Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1994 USAMO Problems"

m (→‎Resources)
(Added Problem 5)
 
(2 intermediate revisions by 2 users not shown)
Line 27: Line 27:
 
[[1994 USAMO Problems/Problem 4|Solution]]
 
[[1994 USAMO Problems/Problem 4|Solution]]
  
−
== Resources ==
+
==Problem 5==
 +
Let <math>\, |U|, \, \sigma(U) \,</math> and <math>\, \pi(U) \,</math> denote the number of elements, the sum, and the product, respectively, of a finite set <math>\, U \,</math> of positive integers.  (If <math>\, U \,</math> is the empty set, <math>\, |U| = 0, \, \sigma(U) = 0, \, \pi(U) = 1</math>.) Let <math>\, S \,</math> be a finite set of positive integers. As usual, let <math>\, \binom{n}{k} \,</math> denote <math>\, n! \over k! \, (n-k)!</math>. Prove that <cmath> \sum_{U \subseteq S} (-1)^{|U|} \binom{m - \sigma(U)}{|S|} = \pi(S)  </cmath> for all integers <math>\, m \geq \sigma(S)</math>.
 +
 
 +
[[1994 USAMO Problems/Problem 5|Solution]]
 +
 
 +
== See Also ==
 
{{USAMO box|year=1994|before=[[1993 USAMO]]|after=[[1995 USAMO]]}}
 
{{USAMO box|year=1994|before=[[1993 USAMO]]|after=[[1995 USAMO]]}}
 +
{{MAA Notice}}

Latest revision as of 06:58, 19 July 2016

Problems of the 1994 USAMO.

Problem 1

Let $\, k_1 < k_2 < k_3 < \cdots \,$ be positive integers, no two consecutive, and let $\, s_m = k_1 + k_2 + \cdots + k_m \,$ for $\, m = 1,2,3, \ldots \; \;$. Prove that, for each positive integer $\, n, \,$ the interval $\, [s_n, s_{n + 1}) \,$ contains at least one perfect square.

Solution

Problem 2

The sides of a 99-gon are initially colored so that consecutive sides are red, blue, red, blue, ... red, blue, yellow. We make a sequence of modifications in the coloring, changing the color of one side at a time to one of the three given colors (red, blue, yellow), under the constraint that no two adjacent sides may be the same color. By making a sequence of such modifications, is it possible to arrive at the coloring in which consecutive sides are red, blue, red, blue, red, blue, ... red, yellow, blue?

Solution

Problem 3

A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB = CD = EF$ and diagonals $AD$, $BE$, and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $CP/PE = (AC/CE)^2$.

Solution

Problem 4

Let $\, a_1, a_2, a_3, \ldots \,$ be a sequence of positive real numbers satisfying $\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,$ for all $\, n \geq 1$. Prove that, for all $\, n \geq 1, \,$

\[\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).\]

Solution

Problem 5

Let $\, |U|, \, \sigma(U) \,$ and $\, \pi(U) \,$ denote the number of elements, the sum, and the product, respectively, of a finite set $\, U \,$ of positive integers. (If $\, U \,$ is the empty set, $\, |U| = 0, \, \sigma(U) = 0, \, \pi(U) = 1$.) Let $\, S \,$ be a finite set of positive integers. As usual, let $\, \binom{n}{k} \,$ denote $\, n! \over k! \, (n-k)!$. Prove that \[\sum_{U \subseteq S} (-1)^{|U|} \binom{m - \sigma(U)}{|S|} = \pi(S)\] for all integers $\, m \geq \sigma(S)$.

Solution

See Also

1994 USAMO (Problems • Resources)
Preceded by
1993 USAMO 		Followed by
1995 USAMO
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_USAMO_Problems&oldid=79532"