[u]level 1[/u]

[b]p1.[/b] Given a circle $\omega$ and two points $A$ and $M$ on $\omega$. Take a point $B$ on $\omega$ and construct a point $C$ such that the segment $AM$ is the median of $ABC$. Find $B$ so that the area of ​​triangle $ABC$ is maximum.


[b]p2.[/b] A triangle $ABC$ and a point $P$ are given. Let $S_1$ be the symmetric point of $P$ wrt $A$, $S_2$ the symmetric point of $S_1$ wrt $B$ and $S_3$ the symmetric point of $S_2$ wrt $C$. Find the locus of $S_3$ as $P$ moves.


[b]p3.[/b] Given a segment $AB$ and a point $X$ on $AB$. The line $ r$, perpendicular to $AB$, is drawn at $X$. Let $O$ be a point on the line $ r$ and $\omega$ the circle with center $O$ passing through $A$. The tangents to $\omega$  are traced through $B$, touching $\omega$  at $T_1$ and $T_2$. Find the locus of $T_1$ and $T_2$ as $O$ moves on $ r$.

[u]level 2[/u]


[b]p4. [/b]Given a circle $\omega$ and a chord $DA$ on the circle. Take another point $B$ on $\omega$ (different from $A$ and $D$) and draw the line $s$ passing through $A$ perpendicular on $DB$. The intersection point of line $s$ and line $DB$ is $C$. Find point $B$ so that the area of ​​triangle $ABC$ is maximum.


[b]p5.[/b] Given a triangle $ABC$, let $P$ be any point in the plane. The perpendicular bisectors of $AP$, $BP$ and $CP$ that intersect at three points, $A'$, $B'$ and $C'$, are drawn. Find a point $P$ so that the triangle $ABC$ and $A'B'C'$ are similar.


[b]p6.[/b] Let $C_1$ and $C_2$ be two intersecting circles at $A$ and $B$ where $AB$ is the diameter of $C_1$. Let $P$ be a point on $C_2$, in the interior of $C_1$. Construct two points $C$ and $D$ on $C_1$ such that $CD$ is perpendicular to $AB$ and $\angle CPD$ is right, using only the constructions of:
$\bullet$ a line given by two points
$\bullet$ intersection of $2$ objects 
$\bullet$ line perpendicular on a line.
Given the problem, we need to find the locus of points \( T_1 \) and \( T_2 \) as \( O \) moves along the line \( r \) perpendicular to \( AB \) at point \( X \).

1. **Define the coordinates:**
   - Let \( A = (0, 0) \) and \( B = (b, 0) \).
   - Let \( X = (t, 0) \) where \( t < b \).
   - Let \( O = (t, \lambda) \) be a point on the line \( r \) perpendicular to \( AB \) at \( X \).

2. **Equation of the circle:**
   - The circle \(\omega\) with center \( O \) and passing through \( A \) has the equation:
     \[
     (x - t)^2 + (y - \lambda)^2 = t^2 + \lambda^2
     \]
     Simplifying, we get:
     \[
     x^2 + y^2 - 2tx - 2\lambda y = 0
     \]

3. **Polar line of \( B \):**
   - The polar line of \( B \) with respect to the circle \(\omega\) is given by:
     \[
     (t - b)x + \lambda y + bt = 0
     \]
     This is derived from the general formula for the polar of a point \((x_1, y_1)\) with respect to the circle \(x^2 + y^2 + Dx + Ey + F = 0\):
     \[
     xx_1 + yy_1 + \frac{D}{2}(x + x_1) + \frac{E}{2}(y + y_1) + F = 0
     \]
     Substituting \( (x_1, y_1) = (b, 0) \) and the circle equation \( x^2 + y^2 - 2tx - 2\lambda y = 0 \), we get:
     \[
     (t - b)x + \lambda y + bt = 0
     \]

4. **Eliminate the parameter \(\lambda\):**
   - To find the locus of \( T_1 \) and \( T_2 \), we need to eliminate \(\lambda\) from the equation of the polar line.
   - The polar line equation can be rewritten as:
     \[
     \lambda y = - (t - b)x - bt
     \]
     Solving for \(\lambda\):
     \[
     \lambda = -\frac{(t - b)x + bt}{y}
     \]

5. **Locus of \( T_1 \) and \( T_2 \):**
   - Since \( T_1 \) and \( T_2 \) are the points where the tangents from \( B \) touch the circle, they lie on the circle with center \( B \) and radius equal to the distance from \( B \) to the circle \(\omega\).
   - The locus of these points is a circle centered at \( B \) with radius \( bt \):
     \[
     (x - b)^2 + y^2 = (bt)^2
     \]
     Simplifying, we get:
     \[
     x^2 + y^2 - 2bx + b^2 = b^2t^2
     \]
     \[
     x^2 + y^2 - 2bx + b^2(1 - t^2) = 0
     \]

The final answer is \( \boxed{ (x - b)^2 + y^2 = (bt)^2 } \).