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.

A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point inside and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators.

Click for solution Cute problem . First of all, let's observe that for any $i,j\in\overline{1,5}$, there is a separator through $A_i,A_j$. This is because there are three circles through $A_i,A_j$, and there is at most one circle containing the other two points and at most one leaving the other two points outside, so there is at least one which is a separator. This, in turn, means that there are at least $4$ separators because if there are $\le$, then these have at most $3\cdot 3=9$ chords with endpoints $A_i$, but there are $10$ segments $A_iA_j$, a contradiction. Another observation: through a point (let's call it $A_1$) there are $2$ or $3$ separators: after the inversion of pole $A_1$ the quadrilateral $\mathcal Q=A_2'A_3'A_4'A_5'$ (where, in general, $X'$ is the image of $X$) is either convex or concave. A separator is the preimage of a line $A_l'A_k'$ which separates the other two images of points, and there are two such lines (the diagonals) if $\mathcal Q$ is convex, and three (concurrent) if $\mathcal Q$ is concave. From this we have also obtained the fact that if through $A_1$ there are $3$ separators, then there is an $i\ne 1$ s.t. all the three separators also pass through $A_i$. Assume now that through each point there are $2$ separators. It's easy to get a contradiction based only on the fact that we can't find at least $4$ subsets of $\{A_i\}$ s.t. each point belongs to exactly two sets. This means that there is at least a pair (call it $(A_1,A_2)$) of points with $3$ separators each. Then $A_1A_2A_3,A_1A_2A_4,A_1A_2A_5$ are separators, and it's pretty clear that $A_3A_4A_5$ is also a separator. Any attempt to add another separator will cause one of $A_1,A_2$ to have more than $3$ separators passing through it, and this is a contradiction. Sorry if it appears long and boring..

A set $ S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.

For a triangle $T = ABC$ we take the point $X$ on the side $(AB)$ such that $AX/AB=4/5$, the point $Y$ on the segment $(CX)$ such that $CY = 2YX$ and, if possible, the point $Z$ on the ray ($CA$ such that $\widehat{CXZ} = 180 - \widehat{ABC}$. We denote by $\Sigma$ the set of all triangles $T$ for which $\widehat{XYZ} = 45$. Prove that all triangles from $\Sigma$ are similar and find the measure of their smallest angle.

Let $ABC$ be a triangle, $\Omega$ its incircle and $\Omega_{a}, \Omega_{b}, \Omega_{c}$ three circles orthogonal to $\Omega$ passing through $(B,C),(A,C)$ and $(A,B)$ respectively. The circles $\Omega_{a}$ and $\Omega_{b}$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of the circumcircle of $A'B'C'$ is half the radius of $\Omega$.

Click for solution I remember posting this a long time ago . Consider the inversion wrt the incircle. It invariates the incircle and $\Omega_a,_b,_c$, so it turns $X$ into $X'$, for any $X\in\{A,B,C\}$. At the same time, if the incircle touches $BC,CA,AB$ in $D,E,F$, then the inversion turns $A$ into the midpoint of $EF$ and the like, so $A',B',C'$ are the midpoints of $EF,FD,DE$ respectively, meaning that the circumcircle of $A'B'C'$ is the nine-point circle of $D EF$, thus having half its circumradius, which is the inradius of $ABC$.

Two circles $\Omega_{1}$ and $\Omega_{2}$ touch internally the circle $\Omega$ in M and N and the center of $\Omega_{2}$ is on $\Omega_{1}$. The common chord of the circles $\Omega_{1}$ and $\Omega_{2}$ intersects $\Omega$ in $A$ and $B$. $MA$ and $MB$ intersects $\Omega_{1}$ in $C$ and $D$. Prove that $\Omega_{2}$ is tangent to $CD$.

