IMO2023 dan P2 ni ko'rib chiqamiz:
▻ $ABC$ uchburchakda $AB<AC$ va $S$ nuqta $BAC$ katta yoyning o'rtasi bo'lsin. Aytaylik, uchburchakning $A$ dan tushirilgan balandligi tashqi aylanani ikkinchi bor $E$ nuqtada kesadi. $BS$ va $AE$ lar $D$ nuqtada kesishsin. $D$ nuqtadan $BC$ ga parallel qilib o'tkazilgan t/ch $BE$ bilan $L$ nuqtada uchrashsin. $(LBD)$ va $(ABC)$ aylanalar esa ikkinchi bor $P$ nuqtada kesishadi deb qaraylik. U holda $(LBD)$ aylanaga $P$ nuqtadan o'tkazilgan urinmaning $BS$ bilan kesishish nuqtasi bo'lmish $K$ nuqta, $BAC$ burchak bissektrisasida yotishini isbotlang.
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| 1-chizma |
Masalaga turlicha yondashuvlardan turlicha yechimlarni beramiz.
1-yechim. (sintetik yondashuv) Ma'lum-ki, $\angle BPD = \angle BLD = 90^{\circ} - \angle C$ dan $PD$ ning $A$ nuqtaning diametral qarama-qarshi nuqtasi bo'lmish $A'$ dan o'tadigan bo'ladi (2-chizma ga qarang). Endi, ko'rish qiyin emas-ki, $S,P,L$ nuqtalar collinear bo'ladi. Bundan tashqari, $(LBD)$ ning $AE$ ni ikkinchi bor kesishish nuqtasini $D'$ nuqta desak, $\angle LDD'=90^{\circ}$ dan $\angle LPD' = 90^{\circ}$ va $P,D',S'$ larning collinear ekanligi keladi, bu yerda $S'$ nuqta $BC$ kichik yoyning o'rtasi. $D'$ nuqta shunday nuqtaki, $\angle D'BC=\angle C$ va $\angle ABD=\angle D'BD = \frac{\angle B - \angle C}{2}$ tengliklar bajariladi.
Faraz qilaylik, $BS$ kusma $\angle BAC$ ning bissektrisasini $K$ nuqtada kesadi. Biz $KP$ ning $(LBD)$ aylanaga urinma ekanligini ko'rsatamiz. Yuqoridagi topilganlar bo'yicha davom etadigan bo'lsak, \[ \angle KAD' = \angle KBD' = \angle KBA \] va $AKD'B$-cyclic hamda $KA=KD'$ bo'ladi. Bundan tashqari, \[ KD'^2 = KD\cdot KB \] va $KD'$ t/ch $(LBD)$ ga urinema bo'ladi; $KP$ ning ham ushbu aylanaga urinma ekanligini ko'rsatish uchun $KP = KD'$ ni ko'rsatish yetarli.
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| 2-chizma |
Aytaylik, $A'P\cap AS' = T(.)$ bo'lsin. U holda $\angle TPD' = \angle TAD'$ dan $TD'PA$ ning cyclic ekanligi keladi. Bu degani, $\angle AD'T = 90^{\circ}$ va $(TD'PA)$ aylanani $AT$ diametrli aylana deyish mumkin. $KA=KD'$ dan esa, $K$ nuqtaning $AT$ ni o'rtasi ekanligi va $(TD'PA)$ aylananing ham markazi ekanligi keladi. Demak, $KP = KA = KT = KD'$ va kerakli tasdiq isbotlandi. ▢
2-yechim. (analitik yondashuv: Ozodbek Akhtamov tomonidan olimpiada jarayonida berilgan yechim)
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| 3-chizma |
Aytaylik, $AE\cap BC = H(.)$. Bilamiz-ki, $S,P,L$ nuqtalar bir to'g'ri chiziqda yotadi. Aytaylik, $K$ nuqta sifatida $\angle A$ ning bissektrisasi bilan $BS$ ning kesishish nuqtasini, $K'$ nuqta sifatida esa $(BDL)$ ga $P$ nuqtadan o'tkazilgan urinmaning $BS$ bilan kesishish nuqtasi olingan (3-chizma ga qarang). Bizning asosiy maqsadimiz, $\displaystyle \frac{SK}{BK}=\frac{SK'}{BK'}$ nisbatlarni tengligini ko'rsatishdir.
