Fun Geometry Problem with Solution #165
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٨ x = 20°
٨



(1)∠CAD = 50°- x Р∠ABC = 100°

(2) ˹ش O circumcenter ͧ∆ACD ...
AO = CO = DO
∠AOD = 2(∠ACD) ∠AOD = 2x
∠COD = 2(∠CAD) ∠COD = 100°- 2x
∵ AO = CO ∆ACO 繠∆˹Ҩ ՠ∠O (= 100°) ʹ ∠CAO = ∠ACO = 40° ∠BAO = ∠BCO = 20° ☐ACBO öṺǧ ∠ABO =∠ACO ∠ABO = 40°

(3) ˹ش P AC CP = BC
∵ BC = CP Р∠BCP = 60° ∆BCP 繠∆ҹ BP = CP,∠CBP = 60°(∠ABP = 40°) Р∠BPC = 60°
∵ BP = CP Р∠BPC = 2(∠BDC) ش P circumcenter ͧ∆BCD CP = DP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ ∠CDP =∠DCP ∠CDP = x

(4) ѧࡵҠ∆ABO∆ABP ¤ѹẺ -- (∠BAO =∠BAP, AB = AB,∠ABO =∠ABP) AO = AP ∆AOP ∆˹Ҩ ՠ∠A (= 40°) ʹ ∠APO (=∠AOP) = 70°
ѧࡵҠ∆DOP∆COP ¤ѹẺ -- (DO = CO, DP = CP, OP = OP) ∠ODP =∠OCP (= 40°) ∠ODP = ∠OAP ☐ADPO öṺǧ ∠ADO =∠APO ∠ADO = 70°

(5) ԨóҠ∠ADC 110°+ x = 130° x = 20° Q.E.D.




Create Date : 26 ԧҤ 2558
Last Update : 26 ԧҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #164
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ ❀❀❀
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٨ cos36°- cos72°= 1/2 ԸշҧâҤԵ

٨



(1) ҧ∆ABC · AC = BC =LР∠A = 36°
∴ ∆ABC 繠∆˹Ҩ ՠ∠C ʹ ∠ABC =∠BAC ∠ABC = 36°

(2) ˹ش D AB AD = L AD = AC ∆ACD 繠∆˹Ҩ ՠ∠A (= 36°) ʹ ∠ADC (=∠ACD) = 72° ∠BCD = 36° ∠BCD =∠CBD ∆BCD 繠∆˹Ҩ ՠ∠D ʹ BD = CD

ԨóҠ∆ABC 繠∆˹Ҩ ҡǹ٧ CH Ҡ∆ACH Р∆BCH 繠∆ҡ ҡѹءС
¹ͧ cosine 㹠∆ҡ AH = BH = Lcos36° AB = 2Lcos36°
㹷ӹͧǡѹ Ѻ ∆ACD CD = 2Lcos72° BD = 2Lcos72°

∵ AB - BD = AD 2Lcos36°- 2Lcos72°= L cos36°- cos72°= 1/2 Q.E.D.




Create Date : 24 ԧҤ 2558
Last Update : 9 ѹ¹ 2558 22:18:00 .
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Fun Geometry Problem with Solution #163
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ Ԡ❀❀❀
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٨ x= 60°
٨



(1)∠ACB = 10°Р∠ADB = 20°

AB = L

(2) ˹ش P AC BP = L BP = AB ∆ABP 繠∆˹Ҩ ՠ∠B ʹ ∠APB =∠BAP ∠APB = 20° ∠ABP = 140°, ∠CBP = 10° Р∠DBP = 30°
∵ ∠CBP =∠BCP ∆BCP 繠∆˹Ҩ ՠ∠P ʹ CP = BP CP = L

(3) ˹ش Q AD BQ = L BQ = AB ∆ABQ 繠∆˹Ҩ ՠ∠B ʹ ∠AQB =∠BAQ ∠AQB = 50° ∠DBQ = 30°
∵ BP = BQ Р∠PBQ = 60° ∆BPQ 繠∆ҹ PQ = BP PQ = L
͡ҡ ѧҠ∠BPQ = 60° ∠APQ = 40° ∠CPQ = 140°Р∠DQP = 70°

