A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.

Click for solution Assume first that $ABCD$ is cyclic. It's easy then to see that $\angle APB+\angle CPD=\pi$, so $2S[APB]=R^2\sin \angle APB=R^2\sin\angle CPD=2S[CPD]$, where $R=PA=PB=PC=PD$ is the radius of the circumcircle of $ABCD$. Conversely, assume we have $S[APB]=S[CPD]$. Let $M,N$ be the midpoints of $AB,CD$ respectively. We have $TM=AM,TN=DN$, where $T=AC\cap BD$. The hypothesis is equivalent to $PM\cdot AM=PN\cdot DN$, and from here we deduce $PM\cdot TM=PN\cdot TN$. A quick angle chase reveals $\angle MPN=\angle MTN$, so the triangles $TMN,PNM$ are, in fact, congruent, meaning that $TMPN$ is a parallelogram. This, however, means that the triangles $TAM,DTN$ are similar (because they have three pairs of perpendicular sides), so $\angle TAB=\angle TDC$, Q.E.D.

Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.

Let $I$ be the incenter of triangle $ABC$. Let $K,L$ and $M$ be the points of tangency of the incircle of $ABC$ with $AB,BC$ and $CA$, respectively. The line $t$ passes through $B$ and is parallel to $KL$. The lines $MK$ and $ML$ intersect $t$ at the points $R$ and $S$. Prove that $\angle RIS$ is acute.

Click for solution I haven't thought about a generalization, but here's a solution to this problem. The assertion is equivalent to $BI^2>BS\cdot BR$. It's easy to see that triangles $BSK,BMR$ are similar (in this order of the vertices). We only need to employ a simple angle chase. From here we derive $\frac{BS}{BK}=\frac{BM}{BR}\Rightarrow BS\cdot BR=BM\cdot BK=BM^2$, and since in the right triangle $BMI$ $BM$ is a leg, while $BI$ is the hypothenuse, we get $BI>BM\Rightarrow BI^2>BM^2=BS\cdot BR$, Q.E.D.

Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that \[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC. \] Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1. \]

Let $ABC$ be a triangle, $H$ its orthocenter, $O$ its circumcenter, and $R$ its circumradius. Let $D$ be the reflection of the point $A$ across the line $BC$, let $E$ be the reflection of the point $B$ across the line $CA$, and let $F$ be the reflection of the point $C$ across the line $AB$. Prove that the points $D$, $E$ and $F$ are collinear if and only if $OH=2R$.

Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^{\circ }$ and \[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \] Prove that \[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \]

Let $ABC$ be a triangle such that $\angle ACB=2\angle ABC$. Let $D$ be the point on the side $BC$ such that $CD=2BD$. The segment $AD$ is extended to $E$ so that $AD=DE$. Prove that \[ \angle ECB+180^{\circ }=2\angle EBC. \]

Click for solution Here's a sketch (I don't want to go into all the details, but I will if you really want me to ). Try to show that when $B,C$ are fixed, $A$ moves on the hyperbola having $D$ as its center and $B$ as one of its vertices, which has eccentricity $2$. To be precise, $A$ lies on the branch of this hyperbola which doesn't pass through $B$. $E$ also lies on this ellipse, being the "antipode" of $A$. Let $F$ be the symmetric of $E$ wrt $BC$. This means that $F$ is also on the hyperbola, on the other branch (the branch which doesn't contain $A$). From this condition we find that in the triangle $FCB$, if $G$ is the point where the perp bisector of $BC$ cuts $FC$, then $\frac{CG}{GF}=\frac{CB}{CF}$. Let $X$ be the point where the bisector of $\angle FCB$ cuts $FB$. We then get $GX\|BC$. At the same time, from $(*)$, we find $GX$ to be the bisector of $\angle BGF$. Now show that $GX=GB=GC$. What interests us is that $GX=GB$. A quick angle chase gives what we want. I'm sure there's a simpler solution, but I can't see anything right now, sorry.

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$. commentEdited by Orl.