Click for solution {U,V} = o1^o2; E=AN^o2; F=BN^o2;p=DF^CE ; X = centre(o1); Y = centre(o2). AC*AM = AV*AU = AE*AN ==> MNEC cyclic; <ACE = <ANM = <ABM = <CDM ==> CE tangent to o1; <AEC = <AMN = <ABN = <EFN ==> EC tangent to o2; G= o2^XY, H = foot(Y) on CX; YH//CE ==> XH = r1 – r2 = XG ==> CXG == YXH ==> <CGX = 90° XY symmetry axes of CD and EF ==> <DGX = 90°.

The point $M$ is inside the convex quadrilateral $ABCD$, such that \[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \] Prove that $AB \cdot CM = BC \cdot MD$ and $BM \cdot AD = MA \cdot CD.$

Click for solution Let A' = DM^BC, from <CMD = <MCB+<MAB it follows that ABA'M is ciclyc. Let C' = BM^AD, from <AMB = <MAD+<MCD it follows that DMCC' is ciclyc. As supplementary of the same angle, it holds that <DMA = <CBA. From collinearity problem we have that <ACB = <DAM. Then DM/AM = AB/BC and finally DM/CM=AB/BC. As both supplementary of the equal angles <C'MC = <C'DC, it holds <BMC = <ADC. Furthermore, see again collinearity problem, <MCB = <MCA+<ACB = <MAC+<DAM = <DAC. Then BM/MC=CD/AD or BM/MA=CD/AD.

Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.

Click for solution Let $M,N$ be the midpoints of the arcs $CA,CB$ respectively. We have $MO_{1}=MC,\ NO_{2}=NC$, so $\frac{MO_{1}}{NO_{2}}=\frac{MC}{NC}$, which is constant (doesn't depend on $X$). Let $P$ be a point on the arc $AB$ s.t. $\frac{MP}{NP}=\frac{MC}{NC}$. Since we alos have $\angle PMO_{1}=\angle PNO_{2}$, it means that $PMO_{1},PNO_{2}$ are similar, so $\angle O_{1}PO_{2}=\angle MPN=\angle MXN=\angle O_{1}XO_{2}$, so $P$ always belongs to $\Omega$. [Moderator edit: This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=555 .]

Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p-1}$ is a divisor of $ (p-1)^{x}+1$.

Click for solution First of all, if $p=2$, then $x$ is $1$ or $2$. Assume now $p\ge 3$. Clearly, $x$ must be odd (let's say that $x=2k+1$), and let's assume $x>1$. Let $q$ be the smallest prime divisor of $x$. We have $q|((p-1)^2x-1,(p-1)^{q-1}-1)=(p-1)^2-1=p(p-2)$. Assume $q\ne p$. We find that $q|p-2$. At the same time, we have $q|(p-1)^x+1=(p-1)^{2k+1}+1=[(p-1)^{2k}-1](p-1)+p$, so $q|p$, which contradicts the fact that $q|p-2$ and $q$ is odd. This means that $q=p$, so $x\in\{p,2p\}$, and since it's odd, we have $x=p$. We now have to find those odd primes $p$ s.t. $p^{p-1}|(p-1)^p+1$. After expanding the binomial on the right, we find that the expression on the right is $p^2+kp^3$, so $p-1\le 2$, meaning that the only odd prime satisfying this is $p=3$ (it it does satisfy the relation, since $3^2|2^3+1$). The solutions are then $(1,p)$ for any prime $p$, $(2,2)$ and $(3,3)$.

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.

Prove that there exists two strictly increasing sequences $(a_{n})$ and $(b_{n})$ such that $a_{n}(a_{n}+1)$ divides $b^{2}_{n}+1$ for every natural n.

