Uzbekistan NMO 2011 2-round

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Uzbekistan NMO 2011 2-round

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#1 h Jun 21, 2011, 10:02 AM • 1 Y

Y by Adventure10

Find the minimum value of
$|x-y|+\sqrt{(x+2)^2+(y-4)^4}$

This post has been edited 1 time. Last edited by shohvanilu, Jan 26, 2012, 1:35 PM

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pco

#2 h Jun 21, 2011, 3:31 PM • 3 Y

Y by shohvanilu, Adventure10, Mango247

shohvanilu wrote:

Find the minimum value of
$|x-y|+\sqrt{(x+2)^2+(y-4)^4}$

You should change your source of olympiad training problems (or stop just inventing crazy problems)

Replacing $x\to x-2$ and $y\to a+4$ , we can define $f(x)=|x-a-6|+\sqrt{x^2+a^4}$

If $x\ge a+6$ , we get $f'(x)=\frac{x+\sqrt{x^2+a^4}}{\sqrt{x^2+a^4}}\ge 0$ and so minimum is reached when $x=a+6$

If $x\le a+6$ , we get $f'(x)=\frac{x-\sqrt{x^2+a^4}}{\sqrt{x^2+a^4}}\le 0$ and so minimum is reached when $x=a+6$

$f(a+6)=\sqrt{(a+6)^2+a^4}$ and we are looking for $a$ such that this value is minimum

So we are looking for the minimum of $g(x)=x^4+(x+6)^2=x^4+x^2+12x+36$

$g'(x)=4x^3+2x+12$
$g"(x)=12x^2+2>0$

So $g(x)$ is decreasing on $(-\infty,r]$ and increasing on $[r,+\infty)$ and the requested value is $\sqrt{g(r)}$ where $r$ is the unique real root of $x^3+\frac 12x+3=0$

This is a standard basic cubic in Cardano's form and we immediately get $r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}}$ $-\sqrt[3]{\frac{\sqrt{2922}+54}{36}}$

Hence the answer :
Minimum is $\sqrt{r^4+r^2+12r+36}$ where $r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}}$ $-\sqrt[3]{\frac{\sqrt{2922}+54}{36}}$

Numeric application gives :
$r\sim -1.32695628567897...$
$\min_{x,y\in\mathbb R}(|x-y|+\sqrt{(x+2)^2+(y-4)^4}$ $\boxed{\sim 4.99377611842389...}$

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#3 h Jun 21, 2011, 9:40 PM • 2 Y

Y by Adventure10, Mango247

I hope he meant $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$ (the power $4$ was just a typo); then at least it has a nice geometric meaning and a nice, tame solution.

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#4 h Jun 22, 2011, 5:33 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

mavropnevma wrote:

I hope he meant $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$ (the power $4$ was just a typo); then at least it has a nice geometric meaning and a nice, tame solution.

Maybe it was $|x-y|+\sqrt{(x+2)^0+(y-4)^0}$ , which seems easy too.

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#5 h Jan 5, 2013, 8:42 AM • 2 Y

Y by Adventure10, Mango247

is there a solution for $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$

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#6 h Jan 6, 2013, 9:42 AM • 4 Y

Y by jayden94941, Adventure10, Mango247, and 1 other user

You can fix $|x-y|=r>0$ . Then for each $r$ , let $f(r)$ be the minimum distance (in $r$ ) of any line in the curve $|x-y|=r$ to the point $(-2,4)$ . Then, minimize $r+f(r)$ .

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Sardor

#7 h Jun 13, 2014, 3:58 AM • 3 Y

Y by jayden94941, Adventure10, Mango247

Mavropnevma you right it's original version of this problem.

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#9 h Mar 14, 2016, 12:38 AM • 1 Y

Y by Adventure10

Sardor wrote:

Mavropnevma you right it's original version of this problem.

$|x-y|+\sqrt{(x+2)^2+(y-4)^2}\ge|x-y|+\frac{\sqrt{2}}{2}|(x+2)-(y-4)|$
$=|x-y|+|3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y)|\ge |x-y|+3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y)$
$\ge (1-\frac{\sqrt{2}}{2})|x-y|+3\sqrt{2}\ge 3\sqrt{2}$

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#10 h Mar 25, 2017, 2:34 AM • 2 Y

Y by Adventure10, Mango247

CAn we substitute x by cost-2 and y by sint +4?

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