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Difference between revisions of "1962 IMO Problems/Problem 2"

(New page: ==Problem== Determine all real numbers <math>x</math> which satisfy the inequality: <center> <math>\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}</math> </center> ==Solution== {{solution}} =...)
 
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==Solution==
 
==Solution==
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{{solution}}
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Obviously we need <math>\sqrt{3-x} \geq \sqrt{x+1}</math> for the outer square root to be defined, <math>x\leq 3</math> for the first inner square root to be defined,
 +
and <math>x\geq -1</math> for the second inner square root to be defined. Solving these we get that the left hand side is defined for <math>x\in \left[ -1,1 \right]</math>.
 +
 
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Now obviously the function <math>f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}</math> is continuous on <math>\left[ -1,1 \right]</math>, with <math>f(-1)=\sqrt 2</math> and <math>f(1)=0</math>.
 +
Moreover, as <math>3-x</math> is a decreasing and <math>x+1</math> an increasing function, both <math>\sqrt{3-x}</math> and <math>-\sqrt{x+1}</math> are decreasing functions, and hence <math>f(x)</math> is a decreasing function. Therefore there is exactly one solution to <math>f(x)=\dfrac{1}{2}</math>.
 +
 
 +
We can now find this solution:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
\sqrt{\sqrt{3-x}-\sqrt{x+1}} &= \dfrac{1}{2}
 +
\\
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\sqrt{3-x}-\sqrt{x+1} &= \dfrac{1}{4}
 +
\\
 +
\sqrt{3-x} &= \dfrac{1}{4} + \sqrt{x+1}
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\\
 +
3-x &= \dfrac 1{16} + x+1 + \dfrac{\sqrt{x+1}}2
 +
\\
 +
2 - 2x - \dfrac 1{16} &= \dfrac{\sqrt{x+1}}2
 +
\\
 +
31 - 32x &= 8\sqrt{x+1}
 +
\\
 +
1024 x^2 - 1984x + 961 &= 64(x+1)
 +
\\
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1024 x^2 - 2048x + 897 &= 0
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\end{align*}
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</cmath>
 +
 
 +
(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)
 +
 
 +
Solving the quadratic equation for <math>x</math>, we get
 +
<cmath>
 +
x_{1,2}=\dfrac{ 2^{11} \pm \sqrt{ 2^{22} - 2^{12}\cdot 897} }{2^{11}} = 1 \pm \dfrac{\sqrt{127}}{32}
 +
</cmath>
 +
 
 +
The reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than <math>1</math> is the additional solution, and <math>x=1-\dfrac{\sqrt{127}}{32}</math> is the root we need.
 +
 
 +
Hence the solutions to the given inequality are precisely the reals in the interval <math>\boxed{ \left[ ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }</math>.
 +
 
 +
<asy>
 +
import graph;
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size(300,300,IgnoreAspect);
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 +
real f(real x) {return sqrt( sqrt(3-x) - sqrt(x+1) );};
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real g(real x) {return 1/2;};
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 +
draw(graph(f,-1,1),blue,"$f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}$");
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draw(graph(g,-1,1),red,"$g(x)=1/2$");
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 +
xaxis("$x$",BottomTop,Ticks);
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yaxis("$y$",LeftRight,Ticks);
 +
 
 +
attach(legend(),point(E),20E,UnFill);
 +
 
 +
</asy>
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1962|num-b=1|num-a=3}}
 
{{IMO box|year=1962|num-b=1|num-a=3}}

Latest revision as of 13:43, 2 February 2009

Problem

Determine all real numbers $x$ which satisfy the inequality:

$\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}$

Solution

Obviously we need $\sqrt{3-x} \geq \sqrt{x+1}$ for the outer square root to be defined, $x\leq 3$ for the first inner square root to be defined, and $x\geq -1$ for the second inner square root to be defined. Solving these we get that the left hand side is defined for $x\in \left[ -1,1 \right]$.

Now obviously the function $f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}$ is continuous on $\left[ -1,1 \right]$, with $f(-1)=\sqrt 2$ and $f(1)=0$. Moreover, as $3-x$ is a decreasing and $x+1$ an increasing function, both $\sqrt{3-x}$ and $-\sqrt{x+1}$ are decreasing functions, and hence $f(x)$ is a decreasing function. Therefore there is exactly one solution to $f(x)=\dfrac{1}{2}$.

We can now find this solution:

\begin{align*} \sqrt{\sqrt{3-x}-\sqrt{x+1}} &= \dfrac{1}{2} \\ \sqrt{3-x}-\sqrt{x+1} &= \dfrac{1}{4} \\ \sqrt{3-x} &= \dfrac{1}{4} + \sqrt{x+1} \\ 3-x &= \dfrac 1{16} + x+1 + \dfrac{\sqrt{x+1}}2 \\ 2 - 2x - \dfrac 1{16} &= \dfrac{\sqrt{x+1}}2 \\ 31 - 32x &= 8\sqrt{x+1} \\ 1024 x^2 - 1984x + 961 &= 64(x+1) \\ 1024 x^2 - 2048x + 897 &= 0 \end{align*}

(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)

Solving the quadratic equation for $x$, we get \[x_{1,2}=\dfrac{ 2^{11} \pm \sqrt{ 2^{22} - 2^{12}\cdot 897} }{2^{11}} = 1 \pm \dfrac{\sqrt{127}}{32}\]

The reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than $1$ is the additional solution, and $x=1-\dfrac{\sqrt{127}}{32}$ is the root we need.

Hence the solutions to the given inequality are precisely the reals in the interval $\boxed{ \left[ ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }$.

[asy] import graph; size(300,300,IgnoreAspect); real f(real x) {return sqrt( sqrt(3-x) - sqrt(x+1) );}; real g(real x) {return 1/2;}; draw(graph(f,-1,1),blue,"$f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}$"); draw(graph(g,-1,1),red,"$g(x)=1/2$"); xaxis("$x$",BottomTop,Ticks); yaxis("$y$",LeftRight,Ticks); attach(legend(),point(E),20E,UnFill); [/asy]

See Also

1962 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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