Fun Geometry Problem with Solution #75



٨ x = 90°
٨



(1)∵ ش I incenter ͧ ∆ABC AI, BI CI 觤觠∠BAC,∠ABC Р∠ACB ӴѺ ∠BAI =∠CAI,∠ABI =∠CBI Р∠ACI =∠BCI

∠ABI =θ ∠CBI = θ

(2) ˹ش P AB AP = AC
∵ AB = AC + CI AP + BP = AC + CI BP = CI
Ҡ∆AIP∆ACI ¤ѹẺ -- (AP = AC,∠IAP =∠CAI, AI = AI) IP = CI IP = BP ∆BIP 繠∆˹Ҩ ՠ∠P ʹ ∠BIP =∠IBP ∠BIP = θ

(3) ѧࡵ CI = IP Р∠CBI =∠IBP ☐BCIP öṺǧ ∠BCP =∠BIP Р∠ICP =∠IBP ∠BCP =θР∠ICP = θ

(4)∵AC = BC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ ∠BAC =∠ABC ∠BAC = 2θ
∵∠ACI =∠BCI ∠ACI = 2θ
ԨóҠ∆ABC Ҡ∠BAC +∠ABC +∠ACB= 180° 2θ+ 2θ+ 4θ= 180° θ= 22.5° 4θ = x = 90° Q.E.D.




Create Date : 03 ѹҤ 2557
Last Update : 3 áҤ 2558 23:23:00 .
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Fun Geometry Problem with Solution #74



٨ x = 60°
٨



(1)∠APC = 150°Р∠BPC = 110°

(2) ˹ش O 繨شٹҧͧǧՠ∆BCP Ṻ ...
ش O ͡∆BCP
BO = CO = OP
∠BOP = 2(∠BCP) ∠BOP = 80°
∠COP = 2(∠CBP) ∠COP = 60°
∵ CO = OP Р∠COP = 60° ∆COP 繠∆ҹ CP = OP
͡ҡ ѧҠ∠CPO = 60° ∠APO = 150°

(3) ѧࡵҠ∆AOP∆ACP ¤ѹẺ -- (OP = CP,∠APO =∠APC, AP = AP) ∠OAP =∠CAP ∠OAP = 20°
͡ҡ ѧҠ∠AOP =∠ACP ∠AOP = 10° ∠AOB = 70°

(4) ѧࡵҠ∆ABO∆ACO ¤ѹẺ -- (BO = CO,∠AOB =∠AOC, AO = AO) ∠BAO =∠CAO ∠BAO = 40° ∠BAP = x = 60° Q.E.D.




Create Date : 30 Ȩԡ¹ 2557
Last Update : 30 Ȩԡ¹ 2557 0:00:01 .
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Fun Geometry Problem with Solution #73



٨ x = 100°
٨



(1) ԨóҠ∆ABC ∠ACB = 20° ∠ACB = ∠BAC ∆ABC 繠∆˹Ҩ ՠ∠B ʹ AB = BC

(2) ˹ش O 繨شٹҧͧǧՠ∆ABD Ṻ ...
ش O ˹ AD
AO = BO = DO
∠BOD = 2(∠BAD) ∠BOD = 60°
∵ BO = DO Р∠BOD = 60° ∆BDO 繠∆ҹ BD = BO
͡ҡ ѧҠ∠DBO = 60° ∠ABO = 40°
∵ AO = BO ∆ABO 繠∆˹Ҩ ՠ∠O ʹ ∠BAO =∠ABO ∠BAO = 40° ∠AOB = 100°

(3) ѧࡵҠ∆BCD∆ABO ¤ѹẺ -- (BC = AB,∠CBD =∠ABO, BD = BO) ∠BDC =∠AOB x = 100° Q.E.D.




