Difference between revisions of "1963 IMO Problems/Problem 5"
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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> | ||
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+ | |||
+ | ===Solution 3=== | ||
+ | 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\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. | ||
+ | |||
+ | ~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 20:40, 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 |