot 45=1035$.