Difference between revisions of "1963 IMO Problems/Problem 5"
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<cmath>\omega\left(\frac{\omega^{14}-1}{\omega^2-1}\right)-\omega^7</cmath> | <cmath>\omega\left(\frac{\omega^{14}-1}{\omega^2-1}\right)-\omega^7</cmath> | ||
But since <math>\omega</math> is a <math>14</math>th root of unity, <math>\omega^{14}=1</math>. The answer is then <math>-\omega^{7}=1</math>, as desired. | But since <math>\omega</math> is a <math>14</math>th root of unity, <math>\omega^{14}=1</math>. The answer is then <math>-\omega^{7}=1</math>, as desired. | ||
+ | |||
+ | ~yofro | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1963|num-b=4|num-a=6}} | {{IMO box|year=1963|num-b=4|num-a=6}} |
Revision as of 19:41, 8 July 2021
Problem
Prove that .
Solutions
Solution 1
Let . We have
Then, by product-sum formulae, we have
Thus .
Solution 2
Let and . From the addition formulae, we have
From the Trigonometric Identity, , so
We must prove that . It suffices to show that .
Now note that . We can find these in terms of and :
Therefore . Note that this can be factored:
Clearly , so . This proves the result.
Solution 3
Let . Thus it suffices to show that . Now using the fact that and , this is equivalent to But since is a th root of unity, . The answer is then , as desired.
~yofro
See Also
1963 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |