IZHO 2023, P2 masalani qaraylik:
▻ Aytaylik, teng yonli bo`lmagan $ABC$ uchburchakning $C$ uchidan uning tashqi aylanasiga o`tkazilgan urinmasi $AB$ tomonni $D$ nuqtada kesadi. Fazaz qilaylik, $D$ nuqtadan o`tuvchi $l$ to`g`ri chiziq olingan va bu to`g`ri chiziq $AC$ va $BC$ tomonlarni mos ravishda $K$ va $L$ nuqtalarda kesib o`tadi. $M$ va $N$ nuqtalar $AB$ tomonda shunday olingan-ki, $AC\parallel NL$ va $BC\parallel KM$ shart o`rinli. Aytaylik, $NL$ va $KM$ to`g`ri chiziqlar uchburchak ichida $P$ nuqtada kesishadi. Agar $CP$ to`g`ri chiziq $MNP$ uchburchakning $\omega$ tashqi aylanasini ikkinchi bor $Q$ nuqtada kesib o`tsa, u holda $DQ$ ning $\omega$ ga urinma bo`lishini ko`rsating.
Uch xil yechim beramiz. Barcha yechimlarda avvalo quyidagilar kerak bo'ladi.
$KA \parallel LN$ va $KM\parallel LB$ lar uchun Fales teoremasini qo'llash orqali \[ \frac{DA}{DN}=\frac{DK}{DL}=\frac{DM}{DB} \quad \quad \Rightarrow \quad \quad DA\cdot DB=DN\cdot DM \] ni topamiz. Demak, $DC^2=DN\cdot DM$ va masala bizdan $DQ=DC$ bo'lishi kerakligini isbotlashni talab qilmoqda, va bu $DQ$ ning $\omega$ ga urinma bo`lishini kafolatlaydi. Boshlanishida $CP$ ning $AB$ tomonni kesib o'tish nuqtasini $X$ deb olamiz. Ko'rish qiyin emaski, $X$ nuqta $NMP$ va $ABC$ uchburchaklar uchun homotetiya markazi bo'ladi (analitik yondashuv, aynan mana shu homotetik o'xshashlikning koeffisienti atrofida aylanadi...).
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1-yechim. Aytaylik, $NL$ va $KM$ lar davom ettirilgan va $DC$ to`g`ri chiziqni mos ravishda $Z$ va $Y$ nuqtalarda kesib o`tishadi (1-chizma ga qarang). U holda $\angle CZL = \angle ABC$ va $\angle CYK=\angle BAC$ dan bizda 3 ta `cyclic' to`rtburchaklar bor; bular $AYCM$, $BZCN$ va $NYZM$. Demak, \[ DC^2 = DA\cdot DB = DN\cdot DM =DY\cdot DZ \] ekan.
Endi, markazi $D$ nuqtada bo'lgan radiusi $R:=DC$ ga teng(musbat yo`nalishli) $\displaystyle I_{D}^{R}$ inversiyani qaraylik. U holda \[ I_D^R(A) = B, \quad I_D^R(B) = A, \quad I_D^R(Y) = Z, \quad I_D^R(Z) = Y, \quad I_D^R(N) = M, \quad I_D^R(M) = N \]
va $C$ nuqtani o'zida saqlab qolishidan, inversiya natijasida $(AYCM)$ aylananing $(BZCN)$ aylanaga o'tishi, shuningdek, $(ABC)$, $\omega$ va $(NYZM)$ aylanalarni o'z o'rnida qolishi keladi.
Bundan tashqari, $\angle NQC=\angle NQP=\angle NMP=\angle NMY=\angle NZC$ dan $Q$ nuqtaning $(BZCN)$ aylanada yotishini topamiz. Xuddi shunday, $Q$ nuqta $(AYCM)$ aylanada ham yotadi. Demak, $Q$ nuqta ham $I_D^R$ inversiya tasir qilinganda qo'zg'almasdan (o'zgarmasdan) qolar ekan. Ya'ni, $DQ=R=DC$, va isbot tugadi. ▢
2-yechim. Analitik yondashuvga qaytadigan bo'lsak, $X$ nuqtaning $NMP$ va $ABC$ uchburchaklar uchun homotetiya markazi ekanligidan foydalanamiz. Faraz qilaylik, $NMP$ va $ABC$ uchburchaklarni $X$ nuqta atrofida homotetik o'xshashlik koeffisienti $k$ bo'lsin.