Click for solution It suffices to find $c_n,d_n$ s.t. $a_n|c_n^2+1,\ a_n+1|d_n^2+1$, because then, since $(a_n,a_n+1)=1$, we can find $b_n\equiv c_n\pmod {a_n},\ b_n\equiv d_n\pmod{a_n+1}$. Let's show that $a_n=5^{2n}$ works. We have $a_n+1=5^{2n}+1=(5^n)^2+1$, so all we need to show is that there is some $c_n$ s.t. $5^{2n}|c_n^2+1$. We can show something more general: if $p$ is a prime of the form $4k+1$, then there is a $t_n$ s.t. $p^n|t_n^2+1$. We construct $t_n$ inductively. $t_1$ clearly exists, and assuming that $p^{n-1}|t_{n-1}^2+1$ we have to find $k\in\overline{0,p-1}$ s.t. $p^n|(t_{n-1}+kp^{n-1})^2+1$. This is reduced to $p|\frac{t_{n-1}^2+1}{p^{n-1}}+2k$, and it's clear that we can find such a $k$.

Denote by S the set of all primes such the decimal representation of $\frac{1}{p}$ has the fundamental period divisible by 3. For every $p \in S$ such that $\frac{1}{p}$ has the fundamental period $3r$ one may write \[\frac{1}{p}=0,a_{1}a_{2}\ldots a_{3r}a_{1}a_{2} \ldots a_{3r} \ldots , \] where $r=r(p)$; for every $p \in S$ and every integer $k \geq 1$ define $f(k,p)$ by \[ f(k,p)= a_{k}+a_{k+r(p)}+a_{k+2.r(p)}\] a) Prove that $S$ is infinite. b) Find the highest value of $f(k,p)$ for $k \geq 1$ and $p \in S$

Let $n,k$ be positive integers such that n is not divisible by 3 and $k \geq n$. Prove that there exists a positive integer $m$ which is divisible by $n$ and the sum of its digits in decimal representation is $k$.

Prove that for every real number $M$ there exists an infinite arithmetic progression such that: - each term is a positive integer and the common difference is not divisible by 10 - the sum of the digits of each term (in decimal representation) exceeds $M$.

Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality \[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C \left(\sum_{i}x_{i} \right)^4\] holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.

The numbers from 1 to $n^2$ are randomly arranged in the cells of a $n \times n$ square ($n \geq 2$). For any pair of numbers situated on the same row or on the same column the ratio of the greater number to the smaller number is calculated. Let us call the characteristic of the arrangement the smallest of these $n^2\left(n-1\right)$ fractions. What is the highest possible value of the characteristic ?

A game is played by $n$ girls ($n \geq 2$), everybody having a ball. Each of the $\binom{n}{2}$ pairs of players, is an arbitrary order, exchange the balls they have at the moment. The game is called nice nice if at the end nobody has her own ball and it is called tiresome if at the end everybody has her initial ball. Determine the values of $n$ for which there exists a nice game and those for which there exists a tiresome game.

Prove that the set of positive integers cannot be partitioned into three nonempty subsets such that, for any two integers $x,y$ taken from two different subsets, the number $x^2-xy+y^2$ belongs to the third subset.

Find all the functions $f: \mathbb{R} \to\mathbb{R}$ such that \[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]for all $x,y \in \mathbb{R} $.

For $n \geq 3$ and $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$ given real numbers we have the following instructions: - place out the numbers in some order in a ring; - delete one of the numbers from the ring; - if just two numbers are remaining in the ring: let $S$ be the sum of these two numbers. Otherwise, if there are more the two numbers in the ring, replace Afterwards start again with the step (2). Show that the largest sum $S$ which can result in this way is given by the formula \[S_{max}= \sum^n_{k=2} \begin{pmatrix} n -2 \\ [\frac{k}{2}] - 1\end{pmatrix}a_{k}.\]

Let $n \geq 1$ be an integer. A path from $(0,0)$ to $(n,n)$ in the $xy$ plane is a chain of consecutive unit moves either to the right (move denoted by $E$) or upwards (move denoted by $N$), all the moves being made inside the half-plane $x \geq y$. A step in a path is the occurence of two consecutive moves of the form $EN$. Show that the number of paths from $(0,0)$ to $(n,n)$ that contain exactly $s$ steps $(n \geq s \geq 1)$ is \[\frac{1}{s} \binom{n-1}{s-1} \binom{n}{s-1}.\]

If a $5 \times n$ rectangle can be tiled using $n$ pieces like those shown in the diagram, prove that $n$ is even. Show that there are more than $2 \cdot 3^{k-1}$ ways to file a fixed $5 \times 2k$ rectangle $(k \geq 3)$ with $2k$ pieces. (symmetric constructions are supposed to be different.)

