Difference between revisions of "1959 IMO Problems/Problem 3"
(The "solution 2" was a complete balooney. The two polynimials are not equivalent by any means) |
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The roots of the first equation are <math> \cos {x} = \frac{-1 \pm \sqrt{5}}{4}</math>, which implies that <math>x</math> is one of two certain multiples of <math>\frac{\pi}{5}</math>. The roots of the second equation are <math>\cos(2x) = \frac{1 \pm \sqrt{5}}{4}</math>. It is straightforward to verify that they result in the same values of <math>x</math>. | The roots of the first equation are <math> \cos {x} = \frac{-1 \pm \sqrt{5}}{4}</math>, which implies that <math>x</math> is one of two certain multiples of <math>\frac{\pi}{5}</math>. The roots of the second equation are <math>\cos(2x) = \frac{1 \pm \sqrt{5}}{4}</math>. It is straightforward to verify that they result in the same values of <math>x</math>. | ||
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Revision as of 09:13, 18 May 2009
Problem
Let be real numbers. Consider the quadratic equation in :
Using the numbers , form a quadratic equation in , whose roots are the same as those of the original equation. Compare the equations in and for .
Solution
Let the original equation be satisfied only for . Then we wish to construct a quadratic with roots .
Clearly, the sum of the roots of this quadratic must be
and the product of its roots must be
Thus the following quadratic fulfils the conditions:
Now, when we let , our equations are
and
The roots of the first equation are , which implies that is one of two certain multiples of . The roots of the second equation are . It is straightforward to verify that they result in the same values of .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1959 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |