Difference between revisions of "1969 IMO Problems/Problem 2"

Revision as of 07:30, 12 March 2019

Problem

Let $a_1, a_2,\cdots, a_n$ be real constants, $x$ a real variable, and $f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).$ Given that $f(x_1)=f(x_2)=0,$ prove that $x_2-x_1=m\pi$ for some integer $m.$

Solution

Because the period of $cos(x)$ is $2\pi$ , the period of $f(x)$ is also $2\pi$ . $f(x_1)=f(x_2)=f(x_1+x_2-x_1)$ We can get $x_2-x_1 = 2k\pi$ for $k\in N^*$ . Thus, $x_2-x_1=m\pi$ for some integer $m.$

@@ Line 3: / Line 3: @@
 ==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>.
+<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>
 ==See Also==
 {{IMO box|year=1969|num-b=1|num-a=3}}

1969 IMO (Problems) • Resources
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Difference between revisions of "1969 IMO Problems/Problem 2"

Revision as of 07:30, 12 March 2019

Problem

Solution

See Also