Difference between revisions of "1963 IMO Problems/Problem 5"
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Ronnie 735 (talk | contribs) m (→Solution 2) |
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Then, by product-sum formulae, we have | Then, by product-sum formulae, we have | ||
− | <cmath>S | + | <cmath>S \cdot 2 \sin{\frac{\pi}{7}} = \sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}-\sin{\frac{2\pi}{7}}+\sin{\frac{6\pi}{7}}-\sin{\frac{4\pi}{7}}=\sin{\frac{6\pi}{7}}=\sin{\frac{\pi}{7}}</cmath> |
Thus <math>S = 1/2</math>. <math>\blacksquare</math> | Thus <math>S = 1/2</math>. <math>\blacksquare</math> | ||
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~yofro | ~yofro | ||
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+ | == Solution 5 == | ||
+ | We let <math>\omega = \mathrm{cis} \ \left(\dfrac{2\pi}{7} \right)</math>. We therefore have <math>w^i</math>, where <math>0 \leq i \leq 6</math>, are the <math>7^{\text{th}}</math> roots of unity. Since <math>\sum_{i = 0}^6 \omega^i = 0</math>, then <math>\sum_{i = 1}^6 \omega^i = -1</math>, so <math>\sum_{i = 1}^3 \mathrm{Re}(\omega^i) = -\dfrac{1}{2}</math>. Therefore, because <math>\mathrm{cis} \ \alpha = \cos \alpha + i \sin \alpha, \mathrm{Re}(\mathrm{cis} \ \alpha) = \cos \alpha</math>, so <cmath>\cos \left(\dfrac{2\pi}{7} \right) + \cos \left( \dfrac{4\pi}{7} \right) + \cos \left(\dfrac{6\pi}{7} \right) = -\dfrac{1}{2}</cmath> | ||
+ | <cmath>\implies -\cos \left(\dfrac{2\pi}{7} \right) - \cos \left(\dfrac{4\pi}{7} \right) - \cos \left(\dfrac{6\pi}{7} \right) = \dfrac{1}{2}</cmath> | ||
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+ | Since <math>\cos \alpha = - \cos(\pi - \alpha)</math>, we have <math>\cos \left(\dfrac{\pi}{7} \right) - \cos \left(\dfrac{2\pi}{7} \right) + \cos \left(\dfrac{3\pi}{7} \right) = \dfrac{1}{2}</math> and we are done <math>\blacksquare</math> | ||
+ | |||
+ | ~[[User: Yiyj1|Yiyj1]] | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1963|num-b=4|num-a=6}} | {{IMO box|year=1963|num-b=4|num-a=6}} |
Latest revision as of 08:05, 1 October 2024
Problem
Prove that .
Solution
Because the sum of the -coordinates of the seventh roots of unity is , we have
Now, we can apply to obtain
Finally, since ,
~mathboy100
Solution 2
Let . We have
Then, by product-sum formulae, we have
Thus .
Solution 3
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 4
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
Solution 5
We let . We therefore have , where , are the roots of unity. Since , then , so . Therefore, because , so
Since , we have and we are done
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 |