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 13:36, 29 January 2021
Problem
Let be real constants,
a real variable, and
Given that
prove that
for some integer
Solution
Because the period of is
, the period of
is also
.
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 |