Difference between revisions of "1994 USAMO Problems"
(→Problem 2) |
(Added Problem 5) |
||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 28: | Line 28: | ||
==Problem 5== | ==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>. | |
| − | 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]] | [[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
Problem 1
Let 
be positive integers, no two consecutive, and let 
for 
. Prove that, for each positive integer 
the interval 
contains at least one perfect square.
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?
Problem 3
A convex hexagon 
is inscribed in a circle such that 
and diagonals 
, 
, and 
are concurrent. Let 
be the intersection of and 
. Prove that 
.
Problem 4
Let 
be a sequence of positive real numbers satisfying 
for all 
. Prove that, for all 
![\[\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).\]](http://latex.artofproblemsolving.com/8/2/9/8293056937b2c93fb3ccfde0a9b743d24fd9660d.png)
Problem 5
Let 
and 
denote the number of elements, the sum, and the product, respectively, of a finite set 
of positive integers. (If is the empty set, 
.) Let 
be a finite set of positive integers. As usual, let 
denote 
. Prove that ![\[\sum_{U \subseteq S} (-1)^{|U|} \binom{m - \sigma(U)}{|S|} = \pi(S)\]](http://latex.artofproblemsolving.com/5/0/e/50ed2758adcf62f11d440689fd8665ae335512b4.png)
for all integers 
.
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. 
