[u]level 1[/u]


[b]p1.[/b] i. Given three not collinear points $ A$, $ B$ and $C$, find points $A'$, $B'$ and $ C'$ on $BC$, $AC$ and $AB$, respectively,  such that the triangles $ABC$ and $A'B'C'$ have the same centroid.


ii. Given three not collinear points $ A$, $ B$ and $C$, find points $A'$, $B'$ and $ C'$ on $BC$, $AC$ and $AB$, respectively, such that the triangles $ABC$ and $A'B'C'$ have the same circumcenter.


[b]p2.[/b] Given a parallelogram $ABCD$, find four points $A'$, $B'$, $C'$ and $D'$ all interior to the parallelogram such that the triples of points $AA'D'$, $BB'A'$, $CC'B'$ and $DD'C'$ are collinear  and the polygons $ADD'$, $DC'C$, $CB'B$, $AA'B$ and $A'B'C'D'$ have equal areas.


[b]p3. [/b]Construct a circle with center $O$. Let $AB$ and $SJ$ be two perpendicular diameters, in the minor arc $BS$ take a variable point $H$ . For each $H$ we determine $F$, the intersection point between $AH$ and $SB$ and $T$ the intersection point between $SA$ and $BH$.
i. Show that $TF$ and $AB$ are orthogonal.
ii. Find the locus of the circumcenter of $FSH$.

 
[u]level 2[/u]


[b]p4.[/b] Given a triangle $ABC$, find the locus of the points $ P$ inside the triangle such that $d (P,BC) = d (P,AC) + d (P,AB)$.

Note: $d (P, BC)$ is the distance from $P$ from line $BC$.


[b]p5.[/b] Construct a circle and take a fixed point $A$ on it. Draw a variable line $ r$ through $A$ that is secant to the circle at $A$ and $ B$. Construct an isosceles trapezoid $ABCD$ of bases $AB$ and $CD$ such that $AD = DC = 1/2 AB$.
i. Show that line $BC$ passes through a fixed point.
ii. Find the locus of $C$.
iii. A line $t$ perpendicular to $AC$ is drawn through $D$. Let $ P$ be the intersection point between $t$ and $AB$. Take $J$ on the ray $AB$ and $X$ on the ray $PD$ such that $AJ = PX$. Show that the perpendicular bisector of $JX$ passes through the incenter of triangle $APD$.


[b]p6.[/b] Construct a triangle $ABC$, and then find points $A'$, $B'$ and $ C'$ on $BC$, $AC$ and $AB$, respectively, such that $A'B'$ is perpendicular to $BC$, $B'C'$ is perpendicular to $AC$ and $C'A'$ is perpendicular to $AB$.
### Problem 3

Given a circle with center \( O \). Let \( AB \) and \( SJ \) be two perpendicular diameters. In the minor arc \( BS \), take a variable point \( H \). For each \( H \), we determine \( F \), the intersection point between \( AH \) and \( SB \), and \( T \), the intersection point between \( SA \) and \( BH \).

#### Part (i)

**Show that \( TF \) and \( AB \) are orthogonal.**

1. **Define Coordinates:**
   - Let the circle have radius \( r \).
   - Place the center \( O \) at the origin \((0,0)\).
   - Let \( A(-r,0) \), \( B(r,0) \), \( S(0,r) \), and \( J(0,-r) \).
   - Let \( H \) be a point on the minor arc \( BS \) with coordinates \( (r \cos \alpha, r \sin \alpha) \).

