ប្រព័ន្ធសមីការ IMO Shortlist 1995

តាង $a,b,c$ ជាចំនួនពិតវិជ្ជមាន។ ចូរកំណត់គ្រប់ចំនួនពិតវិជ្ជមាន $x,y,z$ ដែល

$\left\{\begin{array}{ll}x+y+z=a+b+c\\4xyz-\left(a^2 x+b^2 y+c^2 z\right)=abc\end{array}\right.$

ចម្លើយ

វិធីទី ១

សមីការទីពីរសមមូលនឹង

$\displaystyle \frac{a^2}{yz}+\frac{b^2}{zx}+\frac{c^2}{xy}+\frac{abc}{xyz}=4$

តាង

$\displaystyle x_1=\frac{a}{\sqrt{yz}}, y_1=\frac{b}{\sqrt{zx}}, z_1=\frac{c}{\sqrt{xy}}$

ដូច្នេះ

$x_1^2+y_1^2+z_1^2+x_1 y_1 z_1=4$ (*)

ដូច្នេះ $0<x_1, y_1, z_1<2$ ។ យើងចាត់ទុកសមីការ (*) ជាសមីការដឺក្រេទី២អញ្ញាត $z_1$ ។ យើងមាន

$\Delta =(x_1 y_1 )^2-4(x_1^2+y_1^2-4)=16-4x_1^2-4y_1^2+(x_1 y_1 )^2=(4-x_1^2 )(4-y_1^2 )$

តាង $x_1=2\sin{u}, y_1=2\sin{v}$ ដូច្នេះ $0<u,v<\pi/2$ ។ ដូច្នេះ

$\Delta=16 \cos^2u \cos^2v$

$\implies \displaystyle z_1=\frac{-x_1 y_1\pm \sqrt{\Delta}}{2}=\frac{-4\sin{u}\sin{v}\pm 4 \cos{u}\cos{v}}{2}$

តែ $z>0$ ដូច្នេះ

$\displaystyle z_1=\frac{-4 \sin{u}\sin{v}+4 \cos{u}\cos{v}}{2}=2(\cos{u} \cos{v}-\sin{u}\sin{v})$

ដូច្នេះ

$a=x_1\sqrt{yz}=2\sqrt{yz} \sin{u}$
$b=y_1\sqrt{zx}=2\sqrt{zx}\sin{v}$
$c=z_1 \sqrt{yx}=2\sqrt{yx} (\cos{u} \cos{v}-\sin{u}\sin{v})$

តាមសម្មតិកម្ម

$x+y+z=a+b+c$
$\Leftrightarrow x+y+z=2\sqrt{yz} \sin{u}+2\sqrt{zx} \sin{v}+2\sqrt{yx} (\cos{u} \cos{v}-\sin{u} \sin{v})$
$\Leftrightarrow x(\cos^2v+\sin^2v)+y(\cos^2u+\sin^2u)+z=2\sqrt{yz} \sin{u}+2\sqrt{zx}\sin{v}+2\sqrt{yx}(\cos{u} \cos{v}-\sin{u}\sin{v} )$
$\Leftrightarrow x \cos^2v+2\sqrt{xy} \cos{u} \cos{v}+y \cos^2u+x \sin^2v+y \sin^2u+z-2\sqrt{yz} \sin{u}-2\sqrt{zx} \sin{v}+2\sqrt{yx} \sin{u} \sin{v}=0$
$\Leftrightarrow (\sqrt{x} \cos{v}-\sqrt{y} \cos{u} )^2+(\sqrt{x} \sin{v}+\sqrt{y} \sin{u}-\sqrt{z})^2=0$
ដូច្នេះ

$\left\{\begin{array}{ll}\sqrt{x} \cos{v}=\sqrt{y} \cos{u} &(1)\\\sqrt{z}=\sqrt{x} \sin{v}+\sqrt{y} \sin{u} &(2)\end{array}\right.$