A biologist watches a chameleon. The chameleon catches flies and rests after each catch. The biologist notices that: the first fly is caught after a resting period of one minute; the resting period before catching the $2m^\text{th}$ fly is the same as the resting period before catching the $m^\text{th}$ fly and one minute shorter than the resting period before catching the $(2m+1)^\text{th}$ fly; when the chameleon stops resting, he catches a fly instantly. How many flies were caught by the chameleon before his first resting period of $9$ minutes in a row? After how many minutes will the chameleon catch his $98^\text{th}$ fly? How many flies were caught by the chameleon after 1999 minutes have passed?

Let $A$ be a set of $N$ residues $\pmod{N^{2}}$. Prove that there exists a set $B$ of of $N$ residues $\pmod{N^{2}}$ such that $A + B = \{a+b|a \in A, b \in B\}$ contains at least half of all the residues $\pmod{N^{2}}$.

Let $n$ be an even positive integer. We say that two different cells of a $n \times n$ board are neighboring if they have a common side. Find the minimal number of cells on the $n \times n$ board that must be marked so that any cell (marked or not marked) has a marked neighboring cell.

Suppose that every integer has been given one of the colours red, blue, green or yellow. Let $x$ and $y$ be odd integers so that $|x| \neq |y|$. Show that there are two integers of the same colour whose difference has one of the following values: $x,y,x+y$ or $x-y$.

Click for solution Let's assume $x,y>0$ (clearly, we can do this, since if what we want to prove doesn't hold, then it doesn't hold if we replace $x$ with $-x$ and/or $y$ with $-y$). Let's work with non-negative integers only. The negation of what we want to prove states that there is a set $S\subset \mathbb N$ s.t. $S,S+x,S+y,S+x+y$ are mutually disjoint, and their union is $\mathbb N$. This means that, working with formal power series, $1+t+t^2+\ldots=\left(\sum_{s\in S}t^s\right)(1+t^x)(1+t^y)$. Assume now that $y<x$. We have $\frac{1+t+t^2+\ldots}{1+t^y}=(1+t+\ldots+t^{y-1})+(t^{2y}+t^{2y+1}+\ldots+t^{3y-1})+\ldots=\mathcal E$. When we divide $\mathcal E$ by $1+t^x$ we have to get a series whose only coefficients are $0$ and $1$, and this will yield the contradiction: our series contains $1+t+\ldots+t^{y-1}$, because $y<x$. There must be a $k$ s.t. $x\in(2ky,(2k+1)y-1)$ (the interval is open because the endpoints are even, but $x$ is odd). However, there is an $\alpha\in\overline{0,y-1}$ s.t. $x+\alpha=(2k+1)y$, and this means that if our power series has no negative terms (to get rid of $t^{(2k+1)y}$, which does not appear in $\mathcal E$), when multiplied by $1+t^x$ contains $t^{(2k+1)y}$, but $\mathcal E$ doesn't have this term, so we have a contradiction.

Let $p >3$ be a prime number. For each nonempty subset $T$ of $\{0,1,2,3, \ldots , p-1\}$, let $E(T)$ be the set of all $(p-1)$-tuples $(x_1, \ldots ,x_{p-1} )$, where each $x_i \in T$ and $x_1+2x_2+ \ldots + (p-1)x_{p-1}$ is divisible by $p$ and let $|E(T)|$ denote the number of elements in $E(T)$. Prove that \[|E(\{0,1,3\})| \geq |E(\{0,1,2\})|\] with equality if and only if $p = 5$.