4. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the smallest area that such a quadrilateral can have.
Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\frac{1}{2} x_{1} y_{2} \sin \alpha = \frac{1}{2} x_{2} y_{1} \sin \alpha$. Hence, $\frac{x_{1}}{x_{2}} = \frac{y_{1}}{y_{2}}$, and,

![](https://cdn.mathpix.com/cropped/2024_05_06_6f7f77b10b58f29308dag-2.jpg?height=283&width=731&top_left_y=1617&top_left_x=1187)
consequently, triangles $BOC$ and $AOD$ are similar by the first similarity criterion: two sides ($x_{1}$ and $y_{1}$) of triangle $BOC$ are proportional to two sides ($x_{2}$ and $y_{2}$) of triangle $AOD$, and the angles formed by these sides ($\angle BOC$ and $\angle AOD$) are equal. Let $k = \frac{x_{1}}{x_{2}} = \frac{y_{1}}{y_{2}}$ be the similarity coefficient of triangles $BOC$ and $AOD$. Denote by $S$ the areas of triangles $ABO$ and $CDO$ (by the condition $S = \frac{3}{2}$). Then $S_{BOC} = k \cdot S$ and $S_{AOD} = S / k$. In the end, the area of quadrilateral $ABCD$ can be represented as:

$$
S_{ABCD} = S_{AOD} + S_{CDO} + S_{BOC} + S_{ABO} = 2S + S\left(k + \frac{1}{k}\right)
$$

It is known that for $k > 0$ the minimum value of the expression $k + \frac{1}{k}$ is achieved at $k = 1$. Therefore, $x_{1} = x_{2}$ and $y_{1} = y_{2}$, which means that the diagonals of the quadrilateral are bisected by their point of intersection, so $ABCD$ is a parallelogram. Its area $S_{ABCD} = 4S = 6$.

Answer: 6.

## Interregional School Olympiad in Mathematics for Students of Vedomstvennye Educational Organizations

5. Find the sum of all prime numbers whose decimal representation has the form 101010...01.

Solution. Let $2n+1$ be the number of digits in the number $A = 101010 \ldots 101$. Let $q = 10$ be the base of the number system. Then $A = q^{0} + q^{2} + \cdots + q^{2n} = \frac{q^{2n+2} - 1}{q^{2} - 1}$. Consider the cases of even and odd $n$.

- $n = 2k \Rightarrow A = \frac{q^{2n+2} - 1}{q^{2} - 1} = \frac{q^{2k+1} - 1}{q - 1} \cdot \frac{q^{2k+1} + 1}{q + 1}$. Thus, the number $A$ is represented as the product of two integers (by the theorem of Bezout, the polynomial $q^{2k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, for even $n$, the number $A$ is not prime.
- $n = 2k-1 \Rightarrow A = \frac{q^{2n+2} - 1}{q^{2} - 1} = \frac{q^{2k} - 1}{q^{2} - 1} \cdot (q^{2k} + 1)$. For $k > 1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k = 1$, we get the prime number $A = q^{0} + q^{2} = 101$.

## Answer: 101.