Agar $AX$ to'g'ri chiziq $(ABC)$ aylanani ikkinchi bor $R$ nuqtada kesadi desak, $\omega$ ga $Q$ nuqtada o'tkazilgan urinma $(ABC)$ ga $R$ nuqtada o'tkazilgan urinmaga parallel bo'ladi. Bundan $DQ$ ning $\omega$ ga urinma bo`lishini ko'rsatish uchun $DQ\parallel TR$ yoki $\frac{TX}{DX}=\frac{RX}{QX}=k$ ni isbotlash yetarli, bu yerda $T$ nuqta $(ABC)$ ga $R$ nuqtadan o'tkazilgan urinmaning $AB$ bilan kesishish nuqtasidir (2-chizma ga qarang).
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| 2-chizma |
Aytaylik, $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle ACX=x$, $\angle BCX=y$ va $\angle KDM = \varphi$ bo'lsin. Shartli ravishda, 2-chizma bo'yicha davom etamiz va $\alpha>\beta$, $y>x$ deb faraz qilamiz. U holda $\angle CDA = \alpha-\beta$ va $\angle RTA = y-x$ bo'ladi.
Birinchidan, $TXR$ va $DXC$ uchburchaklarda sinuslar teoremasini qo'llashdan \[ \frac{TX}{RX} = \frac{\sin (\beta+x)}{\sin (y-x)}, \quad \quad \quad \frac{DX}{CX} = \frac{\sin (\beta+x)}{\sin (\alpha - \beta)} \] va \[ \frac{TX}{DX} = \frac{RX}{CX}\cdot \frac{\sin (\alpha -\beta)}{\sin (y-x)} \] kelib chiqadi. Qo'shimchasiga $\displaystyle \frac{RX}{CX} = \frac{\sin y \, \sin x}{\sin \alpha \, \sin \beta}$ dan \[ \frac{TX}{DX} = \frac{\sin y \, \sin x \, \sin (\alpha -\beta)}{\sin \alpha \, \sin \beta \, \sin (y-x)} \]
ekan.
Ikkinchidan, $k$ ni topadigan bo'lsak, \[ k=\frac{CX}{PX} = \frac{1}{1-\frac{CP}{CX}} \] dan biz oldiniga $\displaystyle \frac{CP}{CX}$ nisbatni topamiz. Yana ba'zi uchburchaklar uchun sinuslar teoremasidan \[ \frac{CP}{CX} = \frac{CP}{CK}\cdot \frac{CK}{CD}\cdot \frac{CD}{CX} = \frac{\sin (\alpha-\beta -\varphi) \, \sin (x+y) \, \sin (\beta +y)}{\sin (\alpha-\varphi) \, \sin y \, \sin (\alpha-\beta)} \] tenglik keladi. Demak, $k$ ni ham topa olamiz va oxir-oqibatda $\frac{TX}{DX}=k$ ni ko`rsatish uchun \[ \frac{\sin (\alpha-\varphi) \, \sin y \, \sin (\alpha-\beta)}{\sin (\alpha-\varphi) \, \sin y \, \sin (\alpha-\beta) - \sin (\alpha-\beta -\varphi) \, \sin (x+y) \, \sin (\beta +y)} = \frac{\sin y \, \sin x \, \sin (\alpha -\beta)}{\sin \alpha \, \sin \beta \, \sin (y-x)} \] tenglikni, yoki ba'zi hadlarni qisqartirib soddalashtiradigan bo'lsak \[ \frac{\sin (\alpha -\beta -\varphi)}{\sin (\alpha - \varphi)} = \frac{\sin x \, \sin y \, \sin (\alpha-\beta) - \sin \alpha \, \sin \beta \, \sin (y-x)}{\sin x\, \sin (x+y)\, \sin (\beta+y)} \] ni isbotlash qoladi.
