Difference between revisions of "1963 IMO Problems/Problem 5"
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Prove that <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}</math>. | Prove that <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}</math>. | ||
− | == Solution 1 == | + | ==Solution== |
+ | |||
+ | Because the sum of the <math>x</math>-coordinates of the seventh roots of unity is <math>0</math>, we have | ||
+ | |||
+ | <cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} + \cos{\frac{14\pi}{7}} = 0</cmath> | ||
+ | <cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} = -1</cmath> | ||
+ | <cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\left(-\frac{6\pi}{7}\right)} + \cos{\left(-\frac{4\pi}{7}\right)} + \cos{\left(-\frac{2\pi}{7}\right)} = -1.</cmath> | ||
+ | |||
+ | Now, we can apply <math>\cos{x} = \cos{\left(-x\right)}</math> to obtain | ||
+ | |||
+ | <cmath>2\left(\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}}\right) = -1</cmath> | ||
+ | <cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2}.</cmath> | ||
+ | |||
+ | Finally, since <math>\cos{x} = -\cos{(\pi-x)}</math>, | ||
+ | |||
+ | <cmath>\cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} = -\frac{1}{2}</cmath> | ||
+ | <cmath>\cos{\frac{\pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{3\pi}{7}} = \frac{1}{2}\textrm{. }\square</cmath> | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | == Solution 2 == | ||
Let <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S</math>. We have | Let <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S</math>. We have | ||
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Thus <math>S = 1/2</math>. <math>\blacksquare</math> | Thus <math>S = 1/2</math>. <math>\blacksquare</math> | ||
− | == Solution | + | == Solution 3 == |
Let <math>a=\sin{\frac{\pi}{7}}</math> and <math>b=\cos{\frac{\pi}{7}}</math>. From the addition formulae, we have | Let <math>a=\sin{\frac{\pi}{7}}</math> and <math>b=\cos{\frac{\pi}{7}}</math>. From the addition formulae, we have | ||
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Clearly <math>b\neq -1</math>, so <math>8b^3-4b^2-4b+1=0</math>. This proves the result. <math>\blacksquare</math> | Clearly <math>b\neq -1</math>, so <math>8b^3-4b^2-4b+1=0</math>. This proves the result. <math>\blacksquare</math> | ||
− | == Solution | + | == Solution 4 == |
Let <math>\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)</math>. Thus it suffices to show that <math>\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1</math>. Now using the fact that <math>\omega^k=\omega^{14+k}</math> and <math>-\omega^2=\omega^9</math>, this is equivalent to | Let <math>\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)</math>. Thus it suffices to show that <math>\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1</math>. Now using the fact that <math>\omega^k=\omega^{14+k}</math> and <math>-\omega^2=\omega^9</math>, this is equivalent to | ||
<cmath>\omega+\omega^3+\omega^5+\omega^7+\omega^9+\omega^{11}+\omega^{13}-\omega^7</cmath> | <cmath>\omega+\omega^3+\omega^5+\omega^7+\omega^9+\omega^{11}+\omega^{13}-\omega^7</cmath> | ||
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~yofro | ~yofro | ||
+ | |||
+ | |||
+ | == 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> | ||
+ | |||
+ | 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 19:34, 14 October 2023
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 |