Difference between revisions of "1977 USAMO Problems/Problem 3"
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== Problem == | == Problem == | ||
− | If <math> a</math> and <math> b</math> are two of the roots of <math> x^4 | + | If <math> a</math> and <math> b</math> are two of the roots of <math> x^4+x^3-1=0</math>, prove that <math> ab</math> is a root of <math> x^6+x^4+x^3-x^2-1=0</math>. |
== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
− | a,b,c,d are roots of equation <math> x^4 | + | a,b,c,d are roots of equation <math> x^4+x^3-1=0</math> then by vietas relation |
− | ab +bc+cd+da+ac+bd= | + | <cmath>ab +bc+cd+da+ac+bd= 0</cmath> |
− | let us suppose ab,bc,cd,da,ac,bd are roots of <math> x^6 | + | let us suppose <math>ab,bc,cd,da,ac,bd</math> are roots of <math> x^6+x^4+x^3-x^2-1=0</math>. |
− | then sum of roots = ab +bc+cd+da+ac+bd=c/a = -b/a=0 | + | then sum of roots <math>= ab +bc+cd+da+ac+bd=c/a = -b/a=0</math> |
− | sum taken two at a time= | + | sum taken two at a time <math>= ab\times bc + bc\times ca +..........=c/a=1</math> |
similarly we prove for the roots taken three four five and six at a time | similarly we prove for the roots taken three four five and six at a time | ||
− | to prove ab,bc,cd,da,ac,bd are roots of second equation | + | to prove <math>ab,bc,cd,da,ac,bd</math> are roots of second equation |
Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>. | Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>. |
Revision as of 23:05, 13 March 2015
Problem
If and
are two of the roots of
, prove that
is a root of
.
Solution
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a,b,c,d are roots of equation then by vietas relation
let us suppose
are roots of
.
then sum of roots
sum taken two at a time
similarly we prove for the roots taken three four five and six at a time
to prove
are roots of second equation
Given the roots of the equation
.
First, .
Then and
.
Remains or
.
Let and
, so
(1).
Second, is a root,
and
is a root,
.
Multiplying: or
.
Solving .
In (1): .
or
.
Conclusion: is a root of
.
See Also
1977 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.