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>. |
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+ | ==Solution== | ||
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>. | ||
− | First, <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1</math>. | + | First, Vieta's relations give <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1</math>. |
Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>. | Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>. | ||
− | + | The other coefficients give <math>ab+(a+b)(c+d)+cd = 0</math> or <math>ab+(a+b)[-1-(a+b)]-\frac{1}{ab}=0</math>. | |
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− | + | Let <math>a+b=s</math> and <math>ab=p</math>. | |
− | + | Thus, <math>0=ab+ac+ad+bc+bd+cd=p+s(-1-s)-\frac{1}{p}</math>. (1) | |
− | + | Also, <math>0=abc+abd+acd+bcd=p(-1-s)-s/p</math>. | |
− | + | Solving this equation for <math>s</math>, <math>s= \frac{-p^2}{p^2+1}</math>. | |
− | <math>p^{ | + | Substituting into (1): <math>\frac{p^{6}+p^{4}+p^{3}-p^{2}-1}{p(p^2+1)^2}=0</math>. |
Conclusion: <math>p =ab</math> is a root of <math>x^{6}+x^{4}+x^{3}-x^{2}-1=0</math>. | Conclusion: <math>p =ab</math> is a root of <math>x^{6}+x^{4}+x^{3}-x^{2}-1=0</math>. |
Latest revision as of 22:46, 20 September 2022
Problem
If and
are two of the roots of
, prove that
is a root of
.
Solution
Given the roots of the equation
.
First, Vieta's relations give .
Then and
.
The other coefficients give or
.
Let and
.
Thus, . (1)
Also, .
Solving this equation for ,
.
Substituting into (1): .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.