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.

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.

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 $\frac{\tau (n^{2})}{\tau (n)}=m$.

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

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.

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!