(4) ѧࡵҠ∆BDP∆BDQ ¤ѹẺ -- (BD = BD,∠DBP =∠DBQ, BP = BQ) ∠BDP =∠BDQ ∠BDP = 20°
ԨóҠ∆DPQ Ҡ∠DPQ = 70° ∠CPD = 70°

(5) ѧࡵҠ∆CDP∆DPQ ¤ѹẺ -- (CP = PQ,∠CPD =∠DPQ, DP = DP) ∠CDP =∠PDQ ∠CDP = 40° ∠BDC = x = 60° Q.E.D.




Create Date : 21 ԧҤ 2558
Last Update : 21 ԧҤ 2558 0:36:00 .
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Fun Geometry Problem with Solution #162
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
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˹ DE = a + b + c, DF = a + b EF = a
٨ x= 20°
٨



(1)∠ACB = 40°

(2) AC ͡ѧش P · CP = a ∠BCP = 140°
∵ BC = CP ∆BCP 繠∆˹Ҩ ՠ∠C (= 140°) ʹ ∠CBP = ∠BPC = 20°
∵ ∠BAP =∠ABP ∆ABP 繠∆˹Ҩ ՠ∠P ʹ BP = AP BP = a + b

(3) CA ͡ѧش Q · AQ = c ∠BAQ = 100°
∵ AB = AQ ∆ABQ 繠∆˹Ҩ ՠ∠A (= 100°) ʹ ∠AQB (=∠ABQ) = 40°
∵ ∠BCQ =∠BQC ∆BCQ 繠∆˹Ҩ ՠ∠B ʹ BQ = BC BQ = a

(4) ѧࡵҠ∆DEF∆BPQ ¤ѹẺ -- (DE = PQ, DF = BP, EF = BQ) ∠EDF =∠BPQ x = 20° Q.E.D.




Create Date : 18 ԧҤ 2558
Last Update : 18 ԧҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #161
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ ❀❀❀
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٨ x= 40°
٨ 1



(1)∠ACB = 110°

(2) BC ͡ѧش Q · BQ = AQ ∆ABQ 繠∆˹Ҩ ՠ∠Q ʹ ∠BAQ =∠ABQ ∠BAQ = 50° ∠PAQ = 40°Р∠AQB = 80°

(3) α= 20°
☐APBQ AQ = BQ,∠B =α,∠A = 2αР∠Q = 120°- 2α AP = AQ (Click ʹԸվ٨⨷ 2) ∆APQ 繠∆˹Ҩ ՠ∠A (= 40°) ʹ ∠AQP (=∠APQ) = 70° ∠BQP = 10°

(4)∵∠CAP =∠CQP ☐APCQ öṺǧ ∠ACP =∠AQP ∠ACP = 70° ∠BCP = x = 40° Q.E.D.

٨ 2 (ɮպǡѺǧ)



(1)∠ACB = 110°

(2) BC ͡ѧش Q · BQ = AQ ∠ACQ = 70°
∵ AQ = BQ ∆ABQ 繠∆˹Ҩ ՠ∠Q ʹ ∠BAQ =∠ABQ ∠BAQ = 50° ∠AQB = 80°

(3) α= 20°
☐APBQ AQ = BQ,∠B =α,∠A = 2αР∠Q = 120°- 2α AP = AQ (Click ʹԸվ٨⨷ 2)

(4) CQ ͡ѧش R · AR = AC ∆ACR 繠∆˹Ҩ ՠ∠A ʹ ∠ARC =∠ACR ∠ARC = 70° ∠QAR = 10°

(5) ѧࡵҠ∆ACP∆AQR ¤ѹẺ -- (AC = AR,∠CAP =∠QAR, AP = AQ) ∠ACP =∠ARQ ∠ACP = 70° ∠BCP = x = 40° Q.E.D.




Create Date : 15 ԧҤ 2558
Last Update : 15 ԧҤ 2558 0:00:00 .
Counter : 491 Pageviews.

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