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Difference between revisions of "1984 USAMO Problems"

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Problems from the '''1984 [[USAMO]].'''
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==Problem 1==
 
==Problem 1==
 
The product of two of the four roots of the quartic equation <math>x^4 - 18x^3 + kx^2+200x-1984=0</math> is <math>-32</math>. Determine the value of <math>k</math>.
 
The product of two of the four roots of the quartic equation <math>x^4 - 18x^3 + kx^2+200x-1984=0</math> is <math>-32</math>. Determine the value of <math>k</math>.
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==Problem 3==
 
==Problem 3==
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<math>P, A, B, C,</math> and <math>D</math> are five distinct points in space such that <math>\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta</math>, where <math>\theta</math> is a given acute angle. Determine the greatest and least values of <math>\angle APC + \angle BPD</math>.
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<math>P</math>, <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are five distinct points in space such that <math>\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta</math>, where <math>\theta</math> is a given acute angle. Determine the greatest and least values of <math>\angle APC + \angle BPD</math>.
  
 
[[1984 USAMO Problems/Problem 3 | Solution]]
 
[[1984 USAMO Problems/Problem 3 | Solution]]
  
 
==Problem 4==
 
==Problem 4==
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A dfficult mathematical competition consisted of a Part I and a Part II with a combined total of <math>28</math> problems. Each contestant solved <math>7</math> problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.
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A difficult mathematical competition consisted of a Part I and a Part II with a combined total of <math>28</math> problems. Each contestant solved <math>7</math> problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.
  
 
[[1984 USAMO Problems/Problem 4 | Solution]]
 
[[1984 USAMO Problems/Problem 4 | Solution]]
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<math>P(x)</math> is a polynomial of degree <math>3n</math> such that
 
<math>P(x)</math> is a polynomial of degree <math>3n</math> such that
  
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\[\begin{eqnarray*}
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<cmath>\begin{eqnarray*}
 
P(0) = P(3) = \cdots &=& P(3n) = 2, \\
 
P(0) = P(3) = \cdots &=& P(3n) = 2, \\
 
P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\
 
P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\
 
P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\
 
P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\
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&& P(3n+1) = 730.\end{eqnarray*}\]
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&& P(3n+1) = 730.\end{eqnarray*}</cmath>
  
 
Determine <math>n</math>.
 
Determine <math>n</math>.
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== See Also ==
 
== See Also ==
 
{{USAMO box|year=1984|before=[[1983 USAMO]]|after=[[1985 USAMO]]}}
 
{{USAMO box|year=1984|before=[[1983 USAMO]]|after=[[1985 USAMO]]}}
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{{MAA Notice}}

Latest revision as of 11:33, 18 July 2016

Problems from the 1984 USAMO.

Problem 1

The product of two of the four roots of the quartic equation $x^4 - 18x^3 + kx^2+200x-1984=0$ is $-32$. Determine the value of $k$.

Solution

Problem 2

The geometric mean of any set of $m$ non-negative numbers is the $m$-th root of their product.

$\quad (\text{i})\quad$ For which positive integers $n$ is there a finite set $S_n$ of $n$ distinct positive integers such that the geometric mean of any subset of $S_n$ is an integer?

$\quad (\text{ii})\quad$ Is there an infinite set $S$ of distinct positive integers such that the geometric mean of any finite subset of $S$ is an integer?

Solution

Problem 3

$P$, $A$, $B$, $C$, and $D$ are five distinct points in space such that $\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta$, where $\theta$ is a given acute angle. Determine the greatest and least values of $\angle APC + \angle BPD$.

Solution

Problem 4

A difficult mathematical competition consisted of a Part I and a Part II with a combined total of $28$ problems. Each contestant solved $7$ problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.

Solution

Problem 5

$P(x)$ is a polynomial of degree $3n$ such that

\begin{eqnarray*} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray*}

Determine $n$.

Solution

See Also

1984 USAMO (Problems • Resources)
Preceded by
1983 USAMO 		Followed by
1985 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

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