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Difference between revisions of "1962 IMO Problems/Problem 2"
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<math>\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}</math>
 
<math>\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}</math>
 
</center>
 
</center>
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(Note that the problem as written in the official PDF https://www.imo-official.org/year_info.aspx?year=1962 does not have the outer <math>\sqrt{\ }</math>)
  
 
==Solution==
 
==Solution==
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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(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
 
Solving the quadratic equation for <math>x</math>, we get
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</asy>
 
</asy>
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==Solution 2==
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We first do some substitution, Let <math>a = \sqrt{3-x}</math>, then <math>x = 3 - a^2</math>, and the expression becomes: <math>\sqrt{a-\sqrt{4-a^2}} > \frac{1}{2}</math>. Squaring both sides, we have:
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<cmath>a-\sqrt{4-a^2} > \frac{1}{4} \Rightarrow 4a-1>4\sqrt{4-a^2} \Rightarrow 16a^2-8a+1>16(4-a^2) \Rightarrow 32a^2-8a-63 > 0</cmath>
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Solving yields <math>a < \frac{1-\sqrt{127}}{8}</math> or <math>a > \frac{1+\sqrt{127}}{8}</math>
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Substituting back for <math>x</math>, <math>x < \frac{1-\sqrt{127}}{32}</math> or <math>x > \frac{1+\sqrt{127}}{32}</math>. This is two solutions, however, note that for the square root to be determinant, we need to have:
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1. <math>x \leq 3</math>
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2. <math>x \geq -1</math>
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3. <math>\sqrt{3-x}-\sqrt{x+1} \geq 0 \Rightarrow 3-x \geq x+1 \Rightarrow x \leq 1</math>
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Obviously, <math>x > \frac{1+\sqrt{127}}{32}</math> conflicted with <math>x \leq 1</math>, and the lower bound for x is <math>x \geq -1</math>, so our final solution is:
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<cmath>\boxed{x \in \left[ ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }</cmath>
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~[[User:IDKHowtoaddsolution|IDKHowtoaddsolution]]
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==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 12:12, 4 September 2025

Problem

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

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

(Note that the problem as written in the official PDF https://www.imo-official.org/year_info.aspx?year=1962 does not have the outer $\sqrt{\ }$)

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]

Solution 2

We first do some substitution, Let $a = \sqrt{3-x}$, then $x = 3 - a^2$, and the expression becomes: $\sqrt{a-\sqrt{4-a^2}} > \frac{1}{2}$. Squaring both sides, we have:

\[a-\sqrt{4-a^2} > \frac{1}{4} \Rightarrow 4a-1>4\sqrt{4-a^2} \Rightarrow 16a^2-8a+1>16(4-a^2) \Rightarrow 32a^2-8a-63 > 0\]

Solving yields $a < \frac{1-\sqrt{127}}{8}$ or $a > \frac{1+\sqrt{127}}{8}$

Substituting back for $x$, $x < \frac{1-\sqrt{127}}{32}$ or $x > \frac{1+\sqrt{127}}{32}$. This is two solutions, however, note that for the square root to be determinant, we need to have:

1. $x \leq 3$

2. $x \geq -1$

3. $\sqrt{3-x}-\sqrt{x+1} \geq 0 \Rightarrow 3-x \geq x+1 \Rightarrow x \leq 1$

Obviously, $x > \frac{1+\sqrt{127}}{32}$ conflicted with $x \leq 1$, and the lower bound for x is $x \geq -1$, so our final solution is:

\[\boxed{x \in \left[ ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }\]

~IDKHowtoaddsolution


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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