Olympiad problems

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Found problems: 12

1998 IMO, 3

Tags: number theory , Divisors , Number theoretic functions , prime factorization , IMO , IMO 1998 , IMO Shortlist

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

1998 IMO Shortlist, 3

Tags: geometry , incenter , trigonometry , Triangle , IMO , IMO 1998

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.

1998 IMO, 1

Tags: geometry , circumcircle , quadrilateral , perpendicular bisector , Charles Leytem , IMO , IMO 1998

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.

1998 IMO Shortlist, 5

Tags: matrix , combinatorics , IMO , Inequality , IMO 1998 , IMO Shortlist , Set systems

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 \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]

1998 IMO Shortlist, 1

Tags: number theory , Divisibility , polynomial , IMO , IMO 1998 , david monk , Hi

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

1998 IMO, 2

Tags: matrix , combinatorics , IMO , Inequality , IMO 1998 , IMO Shortlist , Set systems

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 \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]

1998 IMO Shortlist, 5

Tags: number theory , prime numbers , algebra , functional equation , IMO , IMO 1998 , IMO Shortlist

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

1998 IMO Shortlist, 1

Tags: geometry , circumcircle , quadrilateral , perpendicular bisector , Charles Leytem , IMO , IMO 1998

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.

1998 IMO, 6

Tags: number theory , prime numbers , algebra , functional equation , IMO , IMO 1998 , IMO Shortlist

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

1998 IMO, 4

Tags: number theory , Divisibility , polynomial , IMO , IMO 1998 , david monk , Hi

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

1998 IMO, 5

Tags: geometry , incenter , trigonometry , Triangle , IMO , IMO 1998

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.

1998 IMO Shortlist, 6

Tags: number theory , Divisors , Number theoretic functions , prime factorization , IMO , IMO 1998 , IMO Shortlist

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