Difference between revisions of "1969 IMO Problems/Problem 2"
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<cmath>f(x_1)=f(x_2)=f(x_1+x_2-x_1)</cmath> | <cmath>f(x_1)=f(x_2)=f(x_1+x_2-x_1)</cmath> | ||
We can get <math>x_2-x_1 = 2k\pi</math> for <math>k\in N^*</math>. Thus, <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | We can get <math>x_2-x_1 = 2k\pi</math> for <math>k\in N^*</math>. Thus, <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | ||
| + | ==Solution 2 (longer)== | ||
| + | By the cosine addition formula, | ||
| + | <cmath>f(x)=(\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}cos{a_n})\cos{x}-(\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}\sin{a_n}})\sin{x}</cmath> | ||
| + | This implies that if <math>f(x_1)=0</math>, | ||
| + | <cmath>\tan{x_1}=\frac{\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}cos{a_n}}{\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}\sin{a_n}}}</cmath> | ||
| + | Since the period of <math>\tan{x}</math> is <math>\pi</math>, this means that <math>\tan{x_1}=\tan{x_1+\pi}=\tan{x_1+m\pi}</math> for any natural number <math>m</math>. That implies that every value <math>x_1+m\pi</math> is a zero of <math>f(x)</math>. | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1969|num-b=1|num-a=3}} | {{IMO box|year=1969|num-b=1|num-a=3}} | ||
Revision as of 11:29, 4 April 2024
Contents
[hide]Problem
Let 
be real constants, 
a real variable, and ![\[f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).\]](http://latex.artofproblemsolving.com/a/b/9/ab9533ac04d0ae6529faefae8b78bcc83390deee.png)
Given that 
prove that 
for some integer 
Solution
Because the period of 
is 
, the period of 
is also .
![\[f(x_1)=f(x_2)=f(x_1+x_2-x_1)\]](http://latex.artofproblemsolving.com/2/f/3/2f302f4c389b1c58009e2b2653c4c6958f896af6.png)
We can get 
for 
. Thus, for some integer
Solution 2 (longer)
By the cosine addition formula,
![\[f(x)=(\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}cos{a_n})\cos{x}-(\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}\sin{a_n}})\sin{x}\]](http://latex.artofproblemsolving.com/3/7/d/37d2fd6a7c5efef93fab19ccc3506adf980b90d1.png)
This implies that if 
,
![\[\tan{x_1}=\frac{\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}cos{a_n}}{\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}\sin{a_n}}}\]](http://latex.artofproblemsolving.com/d/5/c/d5c4ad2059433988cb7b196c68d618aba560e0ed.png)
Since the period of 
is 
, this means that 
for any natural number 
. That implies that every value 
is a zero of .
See Also
| 1969 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