Click for solution It's still nice to have a solution posted . Let $F$ be the point on the circle $(ABC)$ s.t. $BF\|AY$. We see that the four lines $BF,BZ,BA,BE$ for a harmonic quadruple, so their intersections with the circle (other than $B$) for a harmonic quadrilateral, i.e. the quadrilateral $AFEZ$ is harmonic. This means that the tangents from $A,E$ to the circle meet on the diagonal $ZF$, so $D,Z,F$ are collinear. We will keep this in mind in what follows. Since $\angle ZAB=\angle BFD$ and $\angle AZB=\angle ACB=\angle FBD$, we find the triangles $ZAB,BFD$ to be similar. At the same time, the triangles $ZCD,BFD$ are similar, so $ZAB,ZCD$ are similar. What we have to prove is now reduced to a simple angle chase: $\angle CDZ=\angle ZBA=\angle ZAD$, and that's it.

Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.

Click for solution Unchecked solution by nhat: find all a,b in positive integer such that: $\frac {a^{2}*b+a+b}{a*b^2+b+7}$ in positive integer I want the another solution which is different to the solution of IMO here my solution for this case a=b we easy to have the solution if a<b then $a*b*(a-b)+a-7$<0 (contradition) hence a>b let a in $(m*b,(m+1)*b)$ (m i positive integer) then we easy to prove that : m<$\frac {a^{2}*b+a+b}{a*b^2+b+7}$<(m+1) (contradition) so a divisible to b give $a=k*b$ (k in positive integer) then $\frac{k^2*b^3+k*b+b}{k*b^3+b+7}$=$\frac{b*{k^2*b^2+k+1}}{k*b^3+b+7}$ if $(a,7)=1$ then $(b,k*b^3+b+7)=1$ so that $\frac {k^2*b^2+k+1}{k*b^3+b+7}$ in positive integer and we esay to prove that k>b so easy to prove that \[k\equiv 0 (mod b)\] so $k=q*b$ (b >= 3and q i positive integer) then $\frac {k^2*b^2+k+1}{k*b^3+b+7}$=${\{q^2*b^4+q*b+1}{q*b^4+b+1}$=$q$+$\frac {1-q}{q*b^4+b+1}$ (contradition) if b divisible to 7 so that $b=7*u$ (u in positive integer) and easy to prove that u=k (it's similar for this case two) so we have the solution $a=7*k^2$,$b=7*k$ my solution hasor hasn't correct

Determine all pairs $(a,b)$ of real numbers such that $a \lfloor bn \rfloor =b \lfloor an \rfloor $ for all positive integers $n$. (Note that $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.)

Click for solution First of all, notice that one of the numbers is $0$, then we have a solution, no matter what the other number is, so let's assume that they're both non-$0$. Another thing worth noticing is that the condition is equivalent to $a\{nb\}=b\{na\},\ \forall n\in\mathbb N^*$. Clearly, from this condition we get $a\in\mathbb Q\iff b\in\mathbb Q$. Let's consider the case when both our numbers are irrational. We have $\frac ba=\frac{\{nb\}}{\{na\}}$. $\{na\}$ is dense in $(0,1)$, so if we choose a sequence $n_k$ s.t. $\{n_ka\}\to 1$, and then we get $\{n_kb\}\to 1$, so $\frac{\{n_kb\}}{\{n_ka\}}\to 1$, meaning that $\frac ba=1\iff a=b$, and we can see that, conversely, all pairs $(a,b)=(a,a)$ satisfy the equation. Assume now one of the numbers is an integer. We quickly find the other one to be an integer too, so this is another type of solution: $(a,b)\in\mathbb Z^2$. Finally, the last case: the numbers are rational but not integral. Let $a=\frac{a'}q,\ b=\frac {b'}{q'}$, where $q,q'\in\mathbb N^*,a',b'\in\mathbb Z$, and $(a',q)=(a',q')=1$. By making $n=q$, and then $n=q'$, we find $q=q'$. We can use the relation $\frac ba=\frac{\{nb\}}{\{na\}}$. We can find $n$ s.t. $\{nb\}=\frac{q-1}q$, so $\frac ba\ge 1$. At the same time, we can show in exactly the same way that $\frac ab\ge 1$, so $a=b$. Combining everything, the solutions are either of the form $(a,a)$ with $a$ being any real number, or $(a,b),\ a,b\in\mathbb Z$, or $(0,b),(a,0)$.