Endi, $CLPK$ ni parallelogramligidan foydalanadigan bo`lsak, bizda \[ \frac{\sin (\alpha-\varphi)}{\sin (\beta+\varphi)} = \frac{\sin x}{\sin y} \] tenglik ham keladi. Bundan biz $\varphi$ ni qisqartirishda foydalanishimiz mumkin. Xususan, yuqoridagi tenglikdan \[ \tan \varphi = \frac{\sin \alpha \, \sin y - \sin \beta \, \sin x}{\cos \, \alpha \sin \, y + \cos \, \beta \, \sin x} \] ni topib olamiz.
Isbotlashimiz kerak bo'lgan tenglikga qaytadigan bo'lsak, oldiniga yuqoridagi $\tan \varphi$ ni qo'llash orqali \[ \frac{\sin (\alpha-\beta -\varphi)}{\sin (\alpha -\varphi)} = \frac{\sin (\alpha -\beta) - \cos (\alpha-\beta) \, \tan\varphi}{\sin \alpha - \cos \alpha \,\tan \varphi} = \frac{\sin \alpha \, \sin x - \sin \beta \, \sin y}{\sin x\, \sin (\alpha+\beta)} \] ni olamiz. Endi, $\alpha+\beta+y+x=\pi$ ligini inobatga olgan holda bizda $\sin (\alpha+\beta)=\sin (x+y)$ va $\sin (\alpha+x) = \sin (\beta + y)$ lar bor va yuqoridagi tenglik quyidagicha soddalashadi \[ \sin \alpha \, \sin x \, \sin (\beta+y)-\sin \beta\, \sin y \, \sin(\alpha+x) =\sin x\, \sin y\, \sin (\alpha-\beta) - \sin \alpha \, \sin \beta \, \sin (y-x), \] qaysikim, albatta o`rinli. Isbot tugadi. ▢
3-yechim. Yana bitta proyektiv yechim beramiz. Bu safar harmonik nuqtalardan foydalanmiz.
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| 3-chizma |
Aytaylik, $CS$ kesma $ACB$ uchburchakning symmedianasi bo'lsin. U holda $\{D,A,S,B\}$ nuqtalar harmonik bo'ladi. Agar $CS\cap KL = Z(.)$ desak (3-chizma ga qarang), $\{D,K,Z,L\}$ to'rtlik ham harmonikdir. U holda $PZ\cap NM = Y(.)$ desak, $\{D,N,Y,M\}$ nuqtalar ham harmonik va biz $QY$ ning $NAM$ uchburchak uchun symmediana bo'lishini ko'rsatish masalasi qoladi, xolos. Demak, \[ \displaystyle \frac{YM}{YN} = \frac{QM^2}{QN^2} = \frac{\sin^2 \angle MPQ}{\sin \angle NPQ^2} = \frac{\sin^2 \angle KPC}{\sin^2 \angle LPC} = \frac{CK^2}{CL^2} \] ni ko'rsatish yetarli.
$\{D,N,Y,M\}$ nuqtalar harmonik ekanligidan, $\displaystyle \frac{DM}{DN} = \frac{CK^2}{CL^2}$ ni ham ko'rsatish yetarlidir.
Yana $KM\parallel BL$ va $AK\parallel NL$ lardan, shuningdek $\{D,K,Z,L\}$ larning harmonik ekanligidan foydalanadigan bo'lsak, \[ \frac{DM}{DN} = \frac{DK^2}{DL^2}\cdot \frac{DB}{DA} = \frac{KZ^2}{LZ^2}\cdot \frac{DB}{DA} \] keladi. Ma'lum-ki, \[ \frac{KZ^2}{LZ^2} = \frac{\sin^2 \angle KCZ}{\sin^2 \angle LCZ} \cdot \frac{CK^2}{CL^2}. \] Bundan, \[ \frac{DM}{DN} = \frac{\sin^2 \angle KCZ}{\sin^2 \angle LCZ} \cdot \frac{CK^2}{CL^2} \cdot \frac{DB}{DA} = \frac{AS}{BS}\cdot \frac{DB}{DA}\cdot \frac{CK^2}{CL^2} = \frac{CK^2}{CL^2} \] keladi. Isbot tugadi. ▢