[u]level 1[/u]


[b]p1.[/b] Given a point $P$ let $C_1$, $C_2$ and $C_3$ be circles of equal radius that pass through $P$. Prove that the centers of the three circles are in a circle of the same radius.


[b]p2. [/b]Given the triangle $ABC$. Let P be any point on the side $BC$. Let $D$ and $E$ belong to $AB$ and $AC$ respectively such that ADPE is a parallelogram. Prove that the semiperimeter of $ADPE$ is between the measurements of $AB$ and $AC$.


[b]p3. [/b]Let $ABC$ be a triangle with sides $AB = 10$, $BC = 6$ such that the area of $​​ABC$ is equal to $18$. Let $M$ be the midpoint of $AB$ and $H_1$ in $BC$ such that $MH_1$ is perpendicular to $BC$. Let $H_2$ in $AB$ such that $H_1H_2$ is perpendicular to $AB$. Calculate $H_1H_2$.


[b]p4.[/b] a) In a convex quadrilateral $ABCD$ it is known that the relation between the areas of the triangles $ABC$, $BCD$, $CDA$ and $DAB$ is $1: 1: 1: x$. Find all possible values ​​of $x$. Build an example for each value of $x.$

   b) Same as a) but for  $1: 2: 3: x$.


[b]p5.[/b] Given a point $P$ let $C_1$, $C_2$ and $C_3$ be circles of equal radius that pass through $P$. Prove that the intersections  (different from $P$) of the three circles in pairs are in a circle of the same radius.


[b]p6.[/b] Given a triangle $ABC$, find the locus of the centers of the equilateral triangles $DEF$ with $D, E$ and $F$ belonging to the lines $AB$, $BC$ and $CA$ respectively.
Given \(\triangle ABC\) with vertices \(A(0,a)\), \(B(-b,0)\), and \(C(c,0)\).

1. **Choose Point \(D\) on Line \(AB\):**
   Let \(D(\lambda, \frac{a(\lambda + b)}{b})\) be a point on \(AB\).

2. **Choose Point \(E\) on Line \(BC\):**
   Let \(E(\mu, 0)\) be a point on \(BC\).

3. **Condition for Equilateral Triangle \(DEF\):**
   For \(\triangle DEF\) to be equilateral, the slope \(m\) of line \(DE\) must satisfy the condition for equilateral triangles. The slope \(m\) is given by:
   \[
   m = -\frac{\sqrt{3}a^2(b + \lambda) + a[c\lambda - b(c - \lambda)] - \sqrt{3}bc\lambda}{b(a + \sqrt{3}c)}
   \]

4. **Equation of the Locus:**
   The locus of the centers of these equilateral triangles is derived from the geometric properties and symmetry of the equilateral triangle. The equation of the locus is:
   \[
   \sqrt{3}a(b - c)x + \sqrt{3}(3bc - a^2)y + a[a(b + c) - 2\sqrt{3}bc] = 0
   \]

### Final Answer:
The final answer is:
\[
\boxed{\sqrt{3}a(b - c)x + \sqrt{3}(3bc - a^2)y + a[a(b + c) - 2\sqrt{3}bc] = 0}
\]

### Self Correction:
The problem is solved completely.