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Difference between revisions of "1985 USAMO Problems"

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Problems from the '''1985 [[USAMO]].'''
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==Problem 1==
 
==Problem 1==
 
Determine whether or not there are any positive integral solutions of the simultaneous equations  
 
Determine whether or not there are any positive integral solutions of the simultaneous equations  
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Determine each real root of
 
Determine each real root of
  
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<math>x^4-(2\cdot10^{10}-1)x^2-x+10^{20}+10^{10}-1=0</math>
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<math>x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0</math>
  
 
correct to four decimal places.
 
correct to four decimal places.

Latest revision as of 14:44, 18 July 2016

Problems from the 1985 USAMO.

Problem 1

Determine whether or not there are any positive integral solutions of the simultaneous equations \[x_1^2+x_2^2+\cdots+x_{1985}^2=y^3, \hspace{20pt} x_1^3+x_2^3+\cdots+x_{1985}^3=z^2\] with distinct integers $x_1,x_2,\cdots,x_{1985}$.

Solution

Problem 2

Determine each real root of

$x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0$

correct to four decimal places.

Solution

Problem 3

Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$. Determine the maximum value of the sum of the six distances.

Solution

Problem 4

There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\lfloor n/2\rfloor -1$ of them, each of whom knows both or else knows neither of the two. Assume that "know" is a symmetrical relation; $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.

Solution

Problem 5

Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$, define $b_m=\min\{n: a_n \ge m\}$, that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$. If $a_{19}=85$, determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$.

Solution

See Also

1985 USAMO (Problems • Resources)
Preceded by
1984 USAMO 		Followed by
1986 USAMO
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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