Balkan Shortlist 2022, G6:
▻ Aytaylik, bizga $ABC$ uchburchak berilgan, bunda $AB<AC$, va $D$ nuqta $\angle A$ ning bissektrisasining uchburchak tashqi aylanasi bilan ikkinchi bor kesishish nuqtasi bo'lsin. $E$ va $F$ nuqtalar mos ravishda $AB$ va $AC$ tomonlarda olingan bo'lib, $AE=AF$ bajariladi. Aytaylik, $AD$ va $EF$ lar $P$ nuqtada kesishadi, $M$ nuqta esa $BC$ toman o'rtasi. U holda $AM$ to'g'ri chiziq $\triangle AEF$ va $\triangle PMD$ uchburchaklar tashqi aylanalarining umumiy kesishish nuqtasidan o'tishini isbotlang.
Ushbu masala Romanian TST 2022 da ham uchradi (3-kun, 2-m). Masalaga nimaga G6 ligi qiziq; chunki, nisbatan ancha soddaroq geodek tuyildi. Lekin, masala Balkan Shortlist 2022 dan olingani uchun, buni tushunsa bo'ladi; ushbu shortlistdagi G5 ning ham o'ziga xos tarixi bor ☺ (UZBTST 2023 da ham o'sha masaladan foydalanilgan).
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Yechish. Aytaylik, $AD$ bissektrisa $(AEF)$ aylanani ikkinchi bor $Q$ nuqtada kesadi, $AM$ esa $(AEF)$ va $EF$ lar bilan mos ravishda $T$ va $N$ nuqtalarda kesishadi (1-chizma ga qarang). Bizdan $P$, $T$, $M$, $D$ nuqtalarning bitta aylanada yotishini isbotlash talab etilmoqda. Buning uchun aynan $AP\cdot AD = AT\cdot AM$ isbotashimiz yetarlidir.
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1-chizma |
Ma'lum-ki, $\angle APN = \angle ATQ = 90^{\circ}$ va \[ AP\cdot AQ = AN\cdot AT. \quad \quad \quad (1) \] Bundan tashqari, \[ \frac{EN}{FN} = \frac{\sin \angle EAN}{\sin \angle FAN} = \frac{\sin \angle BAM}{\sin \angle CAM} = \frac{\sin \angle B}{\sin \angle C} \] nisbatlardan va $\triangle EQF$ ning teng yonli ekanligidan \[ \frac{\sin \angle EQN}{\sin \angle FQN} = \frac{\sin \angle B}{\sin \angle C} \] ham keladi. Demak, $\angle EQN = \angle B$, $\angle FQN = \angle C$ va $QN\perp BC$.
Yuqoridan ko'rishimiz mumkin-ki, $QN\parallel DM$ va $\triangle AQN\sim \triangle ADM$ bo'ladi. Bu esa \[ \displaystyle \frac{AD}{AQ}=\frac{AM}{AN} \quad \quad \quad (2) \] deganidir. $(1)$ va $(2)$ dan biz xohlagan $AP\cdot AD = AT\cdot AM$ tenglik kelib chiqadi. Isbot tugadi. ▢