IMO 1989

IMO 1989 — 5/6 solved.

5 items

IMO 1989

Official IMO 1989 problems  ·  5/6 solved.

# Status Time
1 solved 18m36s
2 solved 4m12s
3 solved 5m32s
4 solved 15m12s
5
6 solved 4m10s

Problem 1   solved · 18m36s · Solution →

Prove that in the set ${1,2, \ldots, 1989}$ can be expressed as the disjoint union of subsets $A_i, {i = 1,2, \ldots, 117}$ such that

i.) each $A_i$ contains 17 elements

ii.) the sum of all the elements in each $A_i$ is the same.

Problem 2   solved · 4m12s · Solution →

$ABC$ is a triangle, the bisector of angle $A$ meets the circumcircle of triangle $ABC$ in $A_1$, points $B_1$ and $C_1$ are defined similarly. Let $AA_1$ meet the lines that bisect the two external angles at $B$ and $C$ in $A_0$. Define $B_0$ and $C_0$ similarly. Prove that the area of triangle $A_0B_0C_0 = 2 \cdot$ area of hexagon $AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ABC$.

Problem 3   solved · 5m32s · Solution →

Let $n$ and $k$ be positive integers and let $S$ be a set of $n$ points in the plane such that

i.) no three points of $S$ are collinear, and

ii.) for every point $P$ of $S$ there are at least $k$ points of $S$ equidistant from $P.$

Prove that: $$ k < \frac {1}{2} + \sqrt {2 \cdot n} $$

Problem 4   solved · 15m12s · Solution →

Let $ABCD$ be a convex quadrilateral such that the sides $AB, AD, BC$ satisfy $AB = AD + BC.$ There exists a point $P$ inside the quadrilateral at a distance $h$ from the line $CD$ such that $AP = h + AD$ and $BP = h + BC.$ Show that: $$ \frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} + \frac {1}{\sqrt {BC}} $$

Problem 5

Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.

Problem 6   solved · 4m10s · Solution →

A permutation ${x_1, x_2, \ldots, x_{2n}}$ of the set ${1,2, \ldots, 2n}$ where $n$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i$ in ${1,2, \ldots, 2n - 1}.$ Show that, for each $n$, there are more permutations with property $T$ than without.