[u]level 2[/u]


[b]p5.[/b] Let $ABC$ be a triangle and $M$ be a variable point on $AB$. $N$ is the point on the prolongation of $AC$ such that $CN = BM$ and that it does not belong to the ray $CA$. The parallelogram $BMNP$ is constructed (in that order). Find the locus of $P$ as $M$ varies.


[b]p6.[/b] Let $ABCD$ be a quadrilateral. Let $C_1, C_2, C_3, C_4$ be the circles of diameters $AB, BC, CD$ and $DA$ respectively. Let $P, Q, R$ and $S$ be the points of intersection (which are not vertices of $ABCD$) of $C_1$ and $C_2$, $C_2$ and $C_3$, $C_3$ and $C_4$, $C_4$ and $C_1$ respectively. Show that the quadrilaterals $ABCD$ and $PQRS$ are similar.


[b]p7.[/b] $M, N$, and P are three collinear points, with $N$ between $M$ and $P$. Let $ r$ be the perpendicualr bisector of $NP$. A point $O$ is taken over $ r$. $\omega$ is the circle with center $O$ passing through $N$. The tangents through $M$ to $\omega$ intersect $\omega$ at $T$ and $T'$. Find the locus of the centroid of the triangle $MTT'$ a $O$ varies  over $r$.


[b]p8. [/b] $P, Q$ and $R$ are the centers of three circles that pass through the same point $O$. Let $A$, $B$ and $C$ be the points (other than $O$) of intersection of the circles. Prove that $A, B$ and $C$ are collinear if and only if $O, Q, P$ and $R$ are in the same circle.


[hide=PS.] p8 might have a typo, as [url=https://www.oma.org.ar/enunciados/2da2da.htm]here [/url] in my source it was incorrect, and I tried correcting it.[/hide]
Given the points \( M(-m,0), N(0,0), P(2p,0) \), we need to find the locus of the centroid of the triangle \( MTT' \) as \( O \) varies over the perpendicular bisector \( r \) of \( NP \).

1. **Determine the coordinates of \( O \):**
   Since \( O \) lies on the perpendicular bisector of \( NP \), its coordinates can be written as \( O(p, \lambda) \).

2. **Equation of the circle \( \omega \):**
   The circle \( \omega \) has center \( O(p, \lambda) \) and passes through \( N(0,0) \). The radius of the circle is the distance from \( O \) to \( N \), which is \( \sqrt{p^2 + \lambda^2} \). The equation of the circle is:
   \[
   (x - p)^2 + (y - \lambda)^2 = p^2 + \lambda^2
   \]
   Expanding and simplifying, we get:
   \[
   x^2 + y^2 - 2px - 2\lambda y = 0
   \]

3. **Equation of the polar line of \( M \):**
   The polar line of \( M(-m,0) \) with respect to the circle \( \omega \) is given by:
   \[
   (m+p)x + \lambda y - mp = 0
   \]

4. **Finding points \( T \) and \( T' \):**
   The points \( T \) and \( T' \) are the points where the tangents from \( M \) touch the circle \( \omega \). These points lie on the polar line of \( M \) and satisfy the circle's equation. Solving the system of equations:
   \[
   (m+p)x + \lambda y - mp = 0
   \]
   \[
   x^2 + y^2 - 2px - 2\lambda y = 0
   \]
   yields the coordinates of \( T \) and \( T' \).

5. **Centroid of \( \triangle MTT' \):**
   The centroid \( G \) of \( \triangle MTT' \) is given by the average of the coordinates of \( M, T, \) and \( T' \):
   \[
   G = \left( \frac{-m + x_T + x_{T'}}{3}, \frac{0 + y_T + y_{T'}}{3} \right)
   \]

6. **Locus of the centroid:**
   To find the locus of \( G \), we need to express \( G \) in terms of \( m, p, \) and \( \lambda \). After some algebraic manipulation, we find that the locus of the centroids of \( \triangle MTT' \) is a circle given by:
   \[
   3(m+p)(x^2 + y^2) + 2m(2m + p)x + m^2(m - p) = 0
   \]

The final answer is \(\boxed{3(m+p)(x^2 + y^2) + 2m(2m + p)x + m^2(m - p) = 0}\)