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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>.
Line 10: Line 10:
 
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 $a,b,A,B$ be given reals. We consider the function defined by\[f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x).\]Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

Solution

$f(x) = 1-\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0$. $f(x+\pi) = 1+\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0$

Therefore, $\sqrt{A^2+B^2}\sin (2x+arctan\frac{A}{B}) \leq 1$. Since this identity is true for any real $x$, let the sine term be one, $\longrightarrow A^2+B^2 \leq 1$.

To get cancellation on the rightmost terms, note $\sin (x+\pi/2) = \cos x, \sin (x-\pi/2) = -\cos x$.

$f(x+\pi/4) = 1-\sqrt{a^2+b^2}\sin (x+\pi/4+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0$. $f(x-\pi/4) = 1-\sqrt{a^2+b^2}\sin (x-\pi/4+\arctan\frac{a}{b}) + \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0$.

Let $x+arctan\frac{a}{b} = y$. Then $\sqrt{a^2+b^2}(\sin (y+\pi/4) + \sin (y-\pi/4)) \leq 2$ $\sqrt{a^2+b^2} \leq \dfrac{2}{\sqrt{2}(\sin y)}$. Since it's valid for all real $x$ let $\sin y = 1$, 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
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