Found problems: 892
1964 IMO, 1
(a) Find all positive integers $ n$ for which $ 2^n\minus{}1$ is divisible by $ 7$.
(b) Prove that there is no positive integer $ n$ for which $ 2^n\plus{}1$ is divisible by $ 7$.
1967 IMO Longlists, 57
Let $a_1,\ldots,a_8$ be reals, not all equal to zero. Let
\[ c_n = \sum^8_{k=1} a^n_k\]
for $n=1,2,3,\ldots$. Given that among the numbers of the sequence $(c_n)$, there are infinitely many equal to zero, determine all the values of $n$ for which $c_n = 0.$
1971 IMO, 1
All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$.
[b]a.)[/b] If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length;
[b]b.)[/b] If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.
1990 IMO Longlists, 19
Given an initial integer $ n_0 > 1$, two players, $ {\mathcal A}$ and $ {\mathcal B}$, choose integers $ n_1$, $ n_2$, $ n_3$, $ \ldots$ alternately according to the following rules :
[b]I.)[/b] Knowing $ n_{2k}$, $ {\mathcal A}$ chooses any integer $ n_{2k \plus{} 1}$ such that
\[ n_{2k} \leq n_{2k \plus{} 1} \leq n_{2k}^2.
\]
[b]II.)[/b] Knowing $ n_{2k \plus{} 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k \plus{} 2}$ such that
\[ \frac {n_{2k \plus{} 1}}{n_{2k \plus{} 2}}
\]
is a prime raised to a positive integer power.
Player $ {\mathcal A}$ wins the game by choosing the number 1990; player $ {\mathcal B}$ wins by choosing the number 1. For which $ n_0$ does :
[b]a.)[/b] $ {\mathcal A}$ have a winning strategy?
[b]b.)[/b] $ {\mathcal B}$ have a winning strategy?
[b]c.)[/b] Neither player have a winning strategy?
1999 IMO Shortlist, 1
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$.
1978 IMO, 2
Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$
1960 IMO Shortlist, 7
An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.
a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;
b) Calculate the distance of $p$ from either base;
c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.
1992 IMO, 3
Consider $9$ points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of $\,n\,$ such that whenever exactly $\,n\,$ edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.
1992 IMO Shortlist, 20
In the plane let $\,C\,$ be a circle, $\,L\,$ a line tangent to the circle $\,C,\,$ and $\,M\,$ a point on $\,L$. Find the locus of all points $\,P\,$ with the following property: there exists two points $\,Q,R\,$ on $\,L\,$ such that $\,M\,$ is the midpoint of $\,QR\,$ and $\,C\,$ is the inscribed circle of triangle $\,PQR$.
2023 IMO, 5
Let $n$ be a positive integer. A [i]Japanese triangle[/i] consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A [i]ninja path[/i] in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.
[asy]
// credit to vEnhance for the diagram (which was better than my original asy):
size(4cm);
pair X = dir(240); pair Y = dir(0);
path c = scale(0.5)*unitcircle;
int[] t = {0,0,2,2,3,0};
for (int i=0; i<=5; ++i) {
for (int j=0; j<=i; ++j) {
filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white);
draw(shift(i*X+j*Y)*c);
}
}
draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5));
path q = (3,-3sqrt(3))--(-3,-3sqrt(3));
draw(q,Arrows(TeXHead, 1));
label("$n = 6$", q, S);
label("$n = 6$", q, S);
[/asy]
In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.
1965 IMO, 2
Consider the sytem of equations
\[ a_{11}x_1+a_{12}x_2+a_{13}x_3 = 0 \]\[a_{21}x_1+a_{22}x_2+a_{23}x_3 =0\]\[a_{31}x_1+a_{32}x_2+a_{33}x_3 = 0 \] with unknowns $x_1, x_2, x_3$. The coefficients satisfy the conditions:
a) $a_{11}, a_{22}, a_{33}$ are positive numbers;
b) the remaining coefficients are negative numbers;
c) in each equation, the sum ofthe coefficients is positive.
Prove that the given system has only the solution $x_1=x_2=x_3=0$.
1961 IMO Shortlist, 3
Solve the equation $\cos^n{x}-\sin^n{x}=1$ where $n$ is a natural number.
1995 IMO Shortlist, 5
Let $ ABCDEF$ be a convex hexagon with $ AB \equal{} BC \equal{} CD$ and $ DE \equal{} EF \equal{} FA$, such that $ \angle BCD \equal{} \angle EFA \equal{} \frac {\pi}{3}$. Suppose $ G$ and $ H$ are points in the interior of the hexagon such that $ \angle AGB \equal{} \angle DHE \equal{} \frac {2\pi}{3}$. Prove that $ AG \plus{} GB \plus{} GH \plus{} DH \plus{} HE \geq CF$.
1971 IMO, 3
Prove that we can find an infinite set of positive integers of the from $2^n-3$ (where $n$ is a positive integer) every pair of which are relatively prime.
1998 IMO Shortlist, 6
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 $\frac{\tau (n^{2})}{\tau (n)}=m$.
1978 IMO Longlists, 46
We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.
1977 Germany Team Selection Test, 2
Determine the polynomials P of two variables so that:
[b]a.)[/b] for any real numbers $t,x,y$ we have $P(tx,ty) = t^n P(x,y)$ where $n$ is a positive integer, the same for all $t,x,y;$
[b]b.)[/b] for any real numbers $a,b,c$ we have $P(a + b,c) + P(b + c,a) + P(c + a,b) = 0;$
[b]c.)[/b] $P(1,0) =1.$