Difference between revisions of "1977 IMO Problems/Problem 4"
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==Problem== | ==Problem== | ||
− | Let <math>a,b</math> be | + | Let <math>a,b,A,B</math> be given reals. We consider the function defined by<cmath> f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). </cmath>Prove that if for any real number <math>x</math> we have <math>f(x) \geq 0</math> then <math>a^2 + b^2 \leq 2</math> and <math>A^2 + B^2 \leq 1.</math> |
==Solution== | ==Solution== | ||
− | + | <math> f(x) = 1-\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0</math>. | |
+ | <math> f(x+\pi) = 1+\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0</math> | ||
− | The above solution was posted and copyrighted by | + | Therefore, <math> \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \leq 1</math>. Since this identity is true for any real <math> x</math>, let the sine term be one, <math> \longrightarrow A^2+B^2 \leq 1</math>. |
+ | |||
+ | To get cancellation on the rightmost terms, note <math> \sin (x+\pi/2) = \cos x, \sin (x-\pi/2) = -\cos x</math>. | ||
+ | |||
+ | <math> f(x+\pi/4) = 1-\sqrt{a^2+b^2}\sin (x+\pi/4+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0</math>. | ||
+ | <math> f(x-\pi/4) = 1-\sqrt{a^2+b^2}\sin (x-\pi/4+\arctan\frac{a}{b}) + \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0</math>. | ||
+ | |||
+ | Let <math> x+\arctan\frac{a}{b} = y</math>. | ||
+ | Then <math> \sqrt{a^2+b^2}(\sin (y+\pi/4) + \sin (y-\pi/4)) \leq 2</math> | ||
+ | <math>\sqrt{a^2+b^2} \leq \dfrac{2}{\sqrt{2}(\sin y)}</math>. Since it's valid for all real <math> x</math> let <math> \sin y = 1</math>, and we are done. | ||
+ | |||
+ | The above solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [https://aops.com/community/p1351208] | ||
== See Also == {{IMO box|year=1977|num-b=3|num-a=5}} | == See Also == {{IMO box|year=1977|num-b=3|num-a=5}} |
Latest revision as of 16:48, 29 January 2021
Problem
Let be given reals. We consider the function defined by
Prove that if for any real number
we have
then
and
Solution
.
Therefore, . Since this identity is true for any real
, let the sine term be one,
.
To get cancellation on the rightmost terms, note .
.
.
Let .
Then
. Since it's valid for all real
let
, and we are done.
The above solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]
See Also
1977 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |