Difference between revisions of "1969 IMO Problems/Problem 2"
Durianaops (talk | contribs) m |
(solution 2) |
||
(4 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>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).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>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).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | ||
+ | |||
+ | ==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> | ||
+ | ==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== | ||
+ | |||
+ | {{IMO box|year=1969|num-b=1|num-a=3}} |
Latest revision as of 11:29, 4 April 2024
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
Solution 2 (longer)
By the cosine addition formula, This implies that if , 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 |