Umumiylikka zarar yetkazmasdan, $\angle A = \alpha$, $\angle B = \beta$, $\angle C = \gamma$ deb olaylik. \[ \frac{SK}{BK} = \frac{SA}{AB}\cdot \frac{\sin \angle SAK}{\sin \angle BAK} = \frac{SA}{AB}\cdot \frac{1}{\sin \frac{\alpha}{2}} \quad \quad (1) \] ekanligi bor. Boshqa tomondan, \[ \frac{SK'}{BK'} = \frac{SP}{BP}\cdot \frac{\sin \angle SPK'}{\sin \angle BPK'} = \frac{SP}{BP} \cdot \frac{PL}{PB} \] ham bor. $\triangle SPB\sim \triangle SDL$ dan $\displaystyle \frac{SP}{BP} = \frac{SD}{DL}$ va \[ \frac{SK'}{BK'} = \frac{SD}{DL}\cdot \frac{PL}{PB} = \frac{SD}{BH\cdot \frac{DE}{HE}}\cdot \frac{PL}{PB} = \frac{SD\cdot HE\cdot PL}{BH\cdot DE\cdot PB} = \frac{EH}{BH}\cdot \frac{AD}{BD}\cdot \frac{PL}{PB}. \quad (2) \] Demak, (1) va (2) dan \[ \frac{SA}{AB\cdot \sin\frac{\alpha}{2}} = \frac{EH}{BH}\cdot \frac{AD}{BD}\cdot \frac{PL}{PB} \] ni isbotlashimiz yetarliligi keladi, boshqacha aytganda \[ \frac{PL}{PB} = \frac{SA}{AB}\cdot \frac{1}{\sin \frac{\alpha}{2}} \cdot \frac{BH}{EH}\cdot \frac{BD}{AD} = \frac{\sin \frac{\beta - \gamma}{2}}{\sin \gamma}\cdot \frac{1}{\sin \frac{\alpha}{2}}\cdot \frac{\sin \gamma}{\cos \gamma}\cdot \frac{\cos \beta}{\sin \frac{\beta - \gamma}{2}} = \frac{\cos \beta}{\sin \frac{\alpha}{2}\cdot \cos \gamma} \] bo'lishi lozim.
Endi, $\frac{PL}{PB}$ ni boshqa tomondan topamiz: \[ \frac{PL}{PB} = \frac{BL\cdot EL}{BS\cdot DL} = \frac{BL}{DL}\cdot \frac{EL}{ED}\cdot \frac{ED}{BS} = \frac{\cos \frac{\alpha}{2}}{\sin \angle DBE} \cdot \frac{1}{\cos \gamma} \cdot \frac{ED}{BE}\cdot \frac{BE}{BS}= \] \[ = \frac{\cos \frac{\alpha}{2}}{\sin \angle DBE}\cdot \frac{1}{\cos \gamma}\cdot \frac{\sin\angle DBE}{\sin \frac{\alpha}{2}} \cdot \frac{\cos \beta}{\cos \frac{\alpha}{2}} = \frac{\cos \beta}{\cos \gamma \sin \frac{\alpha}{2}}. \] Demak, $\displaystyle \frac{SK}{BK} = \frac{SK'}{BK'}$ va $K\equiv K'$. Isbot tugadi. ▢
3-yechim. (analitik yondashuv: yuqoridagiga alternativ ravishda olimpiadadan keyin yechildi) 3-chizma bo'yicha davom etamiz. Aytaylik, $\angle A=\alpha$, $\angle B = \beta$, $\angle C=\gamma$ va $\angle BA'P = x$.
