Difference between revisions of "1969 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | Because the period of <math>cos(x)</math> is <math>2\pi</math>, the period of <math>f(x)</math> is also <math>2\pi</math>. | + | Because the period of <math>\cos(x)</math> is <math>2\pi</math>, the period of <math>f(x)</math> is also <math>2\pi</math>. |
<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> | ||
Revision as of 12:36, 29 January 2021
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
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 | ||
