Difference between revisions of "1977 USAMO Problems/Problem 3"
m |
(→Solution) |
||
(5 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== 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== |
− | {{ | + | Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>. |
+ | |||
+ | First, Vieta's relations give <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1</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>. | ||
+ | |||
+ | Let <math>a+b=s</math> and <math>ab=p</math>, so <math>p+s(-1-s)-\frac{1}{p}=0</math>(1). | ||
+ | |||
+ | Second, <math>a</math> is a root, <math>a^{4}+a^{3}=1</math> and <math>b</math> is a root, <math>b^{4}+b^{3}=1</math>. | ||
+ | |||
+ | Multiplying: <math>a^{3}b^{3}(a+1)(b+1)=1</math> or <math>p^{3}(p+s+1)=1</math>. | ||
+ | |||
+ | Solving <math>s= \frac{1-p^{4}-p^{3}}{p^{3}}</math>. | ||
+ | |||
+ | In (1): <math>\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0</math>. | ||
+ | |||
+ | <math>p^{8}+p^{5}-2p^{4}-p^{3}+1=0</math> or <math>(p-1)(p+1)(p^{6}+p^{4}+p^{3}-p^{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>. | ||
== See Also == | == See Also == | ||
{{USAMO box|year=1977|num-b=2|num-a=4}} | {{USAMO box|year=1977|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
− | [[Category:Olympiad | + | [[Category:Olympiad Algebra Problems]] |
Revision as of 09:52, 28 February 2016
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 , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.