Difference between revisions of "1984 USAMO Problems"
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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 | ||
| − | + | <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 }\\ | ||
| − | && P(3n+1) = 730.\end{eqnarray*} | + | && P(3n+1) = 730.\end{eqnarray*}</cmath> |
Determine <math>n</math>. | Determine <math>n</math>. | ||
Revision as of 13:42, 17 September 2012
Problem 1
The product of two of the four roots of the quartic equation 
is 
. Determine the value of 
.
Problem 2
The geometric mean of any set of 
non-negative numbers is the -th root of their product.
For which positive integers 
is there a finite set 
of distinct positive integers such that the geometric mean of any subset of is an integer?
Is there an infinite set 
of distinct positive integers such that the geometric mean of any finite subset of is an integer?
Problem 3
and 
are five distinct points in space such that 
, where 
is a given acute angle. Determine the greatest and least values of 
.
Problem 4
A dfficult mathematical competition consisted of a Part I and a Part II with a combined total of 
problems. Each contestant solved 
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.
Problem 5
is a polynomial of degree 
such that
Determine .
See Also
| 1984 USAMO (Problems • Resources) | ||
| Preceded by 1983 USAMO |
Followed by 1985 USAMO | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
