Let $n$ be an integer greater than two, and let $A_1,A_2, \cdots , A_{2n}$ be pairwise distinct subsets of $\{1, 2, ,n\}$. Determine the maximum value of \[\sum_{i=1}^{2n} \dfrac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}\] Where $A_{2n+1}=A_1$ and $|X|$ denote the number of elements in $X.$

In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \perp BE$. Prove that $ED \perp FD.$ [asy][asy] defaultpen(fontsize(10)); size(6cm); pair A = (3,10), B = (0,0), C = (6,0), D = (3,0), E = intersectionpoints( Circle(B, 3), C--(C+100*dir(B--A)*dir(90)) )[1], M = midpoint(B--E), F = intersectionpoints(M--(M+50*dir(E--B)*dir(90)), circumcircle(A,B,D))[0]; dot(A^^B^^C^^D^^E^^M^^F); draw(B--C--A--B--E--D--F--M^^circumcircle(A,B,D)); pair point = extension(M,F,A,D); pair[] p={A,B,C,D,E,F,M}; string s = "A,B,C,D,E,F,M"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} d[4] = -50; string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy][/asy]

Prove that for every given positive integer $n$, there exists a prime $p$ and an integer $m$ such that $(a)$ $p \equiv 5 \pmod 6$ $(b)$ $p \nmid n$ $(c)$ $n \equiv m^3 \pmod p$

Let $x_1,x_2,\cdots,x_n$ be real numbers with $x_1^2+x_2^2+\cdots+x_n^2=1$. Prove that \[\sum_{k=1}^{n}\left(1-\dfrac{k}{{\displaystyle \sum_{i=1}^{n} ix_i^2}}\right)^2 \cdot \dfrac{x_k^2}{k} \leq \left(\dfrac{n-1}{n+1}\right)^2 \sum_{k=1}^{n} \dfrac{x_k^2}{k}\] Determine when does the equality hold?

Let $f(x)$ and $g(x)$ be strictly increasing linear functions from $\mathbb R $ to $\mathbb R $ such that $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove that for any real number $x$, $f(x)-g(x)$ is an integer.

In acute triangle $ABC$, $AB > AC$. Let $M$ be the midpoint of side $BC$. The exterior angle bisector of $\widehat{BAC}$ meet ray $BC$ at $P$. Point $K$ and $F$ lie on line $PA$ such that $MF \perp BC$ and $MK \perp PA$. Prove that $BC^2 = 4 PF \cdot AK$. [asy][asy] defaultpen(fontsize(10)); size(7cm); pair A = (4.6,4), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P); draw(K--M--F--P--B--A--C); pair point = I; pair[] p={A,B,C,M,P,F,K}; string s = "A,B,C,M,P,F,K"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy][/asy]

For given integer $n \geq 3$, set $S =\{p_1, p_2, \cdots, p_m\}$ consists of permutations $p_i$ of $(1, 2, \cdots, n)$. Suppose that among every three distinct numbers in $\{1, 2, \cdots, n\}$, one of these number does not lie in between the other two numbers in every permutations $p_i$ ($1 \leq i \leq m$). (For example, in the permutation $(1, 3, 2, 4)$, $3$ lies in between $1$ and $4$, and $4$ does not lie in between $1$ and $2$.) Determine the maximum value of $m$.

Determine the least odd number $a > 5$ satisfying the following conditions: There are positive integers $m_1,m_2, n_1, n_2$ such that $a=m_1^2+n_1^2$, $a^2=m_2^2+n_2^2$, and $m_1-n_1=m_2-n_2.$