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Difference between revisions of "1977 IMO Problems/Problem 4"

(Created page with "==Problem== Let <math>a,b</math> be two natural numbers. When we divide <math>a^2+b^2</math> by <math>a+b</math>, we the the remainder <math>r</math> and the quotient <math>q....")
 
 
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==Problem==
 
==Problem==
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Let <math>a,b</math> be two natural numbers. When we divide <math>a^2+b^2</math> by <math>a+b</math>, we the the remainder <math>r</math> and the quotient <math>q.</math> Determine all pairs <math>(a, b)</math> for which <math>q^2 + r = 1977.</math>
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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==
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Using <math>r=1977-q^2</math>, we have <math>a^2+b^2=(a+b)q+1977-q^2</math>, or <math>q^2-(a+b)q+a^2+b^2-1977=0</math>, which implies <math>\Delta=7908+2ab-2(a^2+b^2)\ge 0</math>. If we now assume Wlog that <math>a\ge b</math>, it follows <math>a+b\le 88</math>. If <math>q\le 43</math>, then <math>r=1977-q^2\ge 128</math>, contradicting <math>r<a+b\le 88</math>. But <math>q\le 44</math> from <math>q^2+r=1977</math>, thus <math>q=44</math>. It follows <math>r=41</math>, and we get <math>a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}</math>. By Jacobi's two squares theorem, we infer that <math>15^2+28^2=1009</math> is the only representation of <math>1009</math> as a sum of squares. This forces <math>\boxed{(a,b)=(37,50) , (7, 50)}</math>, and permutations. <math>\blacksquare</math>
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<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>.
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<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>
  
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The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here:
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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>.
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<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>.
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<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>.
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Let <math> x+\arctan\frac{a}{b} = y</math>.
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Then <math> \sqrt{a^2+b^2}(\sin (y+\pi/4) + \sin (y-\pi/4)) \leq 2</math>
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<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.
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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 15:48, 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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