Quyidagi masalani qaraylik (Poland MO dan):
▻ $ABC$ o'tkir burchakli uchburchakning ichki sohasida $X$ nuqta shunday tanlangan-ki, quyidagilar o'rinli: \[ \angle BAX = 2 \angle XBA, \quad \quad \angle CAX = 2\angle XCA. \] Aytaylik, $ABC$ uchburchak tashqi aylanasining $BAC$ katta yoyining o'rtasi $M$ nuqta bo'lsin. U holda $XM=XA$ ni isbotlang.
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| 1-chizma |
Yechim. (kompleks sonlar metodi) Yuqoridagi chizmada keltirilgandek burchaklarni belgilab olamiz. Masalani quyidagicha kompleks tekislikka o'tkazamiz: tashqi aylanani birlik kompleks aylana deb qaraymiz, $A$ nuqtani kompleks $1$ soniga mos qo'yamiz.
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| 2-chizma |
$BX$ tashqi aylanani $Z$ nuqtada kesib o'tsa, unga mos keladigan kompleks son $e^{-2\pi i \theta}$ bo'ladi va ushbu sonni $\zeta$ deb olamiz. Xuddi shunday, $CX$ ning tashqi aylanani kesish nuqtasini $Y$ desak, unga mos keluvchi kompleks son $e^{2\pi i \varphi}$ bo'ladi va uni $\varepsilon$ deb olaylik. Va, $AX$ ni davom ettirib, tashqi aylana bilan kesishish nuqtasini $W$ deb olsak, ushbu nuqtaga $\omega^2$ kompleks sonni mos qo'yamiz, bunda $\omega$ deganda $B$ nuqtani o'z ichiga olgan $AW$ yoyning o'rtasini nazarda tutayapmiz. U holda $B$ va $C$ nuqtalarga mos keluvchi kompleks sonlar mos ravishda $\omega^2 \zeta^2$ va $\omega^2 \varepsilon^2$ bo'ladi. Bundan tashqari, $M$ nuqtaga $-\omega^2\zeta \varepsilon$ kompleks son mos keladi. Aslida $B$ nuqtani o'z ichiga olgan $AW$ yoyning o'rtasini $\omega$ deb olganimizdan, $C$ nuqtani o'z ichiga olgan $AW$ yoyning o'rtasi $-\omega$, va salira $B$ nuqtada $\omega^2\zeta^2$ , $C$ nuqtada esa $(-\omega)^2 \varepsilon^2$ turibdi deb qarashimiz kerak edi. Bilamiz-ki, $BAC$ yoyning o'rtasini $B$ va $C$ nuqtada turgan kompleks sonlar orqali ifodalash mumkin, bunda mos ravishda $b^2$ va $c^2$ turganida $bc$ bo'lar edi. E'tiborli tomoni, biz allaqachon berilgan burchak tengliklardan foydalanib bo'ldik, aynan $\zeta^2$ va $\varepsilon^2$ lar ushbu tengliklarni ifodalaydi. Xullas, yakunda bizda 2-chizma hosil bo'ladi.
Avvalo, $X$ nuqtaga mos keluvchi kompleks sonni $x$ deb oladigan bo'lsak, $X$ nuqtaning $BZ$ va $CY$ lar kesishmasi ekanligidan \[ x=\frac{\omega^2 \zeta^3 (\omega^2\varepsilon^2 +\varepsilon) - \omega^2 \varepsilon^3 (\omega^2\zeta^2+\zeta)}{\omega^2 \zeta^3 - \omega^2 \varepsilon^3} = \frac{\omega^2 \zeta^2\varepsilon^2 + \zeta \varepsilon (\zeta+\varepsilon)}{\zeta^2+\zeta\varepsilon +\varepsilon^2} \] va \[ \overline{x} = \frac{\omega^2\zeta^2+\zeta - \omega^2 \varepsilon^2-\varepsilon}{\omega^2 \zeta^3 - \omega^2\varepsilon^3} = \frac{\omega^2(\zeta+\varepsilon)+1}{\omega^2 (\zeta^2+\zeta\varepsilon +\varepsilon^2)} \] kelib chiqadi. Bundan tashqari, $X$ nuqtaning $AW$ da yotishidan \[ \frac{x-1}{\overline{x}-1} = \frac{\omega^2-1}{\overline{\omega^{2}}-1} = \frac{\omega^2-1}{\omega^{-2}-1} = -\omega^2 \] ham bor, ya'ni $x+\overline{x}\omega^2 = 1+\omega^2$. Agar ushbu tenglikda $x$ va $\overline{x}$ larni o'rniga qo'yadigan bo'lsak, \[ \frac{\omega^2(\zeta^2\varepsilon^2+\zeta+\varepsilon)+\zeta\varepsilon(\zeta+\varepsilon)+1}{\zeta^2+\zeta\varepsilon + \varepsilon^2} = 1+\omega^2 \] va \[ \left(\omega^2(\zeta\varepsilon +\zeta+\varepsilon) + \zeta+\varepsilon + 1 \right)(\zeta-1)(\varepsilon-1) = 0 \] ekanligini topamiz. Albatta, $\varepsilon$ va $\zeta$ lar $1$ dan farqli, shu sababdan, \[ \omega^2(\zeta\varepsilon +\zeta +\varepsilon)+\zeta+\varepsilon+1 = 0 \quad \quad (*) \] bo'lishi lozim.
Isbotlashimiz talab etilgan tasdiqqa kelsak, $XA=XM$ tenglik \[ (x-1)(\overline{x}-1) = (x+\omega^2\zeta\varepsilon)(x+\omega^{-2}\zeta^{-1}\varepsilon^{-1}), \] ya'ni $x+\overline{x}\omega^2\zeta\varepsilon=0$ ga ekvivalentdir. Bunda ham $x$ va $\overline{x}$ larni o'rniga qo'yish orqali \[ \omega^2\zeta^2 \varepsilon^2 +\zeta\varepsilon(\zeta+\varepsilon)+\zeta\varepsilon \left(\omega^2(\zeta+\varepsilon)-1\right) = 0 \] ga teng kuchli ekanligini topamiz. $\zeta$ va $\varepsilon$ lar noldan farqli ekanligidan ushbu tenglik bizda oldindan bor bo'lgan $(*)$ ga ekvivalentdir. Demak, $(*)$ ning o'zi $XA = XM$ ni kafolatlar ekan. ▢
Ba'zi ifodalarning qisqartirilishlarida $\zeta\neq \varepsilon$, $\zeta\neq 1$, $\varepsilon\neq 1$ lardan, ba'zilarida esa $\omega^2\neq 1$ dan foydalandik.