IZHO 2006 ning 2-masalasini qaraylik:
▻ $ABC$ uchburchakning $AB$ va $AC$ tomonlarida shunday $N$ va $M$ nuqtalar tanlangan-ki, $BN=CM$ bo'ladi. $BM$ va $CN$ kesmalar $P$ nuqtada kesishadi. Faraz qilaylik, $AC$ tomonda shunday $Q$ nuqta tanlangan-ki, $PQ$ to'g'ri chiziq $\angle A$ ning bissektrisasiga parallel bo`ladi. U holda $CQ=AB$ ni isbotlang.
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| 1-chizma |
Ushbu masala nisbatan "troll" masala, qanchalik ko'p o'ylasangiz shunchalik to'liq yechimga borishingiz qiyinlashadi. Yuqoridagi 1-chizma esa ushbu konfiguratsiyada aslida nima bo'layotganini bildiradi. $BN=CM$ konfiguratsiyaga doir boshqa masalalarni ham ko'rganmiz:
➢ bissektrisaga perpendikulyar Gauss t/ch
➢ School 239 MO, 2023 / Senior / P2
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Yechish. Faraz qilaylik, $A(.)$ ning $BC$ tomon o'rtasiga nisbatan simmetrigi $A'(.)$ bo'lsin. U holda $AB=CQ$ ni ko'rsatish uchun $PQ$ ning $A'$ nuqtadan o'tishini ko'rsatish yetarli. Boshqacha qilib aytganda, $A'P$ ning $\angle BA'C$ burchak bissektrisa ekanligini ko'rsatish lozim.
Aytaylik, $\angle PBC=x$, $\angle PCB=y$ and $\angle ABC=\beta$, $\angle ACB=\gamma$ bo'lsin. $\triangle BA'C$ uchun Cheva teoremasining trigonometrik ko'rinishini qo'llaydigan bo'lsak, \[ \frac{\sin \angle BA'P}{\sin \angle CA'P}\cdot \frac{\sin (\beta +y)}{\sin y} \cdot \frac{\sin x}{\sin (\gamma +x)} = 1 \] ni topamiz.
Boshqa tomondan esa, $\triangle BMC$ va $\triangle CNB$ larda sinuslar teoremasini qo'llash orqali \[ \frac{\sin (\gamma+x)}{\sin x} = \frac{BC}{CM} = \frac{BC}{BN} = \frac{\sin (\beta + y)}{\sin y} \] larni ham topish mumkin.
Yuqoridagilardan ko'rish mumkin-ki, $\frac{\sin \angle BA'P}{\sin \angle CA'P} = 1$ va $\angle BA'P=\angle CA'P$. ▢
Nisbatan boshqacharoq yondashuv quyidagicha bo'lishi mumkin:
Alternative yechim. Aytaylik, $PQ$ to'g'ri chiziq $AB$ ni $R$ nuqtada kesadi. U holda albatta bizda $AQ=AR$ bor. Agar $\triangle ABM$ uchun Menelay teoremasini qo'llaydigan bo'lsak (ikki marta): \[ R, Q, P \, - \, \text{t/ch uchun} \, \, \, \, \, \, \frac{BP}{PM}\cdot \frac{MQ}{QA}\cdot \frac{AR}{BR} =1 \, \, \, \Rightarrow \, \, \, \frac{BP}{PM} = \frac{BR}{MQ} \] va \[ C,P,N \, - \, \text{t/ch uchun} \, \, \, \, \, \, \frac{BP}{PM}\cdot \frac{MC}{AC}\cdot \frac{AN}{NB}=1 \, \, \, \Rightarrow \, \, \, \frac{BP}{PM}=\frac{AC}{AN} \] larni topishimiz mumkin. Demak, $\frac{BR}{MQ} = \frac{AC}{AN}$ ekan.
Ma'lumki, $BN+AR=CM+AQ$, ya'ni $BR-AN=AC-QM$. Bundan, $BR+QM=AC+AN$ keladi va yuqoridagi nisbatlar tengligini ham inobatga olsak, \[ \frac{BR}{MQ} = \frac{AC}{AN} \, \, \, \Rightarrow \, \, \, \frac{BR+MQ}{MQ}=\frac{AC+AN}{AN} \, \, \, \Rightarrow \, \, \, MQ=AN \, \, \, \Rightarrow \, \, \, AB=CQ \] ni topamiz.
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Yuqoridagi ikkita yechim bilan cheklanib qolmasdan, boshqa turli yechimlarni ham topish mumkin.