Let $n\ge 2$ be a positive integer. Find the number of functions $f:\{1,2,\ldots ,n\}\rightarrow\{1,2,3,4,5 \}$ which have the following property: $|f(k+1)-f(k)|\ge 3$, for any $k=1,2,\ldots n-1$. Vasile Pop
2000 Romania Team Selection Test
Day 1
Let $n\ge 1$ be a positive integer and $x_1,x_2\ldots ,x_n$ be real numbers such that $|x_{k+1}-x_k|\le 1$ for $k=1,2,\ldots ,n-1$. Prove that \[\sum_{k=1}^n|x_k|-\left|\sum_{k=1}^nx_k\right|\le\frac{n^2-1}{4}\] Gh. Eckstein
Prove that for any positive integers $n$ and $k$ there exist positive integers $a>b>c>d>e>k$ such that \[n=\binom{a}{3}\pm\binom{b}{3}\pm\binom{c}{3}\pm\binom{d}{3}\pm\binom{e}{3}\] Radu Ignat
Let $P_1P_2\ldots P_n$ be a convex polygon in the plane. We assume that for any arbitrary choice of vertices $P_i,P_j$ there exists a vertex in the polygon $P_k$ distinct from $P_i,P_j$ such that $\angle P_iP_kP_j=60^{\circ}$. Show that $n=3$. Radu Todor
Day 2
Prove that the equation $x^3+y^3+z^3=t^4$ has infinitely many solutions in positive integers such that $\gcd(x,y,z,t)=1$. Mihai Pitticari & Sorin RĒdulescu
Let ABC be a triangle and $M$ be an interior point. Prove that \[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.\]
Click for solution This is one problem I had in one TST in 2000. Solution: Consider the following lemma: let $ABC$ be a triangle and let $M$ be a point on the side $AC$ such that $\angle BMC \geq 90^\circ$. Then the following inequality takes place: \[ BM+MC < BA + BC \] If $\angle BAC \leq 90^\circ$ let us consider the point $A'$ on the segment $AM$ such that $BA' \perp AC$. We have $BM \leq A'B+ A'M $ and thus $BM + MC \leq A'B+A'C < BA' +BC $, because $ BC$ is the largest side the triangle $A'BC$. But $BA' \leq BA$, so we are done. If $\angle BAC > 90^\circ$ we have $BM \leq AB+ AM $ and thus $BM + MC \leq AB+AC < BA +BC $, because $ BC$ is the largest side the triangle $ABC$. Back to the problem, we notice that at least two of the angles $\angle AMB$, $\angle BMC$, $\angle CMA$ are larger than $90^\circ$. Indeed, if two of them are less than $90^\circ$, since their sum is $360^\circ$, one of them would be larger than $180^\circ$ which is an obvious contradiction. So, let us suppose WLOG that $\angle AMB \geq 90^\circ$ and $\angle AMC \geq 90^\circ$. Also consider $M' = AM \cap BC$. We apply the lemma for the triangles $ABM'$ and $ACM'$ and we obtain \[ AM + MB < AB + BM' \eqno (1) \] \[ AM + MC < AC + CM' \eqno (2) \] From (1) and (2), by addition, we obtain \[ AM+BM+ CM + AM < AB + BC + CA \] which solves our problem. Also there is the official solution, much complicated which I will post later.
Determine all pairs $(m,n)$ of positive integers such that a $m\times n$ rectangle can be tiled with L-trominoes.
Day 3
Let $a>1$ be an odd positive integer. Find the least positive integer $n$ such that $2^{2000}$ is a divisor of $a^n-1$. Mircea Becheanu
Let $ABC$ be an acute-angled triangle and $M$ be the midpoint of the side $BC$. Let $N$ be a point in the interior of the triangle $ABC$ such that $\angle NBA=\angle BAM$ and $\angle NCA=\angle CAM$. Prove that $\angle NAB=\angle MAC$. Gabriel Nagy
Let $S$ be the set of interior points of a sphere and $C$ be the set of interior points of a circle. Find, with proof, whether there exists a function $f:S\rightarrow C$ such that $d(A,B)\le d(f(A),f(B))$ for any two points $A,B\in S$ where $d(X,Y)$ denotes the distance between the points $X$ and $Y$. Marius Cavachi
Day 4
Let $P_1$ be a regular $n$-gon, where $n\in\mathbb{N}$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. Continuing analogously, we obtain regular $n$-gons $P_3,P_4,\ldots ,P_m$. For $m\ge n^2-n+1$, find the maximum number $k$ such that for any colouring of vertices of $P_1,\ldots ,P_m$ in $k$ colours there exists an isosceles trapezium $ABCD$ whose vertices $A,B,C,D$ have the same colour. Radu Ignat
Let $P,Q$ be two monic polynomials with complex coefficients such that $P(P(x))=Q(Q(x))$ for all $x$. Prove that $P=Q$. Marius Cavachi
Prove that every positive rational number can be represented in the form $\dfrac{a^{3}+b^{3}}{c^{3}+d^{3}}$ where a,b,c,d are positive integers.