Fun Geometry Problem with Solution #140



٨ x = 30°
٨ 1



(1)∠ACB = 40°
∵ ∠BAC =∠ABC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ AC = BC

(2) ҧǧͺ∆ABC ǵ AP ͡仾ͺǧش Q ☐ABQC öṺǧ ∠CBQ = ∠CAQ, ∠BCQ = ∠BAQ Р∠AQB =∠ACB ∠CBQ = 40°,∠BCQ = 30°Р∠AQB = 40°
͡ҡ ѧҠ∠BPQ = 80°

(3) ԨóҠ∆ABQ Ҡ∠A = 30°,∠Q = 40°ըش P AQ ∠BPQ = 80° AP = BQ (Click ʹԸվ٨⨷ 1-2)

(4) ѧࡵҠ∆ACP∆BCQ ¤ѹẺ -- (AC = BC,∠CAP =∠CBQ, AP = BQ) ∠ACP =∠BCQ x = 30° Q.E.D.

٨ 2



(1)∠ACB = 40°
∵ ∠BAC =∠ABC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ AC = BC

(2) ˹ش Q AP CQ 觤觠∠ACB ∠BCQ =∠ACQ = (∠ACB)/2 = 20°
Ҡ∆BCQ∆ACQ ¤ѹẺ -- (BC = AC,∠BCQ =∠ACQ, CQ = CQ) ∠CBQ =∠CAQ ∠CBQ = 40° ∠ABQ = 30°Р∠PBQ = 20°
ԨóҠ∆ABQ Р∆ACQ Ҡ∠BQP = 60°Р∠CQP = 60°ӴѺ

(3) ԨóҠ∆BCQ ըش P 繨ش㹷 BP PQ 觤觠∠CBQ Р∠BQC ӴѺ ش P incenter ͧ∆BCQ CP 觤觠∠BCQ ∠BCP (=∠PCQ) = (∠BCQ)/2 = 10° ∠ACP = x = 30° Q.E.D.




Create Date : 15 Զع¹ 2558
Last Update : 15 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #139



٨ x = 18°
٨



AD = L BC = L

(1)∠ACB = 138°

(2) ˹ش P Ҿз͹ͧش C ҹ AB ∆ABP∆ABC AP = AC, BP = BC = L,∠BAP =∠BAC = 30° ∠ABP = ∠ABC = 12°
∵ AC = AP Р∠CAP = 60° ∆ACP 繠∆ҹ AC = CP

(3) ˹ش Q ˹ AC AQ = CQ = L
Ҡ∆ACQ∆BCP ¤ѹẺ -- (AC = CP, AQ = BC, CQ = BP) ∠AQC =∠CBP ∠AQC = 24°
∵AQ = CQ ∆ACQ 繠∆˹Ҩ ՠ∠Q (= 24°) ʹ ∠CAQ = 78°Р∠ACQ = 78°(∠BCQ = 144°)

(4) ∵ BC = CQ ∆BCQ 繠∆˹Ҩ ՠ∠C (= 144°) ʹ ∠BQC (=∠CBQ) = 18°
∵ AD = AQ ∆ADQ 繠∆˹Ҩ ՠ∠A (= 108°) ʹ ∠AQD (=∠ADQ) = 36° ∠CQD = 12° ∠CQD =∠CBD ☐BDCQ öṺǧ ∠ADC =∠BQC x = 18° Q.E.D.




Create Date : 12 Զع¹ 2558
Last Update : 12 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #138



˹ BC = AB + AD + BD
٨ x = 12°
٨



AB = a, AD = b BD = c BC = a + b + c

(1)∠ADB = 84°

(2) AB ͡ѧش P · BP = c ∠DBP = 120°
∵ BD = BP ∆BDP 繠∆˹Ҩ ՠ∠B (= 120°) ʹ ∠BPD (=∠BDP) = 30°

(3) ˹ش Q Ҿз͹ͧش A ҹ DP ∆DPQ∆ADP PQ = AP = a + c, DQ = AD = b Р∠DPQ =∠APD = 30°
∵ AP = PQ Р∠APQ = 60° ∆APQ 繠∆ҹ AQ = a + c
͡ҡ ѧҠ∠PAQ = 60° ∠DAQ = 24°

(4) ˹ش R ҧҹ¢ͧ AD DR = a + c Р∠ADR = 24°
Ҡ∆ADR∆ADQ ¤ѹẺ -- (AD = AD,∠ADR =∠DAQ, DR = AQ) AR = DQ AR = b
∵ AD = AR ∆ADR 繠∆˹Ҩ ՠ∠A ʹ ∠ARD =∠ADR ∠ARD = 24° ∠DAR = 132°

(5) BA ͡仾ǹ͢¢ͧ DR ش S ∠RAS = 12°
ԨóҠ∆ARS Ҡ∠ASR = 12° ∠ASR =∠RAS ∆ARS 繠∆˹Ҩ ՠ∠R ʹ RS = AR RS = b

(6) ѧࡵҠ∆BCD∆BDS ¤ѹẺ -- (BC = DS,∠CBD =∠BDS, BD = BD) ∠BCD =∠BSD x = 12° Q.E.D.




Create Date : 09 Զع¹ 2558
Last Update : 9 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #137
(Kadir Altintas)



٨ x = 30°
٨(Kadir Altintas)



(1)∠APB = 140°
∵ ∠BAP =∠ABP ∆ABP 繠∆˹Ҩ ՠ∠P ʹ AP = BP

(2) ˹ش Q ˹ AP AQ = PQ = AP (= BP) ∆APQ 繠∆ҹ ∠APQ = 60°(∠BPQ = 160°) Р∠AQP = 60°
∵ BP = PQ ∆BPQ 繠∆˹Ҩ ՠ∠P (= 160°) ʹ ∠PBQ =∠BQP = 10°

(3) ˹ش R 繨شѴҧ AC BQ
∵ AQ = PQ Р∠PQR = 2(∠PAR) ش Q circumcenter ͧ∆APR ∠ARP = (∠AQP)/2 ∠ARP = 30°
∵ ∠PBR =∠PCR ☐BCRP öṺǧ ∠CBP =∠ARP x = 30° Q.E.D.




Create Date : 06 Զع¹ 2558
Last Update : 6 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #136



٨ x = 48°
٨



(1)∠ACB = 96°

(2) CB ͡ѧش P ·∠APC = 30° ∠BAP = 24°
ԨóҠ∆ACP Ҡ∠C = 96°,∠P = 30°ըش B CP ∠ABC = 54° BP = AC (Click ʹԸվ٨⨷ 2) BP = b

(3) ˹ش Q Ҿз͹ͧش B ҹ AP ∆APQ∆ABP AQ = AB = a, PQ = BP = b,∠PAQ =∠BAP = 24°Р∠APQ =∠APB = 30°
∵ BP = PQ Р∠BPQ = 60° ∆BPQ 繠∆ҹ BQ = b

(4) ѧࡵҠ∆DEF∆ABQ ¤ѹẺ -- (DE = BQ, DF = AB, EF = AQ) ∠DFE =∠BAQ x = 48° Q.E.D.




Create Date : 03 Զع¹ 2558
Last Update : 3 Զع¹ 2558 0:00:00 .
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TIYHz
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