ARMO 2011, Round 3, 11-sinf, 8-masalasini ko'raylik:
▻ Aytaylik, bizga $AB\cdot CD=BC\cdot DA$ bo'ladigan $ABCD$ qavariq to'rtburchak berilgan. U holda $\angle BAC+\angle CBD+\angle DCA+\angle ADB = 180^{\circ}$ bo'lishini isbotlang.
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Masalaga ikki xil yonadashamiz.
1-yechim. (Isogonal qo'shmaning mavjudligi) Bilamiz-ki, to'rtburchaklarda nuqtaning har doim ham isogonal qo'shmasi mavjud bo'lavermaydi. Biroq ushbu masalani biz o'zaro isogonal nuqtalar topish orqali yechamiz.
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| 1-chizma |
To'rtburchak diagonallari $P$ nuqtada kesishsin. Masala shartiga ko'ra $\displaystyle \frac{AB}{BC}=\frac{AD}{CD}$ berilgan, bundan $ABC$ va $ADC$ uchburchaklarning Apollon aylanalari ustma-ust tushishi keladi, ya'ni $B$ va $D$ burchaklarninhg ichki va tashqi bissektrisasi asoslari ustma-ust tushadi. Agar $BP$ ning $\angle B$ ga nisbatan isogonal t/ch ini $BX$ desak, u holda $DP$ ning $\angle D$ ga nisbatan isogonali $DX$ bo'ladi. Ya'ni, $\angle PDA=\angle XDC$ va $\angle PBA = \angle XBC$.
Xuddi shunday, $BD$ diagonal $Y$ nuqtani aniqlashimiz mumkin: $\angle YAB=\angle PAD$ va $\angle YCB=\angle PCD$. Xullas, $X$ va $Y$ nuqtalar o'haro isogonal nuqtalardir ($ABCD$ to'rtburchakka nisbatan).
$X$ nuqtaning $ABCD$ to'rtburchakka nisbatan isogonali mavjud ekan, $\angle DXC+\angle BXA=180^{\circ}$ bo'ladi. Ya'ni, \[ \angle BAX+\angle ABX +\angle DCX+\angle CDX=180^{\circ} \] va \[ \angle BAP+\angle CBP+\angle DCP+\angle ADP=180^{\circ}. \] Isbot tugadi. ▢
2-yechim. (Harmoniklik; Apollon aylanalari) Oldingi yechimda aytganimizdek, $ABC$ va $ADC$ uchburchaklarning Apollon aylanalari ustma-ust tushadi. Quyidagi 2-chizma ga qaraylik, bunda $X$ va $X'$ nuqtalar sifatida uchburchaklarning ichki va tashqi bissektrisasi asoslari keltirilgan.
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| 2-chizma |
$BX'DX$ cyclic bo'lar ekan, $\angle XBD=\angle XX'D$ keladi. Ya'ni, \[ \angle CBP - \angle ABP = \angle DAP-\angle DCP. \] Xuddi shunday, \[ \angle CDP-\angle ADP = \angle BAP-\angle BCP \] ham keladi. Bulardan, bizga kerakli tenglikni keltirib chiqaramiz. ▢
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Xuddi shu konfiguratsiyaga tushadigan (P6 ekanligi juda g'alati) yaqin orada IMOda ko'rinish bergan quyidagi masala ham bor (IMO 2018, P6):
▻ Aytaylik, bizga $AB\cdot CD=BC\cdot DA$ bo'ladigan $ABCD$ qavariq to'rtburchak berilgan. $M$ nuqta qavariq to'rtburchak ichida yotib, $\angle MAB=\angle MCD$ va $\angle MBC=\angle MDA$ burchak tengliklarini qanoatlantiradi. U holda $\angle BMA+\angle DMC = 180^{\circ}$ ni isbotlang.
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Yechim. Avvalo, takidlash lozim-ki, bunaqa $M$ nuqta to'rtburchak ichida yagonadir: $BDE$ va $ACF$ uchburchaklarning tashqi aylanalarining to'rtburchak ichidagi kesishish nuqtasi bo'ladi, bu yerda $E$ va $F$ nuqtalar to'rtburchak qarama-qarshi tomonlari kesishish nuqtalari (3-chizma ga qarang).
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| 3-chizma |
Yuqoridagi 1-yechimni davom ettiramiz. Demak, bizda o'zaro isogonal bo'lgan $X$ va $Y$ nuqtalar mavjud. $ADX$ va $BCX$ larning tashqi aylanalari ikkinchi bor $M'$ nuqtada kesishsin. U holda \[ \angle BM'D = \angle BM'A+\angle AM'D = \angle BCX +\angle ADX+\angle AXD = 180^{\circ} - \angle BED \] va \[ \angle AM'C = \angle AM'D + \angle DM'C = \angle AXD + \angle DAX+\angle CBX = 180^{\circ} - \angle ADX +\angle CBX = \] \[ =180^{\circ} - CDB + \angle ABD = 180^{\circ} - \angle AFC \] lardan $M'$ nuqtaning $ACF$ hamda $BDE$ uchburchaklarning tashqi aylanasida yotishi keladi, ya'ni ularning to'rtburchak ichidagi kesishish nuqtasi. Xullas, $M'\equiv M$ va \[ \angle BMC+\angle AMD = \angle BXC+\angle AXD = 180^{\circ} \] bo'ladi, chunki $X$ nuqtaning to'rtburchakka nisbatan isogonali mavjud (aynan $Y$ nuqta).
Kerakli tasdiq isbotlandi. ▢