IMO2023 dagi eng qiyin masala bo'lmish (6 ta o'quvchi tomonidangina to'liq yechildi) P6 ni muhokama qilamiz:
▻ $ABC$ muntazam uchburchak ichida $A_1$, $B_1$, $C_1$ nuqtalar olingan bo'lib, $BA_1=CA_1$, $CB_1=AB_1$, $AC_1=BC_1$ va \[ \angle BA_1C + \angle CB_1A+\angle AC_1B = 480^{\circ} \] shartlarni qanoatlantiradi. Aytaylik, $BC_1\cap CB_1 = A_2$, $CA_1\cap AC_1 = B_2$ va $AB_1\cap BA_1=C_2$. Faraz qilaylik, $A_1B_1C_1$ uchburchak turli tomonli. U holda $AA_1A_2$, $BB_1B_2$ va $CC_1C_2$ uchburchaklarning tashqi aylanalari coaxal ekanligini isbotlang.
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| 1-chizma |
Masala IMO2023 da juda muhim rol o'ynadi; garchi 5 ta "perfect score" kuzatilgan bo'lsada, aslida birinchi kundan keyin 50 dan ortiq perfect score kutilayotgan edi!
Yechim. Aytaylik, $\angle CBA_1=\angle BCA_1=\alpha$, $\angle ACB_1=\angle CAB_1=\beta$, $\angle BAC_1=\angle ABC_1=\gamma$ va $AB=BC=CA=a$ bo'lsin. Ko'rish qiyin emas-ki, $\alpha+\beta+\gamma=30^{\circ}$ va $\angle BA_1C = 180^{\circ}-2\alpha$, $\angle BA_2C = 90^{\circ}-\alpha$. Bu degani, $A_1$ nuqta $BA_2C$ uchburchak uchun tashqi aylana markazi vazifasini bajaradi. Shunga o'xshash $B_1$ nuqta $AB_2C$ uchburchak uchun, $C_1$ nuqta esa $AC_2B$ uchburchak uchun tashqi aylana markazi bo'ladi. Bundan \[ \angle AA_1A_2 = |\beta - \gamma|, \] \[ \angle BB_1B_2 = |\gamma - \alpha|, \] \[ \angle CC_1C_2 = |\alpha - \beta| \] ham keladi.
Xuddi, IMOSL 2012/G8 kabi coaxal aylanalarning locusidan foydalanamiz. $(AA_1A_2)$, $(BB_1B_2)$, $(CC_1C_2)$ aylanalarni $\omega_A$, $\omega_B$, $\omega_C$ lar orqali belgilaylik, va $\omega_B$, $\omega_C$ aylanalar berilgan deb qarab, $\omega_A$ aylanani bu aylanalarga coaxial ekanligi ko'rsatamiz. Buning uchun shunday $k$ konstanta mavjud bo'lib, \[ \frac{\mathrm{Pow}_A(\omega_B)}{\mathrm{Pow}_A (\omega_C)} = \frac{\mathrm{Pow}_{A_1} (\omega_B)}{\mathrm{Pow}_{A_1}(\omega_C)}=\frac{\mathrm{Pow}_{A_2}(\omega_B)}{\mathrm{Pow}_{A_2} (\omega_C)}=k \] shart bajarilishini ko'rsatish yetarlidir. Biz barcha powerlarni topib chiqamiz.
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| 2-chizma |
1-qadam: $A(.)$ uchun powerlarni topamiz. 2-chizma bo'yicha harakat qilsak, $\angle BB_1B_2 = \alpha - \gamma$ va $\angle CC_1C_2 = \alpha - \beta$ bo'ladi.
Aytaylik, $AB_2$ t/ch $\omega_B$ ni ikkinchi bor $X$ nuqtada kesib o'tsin. U holda $\angle BXB_2 = \alpha - \gamma $ va $ \angle ABX = 180^{\circ} - \alpha $, \[ \mathrm{Pow}_A(\omega_B) = AB_2\cdot AX = 2\, CB_1 \sin (60^{\circ}-\alpha) \cdot \frac{AB\cdot \sin \alpha}{\sin (\alpha - \gamma)} = \frac{a^2 \sin \alpha \sin (60^{\circ}-\alpha)}{\cos \beta \sin (\alpha-\gamma)} \] keladi. Xuddi shunga o'xshash \[ \mathrm{Pow}_A(\omega_C) = \frac{a^2\sin \alpha \sin (60^{\circ} - \alpha)}{\cos \gamma \sin (\alpha-\beta)} \] ham keladi. Demak, \[ \frac{\mathrm{Pow}_A(\omega_B)}{\mathrm{Pow}_A(\omega_C)} = \frac{\cos \gamma \sin(\alpha - \beta)}{\cos \beta \sin (\alpha-\gamma)}. \]
2-qadam: $A_1(.)$ uchun powerlarni qaraymiz. Aytaylik, $A_1C$ t/ch $\omega_B$ ni ikkinchi bor $Y$ nuqtada ($B_2$ dan tashqari) kesib o'tsin. U holda $\angle BYB_2 = \alpha-\gamma$ va $\angle CBY = \gamma$, \[ \mathrm{Pow}_{A_1}(\omega_B) = A_1B_2 \cdot A_1Y = \frac{AB_2\sin (30^{\circ}-\gamma)}{\cos \alpha}\cdot \frac{BA_1\sin (\alpha+\gamma)}{\sin (\alpha-\gamma)} = \] \[ = \frac{a^2 \sin (60^{\circ}-\alpha) \sin (30^{\circ}-\gamma) \sin (30^{\circ}-\beta)}{2\cos^2 \alpha \cos \beta \sin (\alpha-\gamma)} \] keladi. Xuddi shunday, \[ \mathrm{Pow}_{A_1}(\omega_C) = \frac{a^2\sin (60^{\circ} - \alpha) \sin (30^{\circ}-\gamma) \sin (30^{\circ} - \beta) }{2 \cos^2 \alpha \cos \gamma \sin (\alpha-\beta)}, \] va bundan ko'rish mumkin-ki, \[ \frac{\mathrm{Pow}_{A_1} (\omega_B)}{\mathrm{Pow}_{A_1} (\omega_C)} = \frac{\cos \gamma \sin (\alpha - \beta)}{\cos \beta \sin (\alpha-\gamma)}. \]
3-qadam: $A_2(.)$ uchun powerlarni qarashimiz qoldi, xolos. Aslida ushbu yechimning eng murakkabligi ham shu yerda! Lekin IMOSL 2012/G8 ning 1-yechimini o'rganganlar uchun bu unchalik ham qiyinchilik tug'dirmaydi.