Determine the smallest integer $n\geq 4$ for which one can choose four different numbers $a,b,c$ and $d$ from any $n$ distinct integers such that $a+b-c-d$ is divisible by $20$.

Click for solution I think it should be $9$. Let's work $\pmod{20}$, so when I say distinct, for example, I mean distinct $\pmod{20}$. If we look at systems consisting of distinct numbers, then we see that $7$ suffice, because they form $21$ pairs, so there must be two pairs with the same sum, and two such pairs cannot share one of the members. If a number appears more than once, then no other number can appear more than once, and a certain number cannot appear more than three times, so the number we're looking for is at most $9$ (in the case $a,a,a,b,c,d,e,f,g$). If we find a counterexample for $8$, we will have shown that the answer is $9$. Here are $8$ numbers for which $a+b-(c+d)\ne 0,\ \forall a,b,c,d$ distinct: $0,0,0,1,2,4,7,12$. This shows that the answer is $9$.

A sequence of integers $ a_{1},a_{2},a_{3},\ldots$ is defined as follows: $ a_{1} = 1$ and for $ n\geq 1$, $ a_{n + 1}$ is the smallest integer greater than $ a_{n}$ such that $ a_{i} + a_{j}\neq 3a_{k}$ for any $ i,j$ and $ k$ in $ \{1,2,3,\ldots ,n + 1\}$, not necessarily distinct. Determine $ a_{1998}$.

Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.

For any positive integer $n$, let $\tau (n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\frac{\tau (n^{2})}{\tau (n)}=m$.

Prove that for each positive integer $n$, there exists a positive integer with the following properties: It has exactly $n$ digits. None of the digits is 0. It is divisible by the sum of its digits.

Let $a_{0},a_{1},a_{2},\ldots $ be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form $a_{i}+2a_{j}+4a_{k}$, where $i,j$ and $k$ are not necessarily distinct. Determine $a_{1998}$.

Click for solution I have my doubts about this since I might have rushed into it a bit, but here it goes: Let $f:[0,1),\ f(x):=x^{a_0}+x^{a_1}+x^{a_2}+\ldots$. Since $0$ can be represented in the mentioned way, we must have $a_0=0$ (this will be important a bit later). We have $f(x)f(x^2)f(x^4)=1+x+x^2+\ldots=\frac 1{1-x}$. We do this again for $x\to x^2$ and we get $f(x^2)f(x^4)f(x^8)=\frac 1{1-x^2}$. Divide the first relation by the second one to get $\frac {f(x)}{f(x^8)}=1+x\ (*)$. In the same way we get $\frac{f(x^8)}{f(x^{8^2})}=1+x^8$, and so on. Since $f(0)=1$, for a certain $x$ we have $\lim_{n\to\infty}f(x^{8^n})=f(0)=1$, so, by fixing $x$ and multiplying the relations of the type $(*)$, we get $f(x)=(1+x)(1+x^8)(1+x^{8^2})\ldots$. What we need to find is the $1998$'th positive exponent (in increasing order) from this infinite series. In order to finish it, take $1998$, write it in base $2$, and read the result in base $8$. What we get is $a_{1998}$. I guess that, if I didn't make any mistakes, it should be $8+8^2+8^3+8^6+8^7+8^8+8^9+8^{10}$. The answer looks really strange, and that's what makes me have doubts .

Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that \[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]

Click for solution Put $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. The inequality becomes $\frac{a_1a_2\ldots a_{n+1}}{(1-a_1)(1-a_2)\ldots (1-a_{n+1})}\le \frac 1{n^{n+1}}$, where $a_i$ are positive and $\sum_1^{n+1}a_i=1$. This is a mere application of AM-GM to the denominators $A_i=1-a_i=a_1+a_2+\ldots+a_{i-1}+a_{i+1}+\ldots+a_n+a_{n+1}$. We get $A_i\ge n\sqrt[n]{\frac{\prod a_k}{a_i}}$. After multiplying all of these we get the desired result.

Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that \[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \]

Click for solution Standard Cauchy induction works just fine. We first prove it for two numbers: It takes the form $a,b\ge 1\Rightarrow \frac 1{a^2+1}+\frac 1{b^2+1}\ge \frac 2{ab+1}$. After expanding everything, this reduces to $2ab(ab-1)\le (a^2+b^2)(ab-1)$, which is true by AM-GM $(*)$. The next step is proving it for $n=2^k$. This is done easily by induction: we add two inequalities with $2^{k-1}$ variables each, and then apply the inequality for two variables to get the result $(**)$. After this, we prove that $P(n)\Rightarrow P(n-1)$. Assuming the inequality is true for $n$ variables, take $r_n=\sqrt[n-1]{r_1r_2\ldots r_{n-1}}$. We get $\sum_{i=1}^{n-1}\frac 1{r_i+1}+\frac 1{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}\ge \frac n{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}$. This proves the inequality for $n-1$ variables $(***)$. $(*),(**),(***)$ complete the induction.

Let $x,y$ and $z$ be positive real numbers such that $xyz=1$. Prove that \[ \frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)} \geq \frac{3}{4}. \]

Click for solution Amplify the first, second, and third fraction by $x,y,z$ respectively. The LHS becomes $\sum_{cyc}\frac{x^4}{x(1+y)(1+z)}\ge \frac{(x^2+y^2+z^2)^2}{x+y+z+2(xy+yz+zx)+3}\ge \frac{(x^2+y^2+z^2)^2}{4(x^2+y^2+z^2)}\ge\frac 34$. I used the inequalities: $x^2+y^2+z^2\ge xy+yz+zx$, $x^2+y^2+z^2\ge 3$ and $x^2+y^2+z^2\ge \frac{(x+y+z)^2}3\ge x+y+z$.

For any two nonnegative integers $n$ and $k$ satisfying $n\geq k$, we define the number $c(n,k)$ as follows: - $c\left(n,0\right)=c\left(n,n\right)=1$ for all $n\geq 0$; - $c\left(n+1,k\right)=2^{k}c\left(n,k\right)+c\left(n,k-1\right)$ for $n\geq k\geq 1$. Prove that $c\left(n,k\right)=c\left(n,n-k\right)$ for all $n\geq k\geq 0$.

Determine the least possible value of $f(1998),$ where $f:\Bbb{N}\to \Bbb{N}$ is a function such that for all $m,n\in {\Bbb N}$, \[f\left( n^{2}f(m)\right) =m\left( f(n)\right) ^{2}. \]

Click for solution Denote $f(1)=a$, and put $m=n=1$, therefore $f(f(k))=a^{2}k$ and $f(ak^{2})=f^{2}(k)$, $\forall k \in \mathbb{N}$. Thus now, we have: $f^{2}(x) f^{2}(y)=f^{2}(x)f(ay^{2})=f(x^{2}f(f(ay^{2})))=$ $=f(x^{2}a^{3}y^{2})=f(a(axy)^{2})=f^{2}(axy)$ $\iff f(axy)=f(x)f(y) \Rightarrow f(ax)=af(x)$ $\iff af(xy)=f(x)f(y) , \forall x,y \in \mathbb{N}$. Now we can easily prove that $f(x)$ is divisible by $a$ for each $x$, more likely we have that $f^{k}(x)=a^{k-1}\cdot f(x^{k})$ is divisible by $a^{k-1}$. For proving the above asertion we consider $p^{\alpha}$ and $p^{\beta}$ the exact powers of a prime $p$ that tivide $f(x)$ and $a$ respectively, therefore $k\alpha \geq (k-1)\beta , \forall k \in \mathbb{N}$, therefore $\alpha\geq \beta$, so $f(x)$ is divisible by $a$. Now we just consider the function $g(x)=\frac{f(x)}{a}$. Thus: $g(1)=1, g(xy)=g(x)g(y), g(g(x))=x$. Since $g(x)$ respects the initial condition of the problem and $g(x)\leq f(x)$, we claim that it is enough to find the least value of $g(1998)$. Since $g(1998)=g(2 \cdot 3^{3}\cdot 37) =g(2) \cdot g^{3}(3)\cdot g(37)$, and $g(2),g(3),g(37)$ are disting prime numbers (the proof follows easily), we have that $g(1998)$, is not smaller than $2^{3}\cdot 3 \cdot 5=120$. But $g$ beeing a bijection, the value $120$, is obtained for any $g$, so we have that $g(2)=3, g(3)=2, g(5)=37, g(37)=5$, therefore the answer is $120$, and thus the problem is solved!

