p1. Suppose the number $\sqrt[3]{2+\frac{10}{9}\sqrt3} + \sqrt[3]{2-\frac{10}{9}\sqrt3}$ is integer. Calculate it.
 

p2. A house A is located $300$ m from the bank of a $200$ m wide river. $600$ m above and $500$ m from the opposite bank is another house $B$. A bridge has been built over the river, that allows you to go from one house to the other covering the minimum distance. What distance is this and in what place of the riverbank is at the bridge?
 

p3. Show that it is not possible to find $1992$ positive integers $x_1, x_2, ..., x_{1992}$ that satisfy the equation: $$\sum_{j=1}^{1992}2^{j-1}x_j^{1992}=2^{1992}\prod_{j=2}^{1992}x_j$$


p4. The following division of positive integers with remainder has been carried out. Every cross represents one digit, and each question mark indicates that there are digits, but without specifying how many.  Fill in each cross and question mark with the missing digits.
[img]https://cdn.artofproblemsolving.com/attachments/8/e/5658e77025210a46723b6d85abf5bdc3c7ad51.png[/img]


p5. Three externally tangent circles have, respectively, radii $1$, $2$, $3$. Determine the radius of the circumference that passes through the three points of tangency.


p6. $\bullet$ It is possible to draw a continuous curve that cuts each of the segments of the figure exactly one time? It is understood that cuts should not occur at any of the twelve vertices.
$\bullet$  What happens if the figure is on a sphere?
[img]https://cdn.artofproblemsolving.com/attachments/a/d/b998f38d4bbe4121a41f1223402c394213944f.png[/img]


p7. Given any $\vartriangle ABC$, and any point $P$, let $X_1$, $X_2$, $X_3$ be the midpoints of the sides $AB$, $BC$, $AC$ respectively, and $Y_1$, $Y_2$, $Y_3$ midpoints of $PC$, $PA$, $PB$ respectively. Prove that the segments $X_1Y_1$, $X_2Y_2$, $X_3Y_3$ concur.
1. Denote \( \sqrt[3]{2+\frac{10}{9}\sqrt{3}} = a \) and \( \sqrt[3]{2-\frac{10}{9}\sqrt{3}} = b \). We need to find \( a + b \).

2. First, calculate \( a^3 \) and \( b^3 \):
   \[
   a^3 = 2 + \frac{10}{9}\sqrt{3}, \quad b^3 = 2 - \frac{10}{9}\sqrt{3}
   \]

3. Adding these, we get:
   \[
   a^3 + b^3 = \left(2 + \frac{10}{9}\sqrt{3}\right) + \left(2 - \frac{10}{9}\sqrt{3}\right) = 4
   \]

4. Next, calculate \( ab \):
   \[
   ab = \sqrt[3]{\left(2+\frac{10}{9}\sqrt{3}\right)\left(2-\frac{10}{9}\sqrt{3}\right)}
   \]

5. Simplify the product inside the cube root:
   \[
   \left(2+\frac{10}{9}\sqrt{3}\right)\left(2-\frac{10}{9}\sqrt{3}\right) = 4 - \left(\frac{10}{9}\sqrt{3}\right)^2 = 4 - \frac{100}{81} \cdot 3 = 4 - \frac{100}{27} = \frac{108}{27} - \frac{100}{27} = \frac{8}{27}
   \]

6. Therefore:
   \[
   ab = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}
   \]

7. We need to find \( a + b \). Using the identity for the sum of cubes:
   \[
   a^3 + b^3 = (a+b)(a^2 - ab + b^2)
   \]

8. Substitute the known values:
   \[
   4 = (a+b)\left((a+b)^2 - 3ab\right)
   \]

9. Let \( x = a + b \). Then:
   \[
   4 = x(x^2 - 3ab) = x(x^2 - 2)
   \]

10. Solve the cubic equation:
    \[
    x(x^2 - 2) = 4 \implies x^3 - 2x - 4 = 0
    \]

11. Check for possible rational roots using the Rational Root Theorem. Testing \( x = 2 \):
    \[
    2^3 - 2 \cdot 2 - 4 = 8 - 4 - 4 = 0
    \]

12. Since \( x = 2 \) is a root, factor the cubic polynomial:
    \[
    x^3 - 2x - 4 = (x - 2)(x^2 + 2x + 2)
    \]

13. The quadratic \( x^2 + 2x + 2 \) has no real roots (discriminant \( 2^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \)).

14. Therefore, the only real solution is:
    \[
    x = 2
    \]

Conclusion:
\[
\boxed{2}
\]