Aytaylik, $A_2B_1$ t/ch $\omega_B$ ni ikkinchi bor $Z$ nuqtada kesib o'tsin. Agar \[ CZ = \frac{CB_2\cdot CY}{CB_1} \] dan foydalanadigan bo'lsak, \[ \mathrm{Pow}_{A_2}(\omega_B) = A_2B_1\cdot A_2Z = A_2B_1\cdot (A_2C + CZ) = \] \[ = A_2B_1 \cdot \left( A_2C + \frac{CB_2\cdot CY}{CB_1} \right) \] kelib chiqadi. Bilamiz-ki, \[ A_2B_1 = \frac{A_2B\sin (30^{\circ}-\gamma)}{\cos \beta} = \frac{a \sin (60^{\circ} - \beta) \sin (30^{\circ} - \gamma)}{\cos \alpha \cos \beta}, \] \[ A_2C = \frac{a \sin (60^{\circ} - \gamma)}{ \cos \alpha}, \quad CY = \frac{a \sin \gamma}{\sin (\alpha - \gamma)} \] va \[ \frac{CB_2}{CB_1} = 2\sin (60^{\circ} - \gamma). \] Bularni o'rni-o'rniga qo'yish orqali \[ \mathrm{Pow}_{A_2} (\omega_B) = \frac{a\sin (60^{\circ}-\beta) \sin (30^{\circ} - \gamma)}{\cos \alpha \cos \beta} \left( \frac{a\sin (60^{\circ}-\gamma)}{\cos \alpha} + \frac{2a \sin \gamma \sin (60^{\circ} - \gamma)}{\sin (\alpha - \gamma)} \right) \] \[ = \frac{a^2 \sin (60^{\circ} - \beta) \sin (30^{\circ} -\gamma) \sin (60^{\circ} - \gamma)}{\cos \alpha \cos \beta} \left( \frac{1}{\cos \alpha} + \frac{2\sin \gamma}{\sin (\alpha - \gamma)} \right) \] \[ = \frac{a^2 \sin (60^{\circ} - \beta) \sin (30^{\circ} -\gamma) \sin (60^{\circ} - \gamma) \sin (30^{\circ} - \beta)}{\cos \alpha \cos \beta \sin (\alpha - \gamma)} \] ni topamiz. Xuddi shunga o'xshash (albatta, simmetrik ravishda), \[ \mathrm{Pow}_{A_2} (\omega_C) = \frac{a^2 \sin (60^{\circ} - \gamma) \sin (30^{\circ} -\beta) \sin (60^{\circ} - \beta) \sin (30^{\circ} - \gamma)}{\cos \alpha \cos \gamma \sin (\alpha - \beta)} \] ham keladi. Ko'rish mumkin-ki, suratlar bir xil va \[ \frac{\mathrm{Pow}_{A_2} (\omega_B)}{\mathrm{Pow}_{A_2} (\omega_C)} = \frac{\cos \gamma \sin (\alpha - \beta)}{\cos \beta \sin(\alpha - \gamma)} \] ni topamiz.
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Yuqoridagi qadamlarni birlashtiradigan bo'lsak, haqiqatdan bizga kerakli power nisbatlar tengligini va \[ k= \frac{\cos \gamma \sin (\alpha - \beta)}{\cos \beta \sin(\alpha - \gamma)} \] bo'lishini ko'ramiz. Demak, $A$, $A_1$, $A_2$ nuqtalardan o'tuvchi $\omega_B$ va $\omega_C$ ga coaxal bo'lgan aylana mavjud va bu $\omega_A$ dir; uchta aylana coaxal bo'ladi. ▢
Izoh. E'tibor berish kerak, $A_1B_1C_1$ ning teng yonli emasligi bizga faqat $A$, $A_1$, $A_2$ (yoki shunga o'xshash boshqalari) larning collinear bo'lib qolmasligida yordam beradi. Aslida collinear bo'lganda ham proyektiv ma'noda t/ch ni o'zini aylana deb qarashimiz mumkin.