Problem 1: There are 5 points on plane. Prove that you can chose some of them and shift them such that distances between shifted points won't change and as a result there will be symetric by some line set of 5 points. Spoiler: Xem lời giải Without loss of generality, let no three points be collinear. The other cases are just limit cases of the following cases. Let the points be $A, B, C, D, E$, with the distance between $D, E$ being the maximum possible between any two points in the configuration. The triangle $ABC$ has a greatest angle, let's say it's $A$. We make two cases: 1. $ABC$ is not isosceles. wlog $AB<AC$. Then shift the point $E$ to a location such that $AEBC$ is an isosceles trapezoid with $AE || BC$. Now note that $AE <BC < DE$ because $B, C$ are necessarily acute. So we can put $D$ on the perpendicular bisector of $BC$, which will be the line of symmetry of the configuration. 2. $ABC$ is isosceles at $A$. In that case, shift $E$ to the location such that $EBAC$ is an isosceles trapezoid, with $AB || EC$. Then note that the distance from $E$ to the perpendicular bisector of $AB$ is $0.5 EC < 0.5(AE + AC) = 0.5(BC + AC) \le 0.5(DE + DE) = DE$, so we can place $D$ on the perpendicular bisector of $AB$ and we will be done. Problem 2: Find minimal natural $n$ for which there exist integers $a_1, a_2,\ldots, a_n$ such that quadratic trinom $$x^2-2(a_1+a_2+\cdots+a_n)^2x+(a_1^4+a_2^4+\cdots+a_n^4+1)$$ has at least one integral root. Spoiler: Xem lời giải We need that the discriminant of the quadratic should be a perfect square, i.e., $(a_1+a_2+\cdots+a_n)^4 = x^2 + (a_1^4+a_2^4+\cdots+a_n^4+1)$. Note that $x$ is odd, looking at the parities of all the $a_i$s. Consider the equation modulo $8$. Then the LHS is $0$ or $1$ mod $8$, and RHS is in the range $2$ to $r+2$ where $r$ is the number of odd elements in the set of all $a_i$s. Clearly, we must have $r\ge 6$ and thus $n\ge 6$. For the construction for $n=6$, let the first 4 numbers be 1 and the last two be -1. This concludes the proof. Problem 3: Circle $\Omega$ with center $O$ is the circumcircle of an acute triangle $\triangle ABC$ with $AB<BC$ and orthocenter $H$. On the line $BO$ there is point $D$ such that $O$ is between $B$ and $D$ and $\angle ADC= \angle ABC$ . The semi-line starting at $H$ and parallel to $BO$ wich intersects segment $AC$ , intersects $\Omega$ at $E$. Prove that $BH=DE$. Spoiler: Xem lời giải Let $G,F$ be the feet of the perpendicular from $A$ and $B$ respectively. Let $O'$ be the reflection of $O$ with respect to $AC$. $\angle ADC=\angle B$ and $\angle AHC=\angle 180-\angle B \implies H,D,A,C$ cyclic. Now $O'A=O'D$ and $\angle AO'C=2\angle ADC$ which means $O'$ is the circumcircle of $\triangle ADC$. It is well known that $BH=OO'$ which implies $BHOO'$ is $\parallel gm \implies O'$ lies on $HE$. $BO=OE=O'H=O'D$ so $OD$ and $O'E$ have the same perpendicular bisector and this $\implies$ that $DE=O'O=BH$. Problem 4: 10000 children came to a camp; every of them is friend of exactly eleven other children in the camp (friendship is mutual). Every child wears T-shirt of one of seven rainbow's colours; every two friends' colours are different. Leaders demanded that some children (at least one) wear T-shirts of other colours (from those seven colours). Survey pointed that 100 children didn't want to change their colours [translator's comment: it means that any of these 100 children (and only them) can't change his (her) colour such that still every two friends' colours will be different]. Prove that some of other children can change colours of their T-shirts such that as before every two friends' colours will be different. Spoiler: Xem lời giải Let $V_1, V_2, ... , V_7$ be the set of children with each colors. Define $G_{i,j}$ as the graph with vertex $V_i \cup V_j$ with two vertex connected with edge if and only if there are friend. Let $c_{i,j}$ be the number of the connected components of $G_{i,j}$. If $c_{i,j}$ is greater than $100$ for some $i<j$, one of the component does not contains the '100 children'. Thus, we can let children with cloth $i$ to $j$, and $j$ to $i$ and still friends wear different clothes. Assume $c_{i,j} \le 100$ for every $i<j$. Then the number of edge of $G_{i,j}$ is at least $|V_i |+|V_j | - c_{i,j}$. If we sum over every $i<j$, total number of edge is at least $6 \times (|V_1 |+...+|V_7 |)- \sum c_{i,j} \ge 60000-2100>55000$. However, the total number of edges are $55000$ since each child has 11 friends. Therefore, $c_{i,j}>100$ for some $i<j$ and we can always find some children not containing the "100 children" who can change their clothes preserving the property.
