Difference between revisions of "1977 IMO Problems/Problem 4"
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==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) = 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> | + | <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> |
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>. | 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>. | ||
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To get cancellation on the rightmost terms, note <math> \sin (x+\pi/2) = \cos x, \sin (x-\pi/2) = -\cos x</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>. |
− | <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>. | Let <math> x+arctan\frac{a}{b} = y</math>. |
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 |