A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)

Let $n$ be an integer greater than 2. A positive integer is said to be attainable if it is 1 or can be obtained from 1 by a sequence of operations with the following properties: 1.) The first operation is either addition or multiplication. 2.) Thereafter, additions and multiplications are used alternately. 3.) In each addition, one can choose independently whether to add 2 or $n$ 4.) In each multiplication, one can choose independently whether to multiply by 2 or by $n$. A positive integer which cannot be so obtained is said to be unattainable. a.) Prove that if $n\geq 9$, there are infinitely many unattainable positive integers. b.) Prove that if $n=3$, all positive integers except 7 are attainable.

Cards numbered 1 to 9 are arranged at random in a row. In a move, one may choose any block of consecutive cards whose numbers are in ascending or descending order, and switch the block around. For example, 9 1 $\underline{6\ 5\ 3}$ $2\ 7\ 4\ 8$ may be changed to $9 1$ $\underline{3\ 5\ 6}$ $2\ 7\ 4\ 8$. Prove that in at most 12 moves, one can arrange the 9 cards so that their numbers are in ascending or descending order.

Let $U=\{1,2,\ldots ,n\}$, where $n\geq 3$. A subset $S$ of $U$ is said to be split by an arrangement of the elements of $U$ if an element not in $S$ occurs in the arrangement somewhere between two elements of $S$. For example, 13542 splits $\{1,2,3\}$ but not $\{3,4,5\}$. Prove that for any $n-2$ subsets of $U$, each containing at least 2 and at most $n-1$ elements, there is an arrangement of the elements of $U$ which splits all of them.

In a contest, there are $m$ candidates and $n$ judges, where $n\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]

Click for solution Apparently, I have the same proof as the one on Kalva.. We count the cases when two judges agree on the result of a certain contestant. Let this number be $N$. On the one hand, we have $N\le k\cdot \binom n2=k\cdot\frac{n(n-1)}2$. On the other hand, if we show that $m\cdot \left(\frac{n-1}2\right)^2\le N$, we're done. Consider the whole thing a $0,1$-matrix having the judges as rows and the contestants as columns, and having $a_{ij}=0$ is judge $j$ fails contestant $i$ and $a_{ij}=1$ otherwise. It suffices to show now that for each of the $m$ columns there are at least $\left(\frac{n-1}2\right)^2$ pairs of equal numbers. Assume on a column there are $a$ $0$s and $b$ $1$s with $a+b=n=2k+1$. In this case the number of matching pairs on this column is $\frac{a(a-1)}2+\frac{b(b-1)}2=\frac{a^2+b^2}2-\frac n2$, which, knowing that $a+b$ is constant, is minimal when $|a-b|$ is minimal (i.e. when $a,b$ are as close to each other as possible), and this happens when $\{a,b\}=\{k,k+1\}$. If we compute the number of matching pairs on this column in this particular case, we get precisely $k^2=\left(\frac{n-1}2\right)^2$, so this is the minimal value. That is, on every column the number of matching pairs is $\geq \left(\frac{n-1}2\right)^2$. Thus, $N\geq m\cdot \left(\frac{n-1}2\right)^2$, Q.E.D.

Ten points are marked in the plane so that no three of them lie on a line. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors, in such a way that for any $k$ of the ten points, there are $k$ segments each joining two of them and no two being painted with the same color. Determine all integers $k$, $1\leq k\leq 10$, for which this is possible.

A solitaire game is played on an $m\times n$ rectangular board, using $mn$ markers which are white on one side and black on the other. Initially, each square of the board contains a marker with its white side up, except for one corner square, which contains a marker with its black side up. In each move, one may take away one marker with its black side up, but must then turn over all markers which are in squares having an edge in common with the square of the removed marker. Determine all pairs $(m,n)$ of positive integers such that all markers can be removed from the board.