Difference between revisions of "1977 USAMO Problems/Problem 3"
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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>. | 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> | + | 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>. | 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>. |
Revision as of 22:22, 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
, 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.