Create Date : 27 Ȩԡ¹ 2557
Last Update : 27 Ȩԡ¹ 2557 0:00:02 .
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Fun Geometry Problem with Solution #72



˹∠BDC ҹ
٨ x = 24°
٨



(1)∠CAD = 120°- x Р∠BDC = 150°- x

(2) ˹ش P BC BP = CD
Ҡ∆BDP∆ACD ¤ѹẺ -- (BD = AC,∠DBP =∠ACD, BP = CD) ∠BDP =∠CAD ∠BDP = 120°- x ∠CDP = 30°

(3) ˹ش O 繨شٹҧͧǧՠ∆BCD Ṻ ...
ش O ˹ BC
BO = CO = DO
∠BOD = 2(∠BCD) ∠BOD = 2x
∠COD = 2(∠CBD) ∠COD = 60°
∵ CO = DO Р∠COD = 60° ∆CDO 繠∆ҹ CD = CO = DO
͡ҡ ѧҠ∠DCO = 60°Р∠CDO = 60° ∠BCO = 60°- x Р∠ODP = 30°
∵ BO = CO ∆BCO 繠∆˹Ҩ ՠ∠O ʹ ∠CBO =∠BCO ∠CBO = 60°- x

(4) ѧࡵҠ∆DOP∆CDP ¤ѹẺ -- (DO = CD,∠ODP =∠CDP, DP = DP) ∠DOP =∠DCP ∠DOP = x

(5)∵CD = CO BP = BO ∆BOP 繠∆˹Ҩ ՠ∠B ʹ ∠BPO =∠BOP ∠BPO = 3x
ԨóҠ∆BOP Ҡ∠OBP +∠BOP +∠BPO = 180° (60°- x) + 3x + 3x = 180° x = 24° Q.E.D.




Create Date : 24 Ȩԡ¹ 2557
Last Update : 24 Ȩԡ¹ 2557 0:00:00 .
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Fun Geometry Problem with Solution #71



˹∠ACD ҹ
٨ x = 6°
٨



(1) ԨóҠ∆BCD Ҡ∠ADC = 3x

(2) ˹ش P BC DP = BD BDP 繠˹Ҩ ՠD ʹ BPD =DBP BPD = 2x CDP = x CDP =DCP CDP 繠˹Ҩ ՠP ʹ ⇔ CP = DP CP = BD

(3) ˹ش Q ˹ AB DAQ = x РADQ = 2x CAQ = 30°- x РCDQ = x
ҠADQ BCD ¤ѹẺ -- (DAQ =BCD, AD = BC,ADQ =CBD) ⇒ AQ = CD DQ = BD
ҠCDQ CDP ¤ѹẺ -- (DQ = DP,CDQ =CDP, CD = CD) ⇒ CQ = CP РDCQ =DCP ⇔ CQ = BD РDCQ = x
ԨóҠADCQ ҠAQC (˭) = 360°- 5x AQC () = 5x

(4) ˹ش O 繨شٹҧͧǧՠ∆ACD Ṻ ⇒ ...
ش O AD
AO = CO = DO
∠AOC = 2(∠ADC) AOC = 6x
COD = 2(CAD) COD = 60°
CO = DO РCOD = 60° CDO 繠ҹ CD = CO = DO ⇔ AQ = AO
͡ҡ ѧҠCDO = 60° ADO = 60°- 3x
AO = DO ADO 繠˹Ҩ ՠO ʹ DAO =ADO DAO = 60°- 3x OAQ = 60°- 2x

(5) ˹ش R Ҿз͹ͧش C ҹ AQ AQR ACQ ⇒ ...
QR = CQ ⇔ QR = BD
QAR =CAQ QAR = 30°- x OAR = 30°- x
AQR =AQC AQR = 5x
ѧࡵҠAOR AQR ¤ѹẺ -- (AO = AQ,OAR =QAR, AR = AR) ⇒ OR = QR ⇔ OR = BD
͡ҡ ѧҠAOR =AQR AOR = 5x COR = x

(6) ѧࡵҠCOR CDQ ¤ѹẺ -- (OR = DQ,COR =CDQ, CO = CD) ⇒ CR = CQ
CQ = CR = QR CQR 繠ҹ CQR = 60° ⇔ 10x = 60° ⇔ x = 6° Q.E.D.

˵öʴҨش Q 㹠∆ACD ѧ
∵∠ACD ҹ 150°- 3x > 90° x < 20° ∠DAQ < 20°
∵∠DAQ <∠CAD Р∠ADQ <∠ADC
∴ ش Q 㹠∆ACD




Create Date : 21 Ȩԡ¹ 2557
Last Update : 21 Ȩԡ¹ 2557 0:00:00 .
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TIYHz
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