O‛zb е kiston r е spublikasi oliy va o‛rta maxsus ta’lim vazirligi

Y echim: ABC uchburchak teng yonli bo’lgani uchun A= C=α deb olish mumkin.
AEC: sinα=AE:AC,
ABF: sinα=BF:AB =>
=>AE:AC =BF:AB=>
=>AC:AB=AE:BF=1:2=>
=>cosA=AF:AB=(AC/2):AB=
=(AC:AB)/2=(1:2)/2=1/4
Javob: C)
29) Trapetsiyaning kichik asosi va ikkala yon tomonlari 5 ga teng va burchaklaridan biri 60⁰ ga teng. Trapetsiyaga tashqi chizilgan aylana radiusini toping.
A) 8 B) 7 C) 6 D) 5
Yechim: Dastlab bu masalaning standart yechimini keltiramiz. Oldin ABE to’gri burchakli uchburchakdan trapetsiyaning BE balandligini va AE kesmani topamiz (chizmaga qarang): BE=AB∙sin60⁰=5∙ /2 , AE=AB∙cos60⁰=5/2 .
Bu holda AD=BC+2AE=5+2∙5/2=10, AK=KD=AD/2=5, NC=BN=BC/2=5/2 bo’ladi. Agar OC=OD=R, ON=x desak, OK=NK−ON=BE−x=5 /2−x. CON va DOK to’g’ri burchakli uchburchaklardan quyidagilarni topamiz:
R²=OC²=ON²+NC²=x²+(5/2)² ,
R²=OD²=OK²+KD²=(5 /2−x)²+5² =>
=> (5 /2−x)²+5² =x²+(5/2)² =>
=> x=15/2 =5 /2 =>
=> R²= x²+(5/2)²=75/4+25/4=25 =>
=> R=5=KD=AK.
Demak, tashqi chizilgan aylana markazi trapetsiyaning katta asosi o’rtasida joylashgan va uning radiusi 5 ga teng.
Endi bu masalaning original, olimpiadacha yechimini ko’rsatamiz. Buning uchun ABCD trapetsiyani AD katta asosiga nisbatan simmetrik tarzda akslantiramiz. Natijada tomoni a=5 bo’lgan ABCDC₁D₁ muntazam oltiburchakka ega bo’lamiz. Bu oltiburchakka tashqi chizilgan aylana ABCD trapetsiya uchun ham tashqi chizilgan aylana bo’ladi va uning radiusi R=a=5 .
Javob: D)
30) To’g’ri burchakli uchburchakning katetlari 3 va 4 bo’lsa, unga ichki chizilgan aylana radiusini toping.
A) 1 B) 2 C) 1.5 D) 1.2
Yechim: Katetlar a=3, b=4 bo’lgani uchun gipotenuza c=5. To’g’ri burchakli uchburchakka ichki chizilgan aylana radiusi uchun r=(a+b−c)/2 formuladan r=1 ekanligi kelib chiqadi.
Javob: A)
Izoh: II bosqich uchun keyingi testlar mumkin bo’gan javoblari ko’rsatilmagan holda berilgan.

31) Muntazam (teng tomonli ) uchburchakning bissektrisasi 12 bo’lsa, unga ichki chizilgan aylana radiusini toping.

Javob: ….
Yechim: Dastlab bu masalaning standart yechimini keltiramiz. ABC
muntazam uchburchakda
A= B= C=60⁰ bo’lib,
AB=AC=BC=a deb belgilaymiz va uni topamiz. Bissektrisa BE=12 bir paytning o’zida balandlik ham bo’ladi va shu sababli ABE to’g’ri burchakli uchburchak bo’lib, undan
a=AB=BE/sin60⁰=12∙2/ =8 ekanligini topamiz.
Bu holda S_ABC=a²sin60⁰/2=8²∙3∙ /4=48 . Ikkinchi tomondan ichki chizilgan aylana radiusi OK=ON=OE=r desak, unda yarimperimetr p=3a/2=12 bo’lgani uchun S_ABC=pr formuladan 48 =12 r => r=4 javobni olamiz.
Endi bu masalaning nostandart, olimpiadaga xos yechimini keltiramiz. Uchburchakka ichki chizilgan aylana markazi O uning bissektrisalari kesishgan nuqtada bo’ladi. Muntazam uchburchakning bissektrisalari uning medianalari bilan ustma-ust tushadi va shu sababli O medianalar kesishga nuqta bo’ladi. Mediana xossasiga asosan O nuqtada ular burchak uchidan qaraganda 2:1 nisbatda bo’linadi. Demak, izlangan radius r=BO=BE/3=12/3=4 bo’ladi.
Javob: r=4 .
32) Agar bo’lsa, soni natural son bo’ladigan barcha natural n larni toping.
Javob: …….
Yechim: Bu test yechimining ikki usulini ko’rsatamiz. I usul.

Oxirgi tenglikning chap tomoni 5 ga bo’linadi va shu sababli m=5k±1 bo’lishi kerak.Bu holda 35(1632+n)=5k(5k±2)=>7(1632+n)=k(5k±2). Bu tenglikning chap tomoni 7 ga bo’linadi va shu sababli k=7t yoki 5k±2=7t bo’lishi kerak.
I hol. k=7t => 1632+n=t(35t±2)=35t²±2t . Bunda, test shartiga asosan ,
1000≤n≤2000 bo’lgani uchun 2632≤35t²±2t≤3632 ekanligi kelib chiqadi. Bu yerdan ushbu tengsizliklar sistemasiga kelamiz:
.
1) 35t²±2t−3632≤0 => D=1+35∙3632=127121 =>
.
2) 35t²±2t−2632≥0 => D=1+35∙2632=92121 =>

Demak, 8.6… ≤ t ≤ 10.4…, , => t=9 yoki t=10 .
A) t=9 =>1632+n=35∙9²±2∙9=2835±18 => n=1203±18 => n₁=1221, n₂=1185 .
Bunda
B) t=10 =>1632+n=35∙10²±2∙10=3500±20 => n=1868±20 => n₃=1848, n₄=1888 .
Bunda II hol.
5k=7t±2



Bunda, test shartiga asosan 1000≤n≤2000 bo’lgani uchun 1000≤35q²±12q−1631≤2000 ekanligi kelib chiqadi. Bu yerdan ushbu tengsizliklar sistemasiga kelamiz:
.
1) 35q²±12q−3631≤0 => D=36+35∙3631=127121 =>
.
2) 35q²±12q−2631≥0 => D=36+35∙2631=92121 =>

Demak, 8, … ≤ q ≤ 10, …, , => q=9 yoki q=10 .
A) q=9 =>n=35∙9²±12∙9−1631=1204±108 => n₁=1312, n₂=1096 .
Bunda
B) q=10 =>n=35∙10²±12∙10−1631=1869±120 => n₃=1989, n₄=1749 .
Bunda
II usul. Bu usul BDU dotsenti Z. Hamdamov tomonidan taklif etilgan va unda 57121=239² ekanligidan foydalaniladi.

.
.
Oxirgi kasr qiymati natural son bo’ladigan hollarni qarab chiqamiz.