Quyidagi ajoyib masalani qaraylik:
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| 1-chizma |
Quyidagi yechim orqali geometrik olimpiada masalalarida kompleks sonlarning (kompleks tekislik va undagi metodlar) qay darajada foydali bo`lishini yana bir-bor ko`rsatamiz.
Yechim. (kompleks sonlar metodi) Ichki aylana bo'lmish $(I)$ ni kompleks tekislikdagi birlik aylana deb qaraymiz; $I$ nuqtaga $0$ kompleks soni (koordinatalar boshi) ni mos qo'yamiz. Albatta, qolgan barcha nuqtalar ushbu aylanadagi $E$, $F$, $G$, $H$ nuqtalar orqali aniqlanadi. Shu boisdan ushbu nuqtalarni 2-chizma da ko'rsatilganidek tartibda, turli $a$, $b$, $c$, $d$ kompleks sonlar bilan belgilab olamiz. U holda urinmalar kesishish nuqtasi haqidagi formuladan $A$, $B$, $C$, $D$ nuqtalarga mos keladigan kompleks sonlarni ham topa olamiz.
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| 2-chizma |
Faraz qilaylik, $XZ\cap IP = U(.)$ va $YW\cap IP = V(.)$; ushbu nuqtalarga mos ravishda $u$ va $v$ kompleks sonlar mos kelsin. U holda ikkita to'g'ri chiziqning kesishish nuqtasini topish formulasidan \[ u = \frac{(x\overline{z} - \overline{x}z)p}{(\overline{z}-\overline{x})p - (z-x)\overline{p}} \] va \[ v = \frac{(y\overline{w} - \overline{y}w)p}{(\overline{t} - \overline{y})p-(t-y)\overline{p}} \] lar keladi. Bizdan so'ralgan tasdiq $u=v$ ekanligini isbotlash bilan teng kuchlidir, ya'ni \[ \frac{x\overline{z} - \overline{x}z}{(\overline{z}-\overline{x})p - (z-x)\overline{p}} = \frac{y\overline{w} - \overline{y}w}{(\overline{t} - \overline{y})p-(t-y)\overline{p}}. \quad \quad (1) \] Yuqoridagi tenglikni ikkala tomoni ham birorta o'zgarmas songa teng chiqishi aniq, chunki, $U$ va $V$ nuqtalar $IP$ da yotadi va $u,v$ kompleks sonlar $p$ ga karrali (ya'ni, bir xil argumentga ega). Bizning maqsadimiz, ushbu sonlar tengligini ko'rsatishdir.
Qulayroq usulda hisoblashni boshlaymiz, avvalo $(1)$ ning chap tomonini soddalashtiramiz (o'ng tomonini ham shunga simmetrik ravishda soddalashtirish mumkin; $x\leftrightarrow y$ va $z\leftrightarrow t$ almashtirishlarni bajarish kerak, xolos).
Avvalo suratdan boshlasak, bizda quyidagilar bor: \[ 9x\overline{z} = \frac{a}{c}+\frac{2a}{a+d} + \frac{2a}{a+b} + \frac{2d}{c+d} +\frac{2b}{b+c} + \] \[ + \frac{4cd}{(c+d)(a+d)}+\frac{4cd}{(c+d)(a+b)}+\frac{4bc}{(b+c)(a+d)}+\frac{4bc}{(b+c)(a+b)}, \] \[ 9\overline{x}z = \frac{c}{a} + \frac{2c}{c+d}+\frac{2c}{b+c}+\frac{2d}{a+d}+\frac{2b}{a+b}+ \] \[ +\frac{4ad}{(a+d)(c+d)} + \frac{4ad}{(a+d)(b+c)} + \frac{4ab}{(a+b)(c+d)}+\frac{4ab}{(a+b)(b+c)} \] va o'xshash hadlari ayirib chiqamiz ($a-c$ ayirmani hosil qilish va qavsdan tashqariga chiqarish uchun) \[ 9x\overline{z} -9\overline{x}z = \frac{a^2-c^2}{ac} + (a-c) \left(\frac{4d}{(a+d)(c+d)}+\frac{4b}{(a+b)(b+c)} \right) + \] \[ +\frac{4cd-4ad}{(a+d)(c+d)}+\frac{4bc-4ab}{(a+b)(b+c)} + \] \[ + \frac{4cd(a+d)(b+c)-4ad(c+d)(a+b)}{(a+b)(b+c)(c+d)(d+a)} + \frac{4bc(a+b)(c+d) - 4ab(b+c)(a+d)}{(a+b)(b+c)(c+d)(d+a)}. \] Ko'rish mumkin-ki, yuqoridagi ifodada qisqaradigan hadlar bor (1- va 2- qatorlarda), shuningdek har bir haddan $(a-c)$ ifodani qavsdan tashqariga chiqarish mumkin: \[ x\overline{z}-\overline{x}z = \frac{a-c}{9}\left( \frac{a+c}{ac} - 4\frac{(b^2+d^2)(a+c)+(ac+bd)(b+d)}{(a+b)(b+c)(c+d)(d+a)} \right). \quad \quad (*) \]
Endi maxrajga o'tsak, bizda quyidagilar bor: \[ 3(z - x) = (c-a)+\frac{2d^2(a-c)}{(a+d)(c+d)}+\frac{2b^2(a-c)}{(a+b)(b+c)} \] \[ = (a-c) \left( -1 +\frac{2b^2}{(b+a)(b+c)} +\frac{2d^2}{(d+a)(d+c)} \right), \] \[ 3(\overline{z}-\overline{x}) = \frac{a-c}{ac} + \frac{2(c-a)}{(a+d)(c+d)} +\frac{2(c-a)}{(b+d)(b+a)} \] \[ = (a-c) \left( \frac{1}{ac} - \frac{2}{(b+a)(b+c)} - \frac{2}{(d+a)(d+c)} \right) \] va \[ (\overline{z} - \overline{x})p - (z-x)\overline{p} = \frac{a-c}{3} \left(\left( \frac{p}{ac}+\overline{p} \right) - \frac{2p+2b^2\overline{p}}{(b+a)(b+c)} - \frac{2p+2d^2\overline{p}}{(d+a)(d+c)} \right). \] Qavs ichkarisini alohida soddalashtiradigan bo'lsak, \[ \frac{p}{ac}+\overline{p} = \frac{\frac{a+c}{ac}bd - b-d}{bd-ac} + \frac{b+d - a-c}{bd-ac} = \frac{a+c}{ac} \] va \[ (bd-ac)(a+b)(b+c)(c+d)(d+a) \left(\frac{p+b^2\overline{p}}{(b+a)(b+c)} + \frac{p+d^2\overline{p}}{(d+a)(d+c)}\right) = \] \[ = ((a+c)bd - (b+d)ac) (d^2+b^2+(b+d)(a+c)+2ac) +\] \[+ (b+d-a-c)(2b^2d^2+bd(b+d)(a+c)+ac(b^2+d^2)) = \] \[ = (a+c) (bd-ac) (b^2+d^2-2bd) + (b+d)(bd-ac)(2bd+2ac+(b+d)(a+c)) = \] \[ = 2(bd-ac) ((b^2+d^2)(a+c)+(b+d)(ac+bd)) \] kelib chiqadi. Endi topilganlarni o'rniga qo'yib, \[ (\overline{z} - \overline{x})p - (z-x)\overline{p} = \frac{a-c}{3} \left( \frac{a+c}{ac} - 4\frac{(b^2+d^2)(a+c)+(ac+bd)(b+d)}{(a+b)(b+c)(c+d)(d+a)} \right) \quad \quad (**) \] ni keltiramiz.
Oxir-oqibatda, $(*)$ va $(**)$ tengliklardan bizga kerak bo'lgan \[ \frac{x\overline{z} - \overline{x}z}{(\overline{z}-\overline{x})p - (z-x)\overline{p}} = \frac{1}{3} = \text{const.} \] ni keltirib chiqaramiz. Xuddi shunday, simmetrik o'xshash ayniyatlar orqali \[ \frac{y\overline{t} - \overline{y}t}{(\overline{t}-\overline{y})p - (t-y)\overline{p}} = \frac{1}{3} = \text{const.} \] ni ham keltirish mumkin. Shu bilan, bizga kerak bo'lgan $(1)$ tenglik kelib chiqadi.
Demak, $XZ$ va $YT$ lar $IP$ da kesishar ekan. ▢
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Yuqoridagi yechim orqali, aslida keragidan ortiqroq natijani keltirib chiqargan bo'ldik. Bizda \[ u=v=\frac{p}{3} \] ham bor ekan. Ya'ni, $XZ$ va $YT$ larning har biri aslida $IP$ kesmani $1:2$ nisbatda bo'ladi. Bu esa "chiroyli sintetik yechimlarni ham keltirib chiqarish mumkindir, balki!" degan umidni uyg'otadi. Biz uchun qiziqish doirasi shu yerda tugagani boisidan, boshqacha yechimlarga o'tishni ma'qul topmadik.