ARMO 2010, 11-sinf, 3-m / UZB NO 2012, 5-m:
▻ $ABCD-$cyclic to'rtburchak berilgan; $AC$ va $BD$ diagonallari $K$ nuqtada kesishadi. Aytaylik, $I_1$, $I_2$, $I_3$, $I_4$ lar mos ravishda $ABK$, $BCK$, $CDK$, $DAK$ uchburchaklarning ichki aylana markazlari, $M_1$, $M_2$, $M_3$, $M_4$ lar esa mos ravishda $ABCD$ ning tashqi aylanasining $AB$, $BC$, $CD$, $DA$ yoylarining o'rtalari. U holda $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ lar concurrent ekanligini isbotlang.
Ushbu masala UZB NO 2012 da ham uchragan (Respublika Olimpiadasi 2012, 5-masala). Masalani juda sodda ishlash mumkin; original yechim homotetik to'rtburchaklarni hosil qilishga asoslangan bo'lsada, biz undan chekingan holatda nisbatan farqli yechim berishga harakat qilamiz (balkim, soddaroqhamdir :-)).
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Yechim. Aytaylik, $O$ nuqta $ABCD$ ning tashqi aylanasi markazi bo'lsin. Ko'rish qiyin emas-ki, $I_1I_3\parallel M_1M_3$ va $I_2I_4\parallel M_2M_4$, bundan tashqari $I_1I_3\perp I_2I_4$ va $M_1M_3\perp M_2M_4$. Biz $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ larning kesishish nuqtasi $OK$ da yotishini ko'rsatamiz. Aniqroq qilib aytadigan bo'lsak, ularning barchasi $OK$ ni bir xil nisbatda bo'lishini ko'rsatamiz (tashqaridan).
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| 1-chizma |
Aytaylik, $M_1I_1$ t/ch $OK$ bilan $P$ nuqtada kesishsin. Biz $\displaystyle \frac{PK}{PO}$ nisbatning o'zgarmas ekanligini ($ABCD$ ga nisbatan) ko'rsatamiz, ya'ni ushbu nisbat $M_2I_2$, $M_3I_3$, $M_4I_4$ lar uchun ham bir xil chiqishini ko'rsatamiz.
$O$ nuqtadan $M_1M_3$ ga parallel qilib o'tkazilgan to'g'ri chiziq $M_1I_1$ bilan $X$ nuqtada kesishsin (1-chizma ga qarang). Shuningdek, $KI_1$ bissektrisa $(AKB)$ aylana ikkinchi bor $N_1$ nuqtada kesin o'tsin. U holda $OM_1$ ham $N_1$ nuqtadan o'tadi va \[ \displaystyle \frac{PK}{PO} = \frac{KI_1}{OX} = \frac{KI_1}{I_1N_1} \cdot \frac{I_1N_1}{OX} = \frac{KI_1}{I_1N_1} \cdot \frac{N_1M_1}{M_1O} \] tenglik bajariladi. Agar $N_1I_1=N_1B_1$ ekanligini inobatga olsak, \[ \frac{PK}{PO} = \frac{KI_1}{N_1B_1}\cdot \frac{N_1M_1}{M_1O} = \frac{\frac{r_1}{\sin \frac{\varphi }{2}}}{2R_1\sin \frac{\varphi}{2}}\cdot \frac{\sin \angle N_1BM_1}{\sin \angle OBM_1}\cdot \frac{\sin \angle BOM_1}{\sin \angle BN_1M_1}= \] \[ = \frac{r_1}{2R_1\sin^2 \frac{\varphi}{2}} \cdot \frac{\sin \frac{\angle KBC}{2}}{\cos \frac{\angle KCB}{2}} \cdot \frac{\sin {\angle KCB}}{\cos \frac{\varphi}{2}} \] \[ = \frac{2 \sin \frac{\angle KAB}{2} \sin \frac{\angle KBA}{2}}{\sin \frac{\varphi}{2}} \cdot \frac{\sin \frac{\angle KBC}{2}}{\cos \frac{\angle KCB}{2}} \cdot \frac{\sin {\angle KCB}}{\cos \frac{\varphi}{2}} \] \[ = \frac{8\sin \frac{\angle KAB}{2} \sin \frac{\angle KBA}{2} \sin \frac{\angle KCB}{2} \sin \frac{\angle KBC}{2}}{\sin \varphi} \] \[ = \frac{M_1A\cdot M_2B\cdot M_3C \cdot M_4D}{2R^4 \, \sin \varphi} \] kelib chiqadi, bu yerda $R_1$ va $r_1$ deganda $\triangle KAB$ ning tashqi va ichki aylana radiuslari, $R$ deganda esa $(ABCD)$ ning radiusi nazarda tutilmoqda. Ko'rishimiz mumkin-ki, ushbu nisbat to'rtburchakka nisbatan o'zgarmas miqdorga teng, a'ni $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ larga bog'liq emas. Shunga o'xshash ravishda, qolganlarining ham $OK$ ni shunday nisbatda bo'lishini ko'rsatish mumkin. Bundan tashqari, $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ larning barchasi $OK$ ni tashqarida (kesma tashqarisida) kesishi ham aniq (boshida keltirilgan parallelliklardan keltirish lozim). Demak, ular $OK$ t/ch da yotadigan bitta nuqtada kesishadi. ▢