p1. The sides of a triangle $ABC$ have lengths $AB = 26$ cm, $BC = 17$ cm, and $CA = 19$ cm. The bisectors of the angles $B$ and $C$ intersect at the point $I$. By $I$ a parallel to $BC$ is drawn that intersects the sides $AB$ and $BC$ at points $M$ and $N$ respectively. Calculate the perimeter of the triangle $AMN$.


p2. Let $n$ be a positive integer. Determine the exact value of the following sum 
$$\frac{1}{1+\sqrt3}+\frac{1}{\sqrt3+\sqrt5}++\frac{1}{\sqrt5+\sqrt7}+...+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}$$


p3. Find the product of all positive integers that are less than $150$ and that have exactly $4$ positive divisors. Express your answer as a product of prime factors.


p4. There is a board of $ n \times n$, where n is a positive integer and the numbers from $ 1$ to $n$ are written in each of its rows in some order, so that the arrangement obtained is symmetric with respect to the diagonal. It is desired that the diagonal of the board does not contain all the numbers from $ 1$ to $n$. Show that it is always possible when $n$ is even and that it is impossible when $n$ is odd.


p5. There are six circles of radius $\frac12$, tangent to each other and to the sides of the rectangle, as shown in the figure. Also the centers $A$, $B$, $C$ and $D$ are collinear. Determine the lengths of the sides of the rectangle.
[img]https://cdn.artofproblemsolving.com/attachments/e/1/c8c23f796a73aaab25b78d8f0539750dc3f159.png[/img]
1. Given the triangle \(ABC\) with sides \(AB = 26\) cm, \(BC = 17\) cm, and \(CA = 19\) cm, we need to find the perimeter of the triangle \(AMN\) where \(M\) and \(N\) are points on \(AB\) and \(AC\) respectively such that \(MN \parallel BC\).

2. Since \(MN \parallel BC\), triangles \(BMN\) and \(BIC\) are similar. The angle bisectors of \(B\) and \(C\) intersect at \(I\), the incenter of \(\triangle ABC\).

3. Let \(BM = x\) and \(NC = y\). Since \(MN \parallel BC\), we have:
   \[
   \frac{BM}{AB} = \frac{IN}{BC} \quad \text{and} \quad \frac{NC}{CA} = \frac{IM}{BC}
   \]
   Given that \(AB = 26\) cm, \(BC = 17\) cm, and \(CA = 19\) cm, we can write:
   \[
   \frac{x}{26} = \frac{y}{19}
   \]
   This implies:
   \[
   x = \frac{26y}{19}
   \]

4. Since \(MN \parallel BC\), the length of \(MN\) is proportional to the length of \(BC\). Let \(k\) be the proportionality constant:
   \[
   MN = k \cdot BC = k \cdot 17
   \]

5. The perimeter of \(\triangle AMN\) is given by:
   \[
   AM + MN + AN
   \]
   Since \(AM = AB - BM = 26 - x\) and \(AN = AC - NC = 19 - y\), we have:
   \[
   AM + AN = (26 - x) + (19 - y)
   \]

6. Substituting \(x = \frac{26y}{19}\) into the equation:
   \[
   AM + AN = 26 - \frac{26y}{19} + 19 - y = 45 - \left(\frac{26y}{19} + y\right)
   \]

7. To find \(MN\), we use the fact that \(MN \parallel BC\) and the similarity of triangles:
   \[
   MN = k \cdot BC = k \cdot 17
   \]

8. Since \(MN\) is parallel to \(BC\), the length of \(MN\) is proportional to the length of \(BC\). Therefore, the perimeter of \(\triangle AMN\) is:
   \[
   AM + MN + AN = 45
   \]

The final answer is \(\boxed{45}\)