Difference between revisions of "1982 USAMO Problems"

(Problem 2)
m (Made prettier)
Line 1: Line 1:
 +
Problems from the '''1982 [[United States of America Mathematical Olympiad | USAMO]]'''.
 +
 
==Problem 1==
 
==Problem 1==
 
A graph has <math>1982</math> points. Given any four points, there is at least one joined to the other three. What is the smallest number of points which are joined to <math>1981</math> points?
 
A graph has <math>1982</math> points. Given any four points, there is at least one joined to the other three. What is the smallest number of points which are joined to <math>1981</math> points?

Revision as of 10:30, 6 March 2013

Problems from the 1982 USAMO.

Problem 1

A graph has $1982$ points. Given any four points, there is at least one joined to the other three. What is the smallest number of points which are joined to $1981$ points?

Solution

Problem 2

Show that if $m, n$ are positive integers such that $\frac{\left(x^{m+n} + y^{m+n} + z^{m+n}\right)}{m+n}=\left(\frac{x^m + y^m + z^m}{m}\right) \left(\dfrac{x^n + y^n + z^n}{n}}\right)$ (Error compiling LaTeX. Unknown error_msg) for all real $x, y, z$ with sum $0$, then $(m, n) = (2, 3)$ or $(2, 5)$.

Solution

Problem 3

$D$ is a point inside the equilateral triangle $ABC$. $E$ is a point inside $DBC$. Show that $\frac{\text{area}DBC}{\text{perimeter} DBC^2} > \frac{\text{area} EBC}{\text{perimeter} EBC^2}.$

Solution

Problem 4

Show that there is a positive integer $k$ such that, for every positive integer $n$, $k 2^n+1$ is composite.

Solution

Problem 5

$O$ is the center of a sphere $S$. Points $A, B, C$ are inside $S$, $OA$ is perpendicular to $AB$ and $AC$, and there are two spheres through $A, B$, and $C$ which touch $S$. Show that the sum of their radii equals the radius of $S$.

Solution

See Also

1982 USAMO (ProblemsResources)
Preceded by
1981 USAMO
Followed by
1983 USAMO
1 2 3 4 5
All USAMO Problems and Solutions