p1. Let $X = \{1, 2, 3, 4, 5\}$. Let $F = \{A_1,A_2,A_3, ..., A_m\}$, with $A_i\subseteq X$ and $2$ members in $A_i$ , for $i = 1,2,...,m$. Determine the minimum $m$ so that for any $B\subseteq X$, with $ B$ having $3$ members, there is a member of $F$ contained in $ B$. Prove your answer. p2. Find all triples of real numbers $(x, y, z)$ that satisfy the system of equations: $(x + 1)^2 = x + y + 2$ $(y + 1)^2 = y + z + 2$ $(z + 1)^2 = z + x + 2$ p3. Given the isosceles triangle $ABC$, where $AB = AC$. Let $D$ be a point in the segment $BC$ so that $BD = 2DC$. Suppose also that point $P$ lies on the segment $AD$ such that: $\angle BAC = \angle BP D$. Prove that $\angle BAC = 2\angle DP C$. p4. Suppose $p_1, p_2,.., p_n$ arithmetic sequence with difference $b > 0$ and $p_i$ prime for each $i = 1, 2, ..., n$. a) If $p_1 > n$, prove that every prime number $p$ with $p\le n$, then $p$ divides $b$ . b) Give an example of an arithmetic sequence $p_1, p_2,.., p_{10}$, with positive difference and $p_i$ prime for $i = 1, 2, ..., 10$. p5. Given a set consisting of $22$ integers, $A = \{\pm a_1, \pm a_2, ..., \pm a_{11}\}$. Prove that there is a subset $S$ of $A$ which simultaneously has the following properties: a) For each $i = 1, 2, ..., 11$ at most only one of the $a_i$ or $-a_i$ is a member of $S$ b) The sum of all the numbers in $S$ is divided by $2015$.
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Tags: algebra, geometry, combinatorics, number theory, system of equations
25.05.2023 20:10
P1 Notice that if we fix $A_i$ then $A_i$ can be contained at most in 3 set of $B$ but we know that $|(B)|=\binom{5}{3}=10$ so it must be $3i\geq 10 \iff i\geq 4$ Therefore the minimum of $m$ is 4 One configuration that satisfy is $A_1=\{1,2\},A_2=\{3,4\},A_3=\{3,5\},A_4=\{4,5\},$
25.05.2023 20:44
P2 By summing all the equation we got $x^2+y^2+z^2=3$, and its clear that $$-\sqrt{3}\leq x,y,z\leq \sqrt{3}$$Notice that we can rewrite the first equation as $x(x+1)=y+1$ then multiply all the equation we got $$xyz(x+1)(y+1)(z+1)=(x+1)(y+1)(z+1)$$If one $x,y$ or $z$ is $-1$, then the rest are also gonna be $-1$, we got the first triple are $(-1,-1,-1)$ Now assume $x,y,z\in \mathbb{R}^+$ then by AM-GM inequality $$\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{xyz} = 1$$equality holds if and only if $x=y=z \Rightarrow (1,1,1)$ All triples of real numbers $(x, y, z)$ that satisfy the system of equations are (-1,-1,-1) and (1,1,1)
26.05.2023 09:12
Extend AD with the segment DE,such that m(PBE)=m(ABC), then m(AEB)=m(ACB)=m(ABC),then the points A,B,E,C are conciclic. We havecm(AEC)=m(ABC), and DE is the bisector of angle BEC, so DC/BD=CE/BE=> CE=BE/2. PE=PB. Apply the sine rule in the triangle PEB: PB/sinB=EB/sinA=>BP/sinB=BE/sin2B=>BP=EB/2cosB=CE/cosB. The sine rule in the triangle PCE leads to: BP=CE*sin(PCE)/sin(CPD). The last two equalities leads to: din(PCE)=1 and cos(ABC)=din(CPD), then PCE is right angle ,and m(CPD)=90°-m(B)=1/2*m(A)
22.10.2024 16:52
P4a. FTSOC there's a prime q≤n that doesn't divide b, then gcd(b, q) =1. Notice that the sets {p_1, p_1+b, p_1+2b,..., p_1+(q-1)b} and {0, 1,...,q-1} are equivalent modulo q. Then there's i, a member of {1,2,...,q} such that p_i congruent zero modulo q, which means q | p_i. But p_i>p_1>n≥q, hence p_i isn't a prime number (contradiction)