សមីការ (2) នាំឱ្យ

$\displaystyle \sqrt{z}=\sqrt{x} \sin{v}+\sqrt{y} \sin{u}=\sqrt{x}\frac{y_1}{2}+\sqrt{y}\frac{x_1}{2}=\frac{1}{2} \left(\sqrt{x}\frac{b}{\sqrt{zx}}+\sqrt{y}\frac{a}{\sqrt{yz}}\right)$
$\displaystyle \implies z=\frac{a+b}{2}$

ដូចគ្នា យើងទាញបាន

$\displaystyle x=\frac{b+c}{2}, y=\frac{c+a}{2}$

ដូច្នេះប្រព័ន្ធសមីការមានឫសតែមួយគត់គឺ

$\displaystyle (x,y,z)=\left\{\frac{b+c}{2},\frac{c+a}{2},\frac{a+b}{2}\right\}$

វិធីទី២

តាង

$\displaystyle x=\frac{b+c}{2}-u;y=\frac{c+a}{2}-v;z=\frac{a+b}{2}-w$

$\displaystyle \implies u<\frac{b+c}{2}; v<\frac{c+a}{2}; w<\frac{a+b}{2}$

តាមសម្មតិកម្ម $x+y+z=a+b+c \implies u+v+w=0$ និង

$abc+a^2 x+b^2 y+c^2 z=4xyz$

$\displaystyle \Leftrightarrow abc+a^2 \left(\frac{b+c}{2}-u\right)+b^2 \left(\frac{c+a}{2}-v\right)+c^2 \left(\frac{a+b}{2}-w\right)$

$\displaystyle =4\left(\frac{b+c}{2}-u\right)\left(\frac{c+a}{2}-v\right)\left(\frac{a+b}{2}-w\right)$

$\displaystyle \Leftrightarrow abc+\frac{1}{2} a^2 b+\frac{1}{2} a^2 c-a^2 u+\frac{1}{2} b^2 c+\frac{1}{2} ab^2-b^2 v+\frac{1}{2} ac^2+\frac{1}{2} bc^2-c^2 w=\frac{1}{2} (b+c-2u)(c+a-2v)(a+b-2w)$

$\displaystyle \Leftrightarrow 2abc+a^2 b+a^2 c-2a^2 u+b^2 c+ab^2-2b^2 v+ac^2+bc^2-2c^2 w=(bc+ab-2bv+c^2+ac-2cv-2cu-2au+4uv)(a+b-2w)$

$\displaystyle \Leftrightarrow 2abc+a^2 b+a^2 c-2a^2 u+b^2 c+ab^2-2b^2 v+ac^2+bc^2-2c^2 w=abc+b^2 c-2bcw+a^2 b+ab^2-2abw-2abv-2b^2 v+4bvw+ac^2+bc^2-2c^2 w+a^2 c+abc-2acw-2acv-2bcv+4cvw-2acu-2bcu+4cuw-2a^2 u-2abu+4auw+4auv+4buv-8uvw$

$\displaystyle \Leftrightarrow 0=-2bcw-2abw-2abv+4bvw-2acw-2acv-2bcv+4cvw-2acu-2bcu+4cuw-2abu+4auw+4auv+4buv-8uvw$

$\displaystyle \Leftrightarrow 0=-2bc(u+v+w)-2ab(u+v+w)+4bvw-2ac(u+v+w)+4cvw+4cuw+4auw+4auv+4buv-8uvw$

$\displaystyle \Leftrightarrow 0=4bvw+4cvw+4cuw+4auw+4auv+4buv-8uvw$

$\displaystyle \Leftrightarrow 0=4bv(u+w)+4cw(u+v)+4au(v+w)-8uvw$

$\displaystyle \Leftrightarrow 0=-4bv^2-4cw^2-4au^2-8uvw$

$\displaystyle \Leftrightarrow 0=au^2+bv^2+cw^2+2uvw$ (*)

សមភាពនេះអាចពិតទៅរួច ទាល់តែ $uvw \le 0$ ។ ដូច្នេះក្នុងចំណោម $u,v,w$ ត្រូវតែមានយ៉ាងណាស់ពីរ ដែលមានតម្លៃមិនអវិជ្ជមាន, សន្មតថា $u,v\ge 0$ ។ ជំនួស $w=-u-v$ ចូលក្នុងសមភាព(*) យើងទាញបាន