U holda $\angle AA'P = \gamma - x$, $\angle ABS = \frac{\beta - \gamma}{2}$, $\angle BPK = \angle BLP = 90^{\circ} - x+\frac{\beta -\gamma}{2}$ va $\angle APK = \frac{\alpha}{2}+x$ bo'ladi, va Cheva teoremasining trigonometrik ko'rinishidan foydalanadigan bo'lsak ($\triangle APB$ uchun) $PK$, $AS'$, $BS$ larni bir nuqtada kesishishini isbotlash uchun \[ \frac{\sin \angle BPK}{\sin \angle APK}\cdot \frac{\sin \angle PAK}{\sin \angle BAK}\cdot \frac{\sin \angle ABK}{\sin \angle PBK} = 1 \] ni ko'rsatish yetarli. Yuqoridagi nisbatlar ko'paytmasini soddalashtiradigan bo'lsak, \[ \frac{\cos (x-\frac{\beta - \gamma}{2})}{\sin (\frac{\alpha}{2}+x)} \cdot \frac{\sin (\frac{\alpha}{2}+x)}{\sin \frac{\alpha}{2}}\cdot \frac{\sin \frac{\beta - \gamma}{2}}{\cos (\frac{\alpha}{2}+x)}=1 \] ga, bu ham esa \[ \cot x = -\tan \beta + \frac{1}{\sin \gamma \, \cos \beta} \] ga keladi.
Endi, yuqoridagi tenglikni avlida to'g'ri bo'lishini boshqa tomondan keltirib chiqaramiz. $\triangle ABA' $ da Cheva teoremasining trigonometrik ko'rinishidan foydalanadigan bo'lsak, $A'P$, $AE$, $BS$ chevianalar bitta $D$ nuqtada kesishar ekan, \[ \frac{\sin \angle BAD}{\sin \angle A'AD}\cdot \frac{\sin \angle AA'D}{\sin \angle BA'D}\cdot \frac{\sin \angle A'BD}{\sin \angle ABD} = 1 \] ekanligi keladi. Demak, \[ \frac{\cos \beta}{\sin (\beta -\gamma)} \cdot \frac{\sin (\gamma -x)}{\sin x} \cdot \frac{\cos \frac{\beta - \gamma}{2}}{\sin \frac{\beta-\gamma}{2}}=1 \] va \[ \frac{\sin (\gamma -x)}{\sin x} = \frac{2\sin^2 \frac{\beta -\gamma}{2}}{\cos \beta} = \frac{1}{\cos \beta} - \cos \gamma - \frac{\sin \beta \sin \gamma}{\cos \beta}, \] \[ \sin \gamma \cot x - \cos \gamma = \frac{1}{\cos \beta} - \cos \gamma - \sin \gamma \, \tan \beta \] keladi. Bu esa \[ \cot x = -\tan \beta + \frac{1}{\sin \gamma \, \cos \beta} \] ni beradi. ▢
4-yechim. (proyektiv yondashuv: Maqsadbek Egamberdiyev tomonidan yechildi) 2-chizma bo'yicha yechamiz. Bizda $P,D,A'$ larning collinear ekanligi bor. Aytaylik, $BS\cap AS' = K(.)$ bo'lsin. Biz $KP$ ning $(LBD)$ ga urinma ekanligini ko'rsatamiz. Buning uchun $KP^2 = KD\cdot KB$ ni ko'rsatish yetarlidir.
Boshqa tomondan, $\displaystyle \angle KAD = \frac{\angle B - \angle C}{2} = \angle KBA$ va $KA^2 = KD\cdot KB$ keladi. Bundan, $KA=KP$ ekanligini ko'rsatish kiloya.
Aytaylik, $AS'\cap A'P = T(.)$ va $AA'\cap BS = Q(.)$ bo'lsin. $\triangle DAQ$ uchburchakda $AK$-ichki bissektrisa, $AS$-tashqi bissektrisa ekanligidan, $\{D,K,Q,S\}$-harmonic bundle hosil qiladi. U holda $A'$ nuqtaga nisbatan pencillarni qarasak, $\{T,K,A,A'S\cap S'A\}$ ning ham harmonik ekanligi keladi. Endi, $AS'A'S$ ning to'g'ri to'rtburchak ekanligi va $AS'\parallel SA'$ dan foydalanadigan bo'lsak, $\{T,K,A,\infty\}$ ning harmonik bo'lishi keladi. Demak, $K$ nuqta $AT$ ning o'rtasi ekan. $\triangle APT$ ning to'g'ri burchakli ekanligidan $KA=KT=KP$ kelib chiqadi, demak, $KP^2 = KA^2 = KD\cdot KB$ va $KP$ t/ch $(LBD)$ aylanaga urinma ekan. ▢
