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

1994 USAMO Problems

Revision as of 05:19, 11 July 2009 by God of Math (talk | contribs)

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


Induct on $n$. When $n = 1$, we are to show that the interval $\, [s_n, s_{n + 1}) \,$ contains at least one perfect square. This interval is equivalent to $\, [k_0, k_0 + k_1) \,$ when $n = 1$. Now for some $a , a^2 \le k_0^2 < (a+1)^2$.Then it suffices to show that the minimal "distance spanned" by the interval $\, [k_0, k_0 + k_1) \,$ is greater than or equal to the maximum distance from $k_0$ to the nearest perfect square. Since the smallest element that can follow $k_0$ is $k_0 + 2$, we have to show the below. Note that we ignore the trivial case where $k_0 = a^2$, which should be mentioned. 

  $2a + 1 \le k_0 + 2$
  $2a \le k_0 + 1$
  $2a \le k_0 + (q + 1)$, where $q$ is a member of $\{1, 2, \ldots, 2\a}$ (Error compiling LaTeX. Unknown error_msg)
  We now prove by contradiction. Assume that

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

Resources

1994 USAMO (Problems • Resources)
Preceded by
1993 USAMO 		Followed by
1995 USAMO
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_USAMO_Problems&oldid=32285"