[b]p1.[/b] Somebody placed $2015$ points in the plane in such a way that any triangle with vertices at those points has the area less or equal than $1$. Prove that all points can be placed inside of a triangle which has area equal to $4$.


[b]p2.[/b] Is the following statement true? [i]For any point inside of a convex quadrilateral the sum of distances from that point to the vertices of the quadrilateral is smaller than the perimeter of the quadrilateral.[/i]


[b]p3.[/b] Prove that the following inequality is satisfied for any $a \ne 0$:
$$1 +\frac{1}{a^2} \ge \frac{2}{a} - \frac{11}{25a^2} + \frac{2}{5a}.$$


[b]p4.[/b] A sequence of $128$ numbers $x_1, x_2, ..., x_{128}$ is given, such that each number is either $+1$ or $-1$. We construct a new sequence of $128$ numbers consisting of products: $$x_1x_2,x_2x_3,x_3x_4,...,x_{127}x_{128},x_{128}x_1.$$
Then we construct a new sequence from this one, using the same rule. Prove that if we repeat the construction sufficiently many times we arrive at the sequence consisting only of $+1$’s.

PS. You should use hide for answers.
**p1.** Somebody placed  $2015$  points in the plane in such a way that any triangle with vertices at those points has the area less or equal than  $1$ . Prove that all points can be placed inside of a triangle which has area equal to  $4$ .

1. Consider the convex hull of the 2015 points. The convex hull is the smallest convex polygon that contains all the points.
2. Suppose the convex hull has $n$ vertices. If $n \leq 3$, then the points are already inside a triangle, and we can scale this triangle to have an area of at most $4$.
3. If $n > 3$, consider any triangle formed by three consecutive vertices of the convex hull. By the given condition, the area of this triangle is at most $1$.
4. By the properties of convex polygons, the sum of the areas of the triangles formed by consecutive vertices is equal to the area of the convex hull.
5. Since each triangle has an area of at most $1$, the total area of the convex hull is at most $n-2$.
6. For $n \leq 6$, the area of the convex hull is at most $4$.
7. Therefore, all points can be placed inside a triangle with an area of $4$.

$\blacksquare$

**p2.** Is the following statement true? *For any point inside of a convex quadrilateral the sum of distances from that point to the vertices of the quadrilateral is smaller than the perimeter of the quadrilateral.*

1. Consider a convex quadrilateral $ABCD$ and a point $P$ inside it.
2. Let the distances from $P$ to the vertices $A, B, C, D$ be $PA, PB, PC, PD$ respectively.
3. By the triangle inequality, $PA + PB > AB$, $PB + PC > BC$, $PC + PD > CD$, and $PD + PA > DA$.
4. Adding these inequalities, we get $(PA + PB) + (PB + PC) + (PC + PD) + (PD + PA) > AB + BC + CD + DA$.
5. Simplifying, we get $2(PA + PB + PC + PD) > AB + BC + CD + DA$.
6. Therefore, $PA + PB + PC + PD < \frac{1}{2}(AB + BC + CD + DA)$.
7. Hence, the sum of distances from $P$ to the vertices is smaller than the perimeter of the quadrilateral.

$\boxed{\text{True}}$

**p3.** Prove that the following inequality is satisfied for any  $a \ne 0$ :  $$ 1 +\frac{1}{a^2} \ge \frac{2}{a} - \frac{11}{25a^2} + \frac{2}{5a}. $$

1. Start with the given inequality:
   \[
   1 + \frac{1}{a^2} \ge \frac{2}{a} - \frac{11}{25a^2} + \frac{2}{5a}
   \]
2. Combine like terms:
   \[
   1 + \frac{1}{a^2} \ge \frac{2}{a} + \frac{2}{5a} - \frac{11}{25a^2}
   \]
3. Multiply through by $25a^2$ to clear the denominators:
   \[
   25a^2 + 25 \ge 50a - 11 + 10a
   \]
4. Simplify the inequality:
   \[
   25a^2 + 25 \ge 60a - 11
   \]
5. Rearrange terms to form a quadratic inequality:
   \[
   25a^2 - 60a + 36 \ge 0
   \]
6. Factor the quadratic expression:
   \[
   (5a - 6)^2 \ge 0
   \]
7. Since the square of any real number is non-negative, the inequality holds for all $a \neq 0$.

$\blacksquare$

**p4.** A sequence of  $128$  numbers  $x_1, x_2, ..., x_{128}$  is given, such that each number is either  $+1$  or  $-1$ . We construct a new sequence of  $128$  numbers consisting of products:   $$ x_1x_2,x_2x_3,x_3x_4,...,x_{127}x_{128},x_{128}x_1. $$  Then we construct a new sequence from this one, using the same rule. Prove that if we repeat the construction sufficiently many times we arrive at the sequence consisting only of  $+1$ ’s.

1. Consider the initial sequence $x_1, x_2, \ldots, x_{128}$ where each $x_i \in \{+1, -1\}$.
2. Construct the new sequence $y_i = x_i x_{i+1}$ for $i = 1, 2, \ldots, 127$ and $y_{128} = x_{128} x_1$.
3. Note that $y_i \in \{+1, -1\}$ and $y_i = +1$ if $x_i = x_{i+1}$ and $y_i = -1$ if $x_i \neq x_{i+1}$.
4. Repeat the construction to form the next sequence $z_i = y_i y_{i+1}$ for $i = 1, 2, \ldots, 127$ and $z_{128} = y_{128} y_1$.
5. Observe that each step reduces the number of $-1$'s in the sequence.
6. Eventually, after sufficiently many iterations, all elements of the sequence will be $+1$.

$\blacksquare$