2. **Find Coordinates of \( F \):**
   - \( F \) is the intersection of \( AH \) and \( SB \).
   - The equation of line \( AH \) can be found using the point-slope form:
     \[
     y = \frac{\sin \alpha}{\cos \alpha + 1}(x + r)
     \]
   - The equation of line \( SB \) is:
     \[
     y = -\frac{r}{r}(x - r) = -x + r
     \]
   - Solving these equations simultaneously:
     \[
     \frac{\sin \alpha}{\cos \alpha + 1}(x + r) = -x + r
     \]
     \[
     x \left( \frac{\sin \alpha}{\cos \alpha + 1} + 1 \right) = r \left( 1 - \frac{\sin \alpha}{\cos \alpha + 1} \right)
     \]
     \[
     x = \frac{r(1 - \frac{\sin \alpha}{\cos \alpha + 1})}{1 + \frac{\sin \alpha}{\cos \alpha + 1}} = \frac{r(\cos \alpha + 1 - \sin \alpha)}{\cos \alpha + 1 + \sin \alpha}
     \]
   - Substituting \( x \) back into the equation of \( SB \):
     \[
     y = -\frac{r(\cos \alpha + 1 - \sin \alpha)}{\cos \alpha + 1 + \sin \alpha} + r = \frac{2r \sin \alpha}{\cos \alpha + 1 + \sin \alpha}
     \]
   - Thus, \( F \left( \frac{r(\cos \alpha + 1 - \sin \alpha)}{\cos \alpha + 1 + \sin \alpha}, \frac{2r \sin \alpha}{\cos \alpha + 1 + \sin \alpha} \right) \).

3. **Find Coordinates of \( T \):**
   - \( T \) is the intersection of \( SA \) and \( BH \).
   - The equation of line \( SA \) is:
     \[
     y = \frac{r}{r}(x + r) = x + r
     \]
   - The equation of line \( BH \) is:
     \[
     y = \frac{\sin \alpha}{\cos \alpha - 1}(x - r)
     \]
   - Solving these equations simultaneously:
     \[
     x + r = \frac{\sin \alpha}{\cos \alpha - 1}(x - r)
     \]
     \[
     x \left( 1 - \frac{\sin \alpha}{\cos \alpha - 1} \right) = r \left( \frac{\sin \alpha}{\cos \alpha - 1} + 1 \right)
     \]
     \[
     x = \frac{r(\cos \alpha - 1 + \sin \alpha)}{\sin \alpha + 1 - \cos \alpha}
     \]
   - Substituting \( x \) back into the equation of \( SA \):
     \[
     y = \frac{r(\cos \alpha - 1 + \sin \alpha)}{\sin \alpha + 1 - \cos \alpha} + r = \frac{2r \sin \alpha}{\sin \alpha + 1 - \cos \alpha}
     \]
   - Thus, \( T \left( \frac{r(\cos \alpha - 1 + \sin \alpha)}{\sin \alpha + 1 - \cos \alpha}, \frac{2r \sin \alpha}{\sin \alpha + 1 - \cos \alpha} \right) \).

4. **Orthogonality of \( TF \) and \( AB \):**
   - Since \( x_T = x_F \), the line \( TF \) is vertical.
   - The line \( AB \) is horizontal.
   - Therefore, \( TF \) and \( AB \) are orthogonal.

\(\blacksquare\)

#### Part (ii)

**Find the locus of the circumcenter of \( \triangle FSH \).**

1. **Circumcenter Coordinates:**
   - The circumcenter \( M \) of \( \triangle FSH \) lies on the perpendicular bisector of each side of the triangle.
   - Given the symmetry and the coordinates of \( F \) and \( T \), the \( x \)-coordinate of \( M \) is the same as \( x_F \) and \( x_T \):
     \[
     x_M = x_F = \frac{r(\cos \alpha + 1 - \sin \alpha)}{\cos \alpha + 1 + \sin \alpha}
     \]
   - The \( y \)-coordinate of \( M \) is the radius \( r \) because it lies on the circle:
     \[
     y_M = r
     \]

2. **Locus of \( M \):**
   - The locus of the circumcenter \( M \) is a horizontal line at \( y = r \) for \( 0 \leq x \leq r \).

\(\blacksquare\)

The final answer is the line \( \boxed{ y = r } \) for \( 0 \leq x \leq r \).