Fun Geometry Problem with Solution #135



٨ x = 20°
٨



(1)∠ACD = 90°- x,∠ADC = 10°+ x Р∠BDC = 170°- x

(2) ˹ش P Ҿз͹ͧش C ҹ AD ∆ADP∆ACD AP = AC, DP = CD,∠DAP =∠CAD = 80° ∠ADP = ∠ADC = 10° + x
∵ AC = AP ∆ACP 繠∆˹Ҩ ՠ∠A (= 160°) ʹ ∠ACP =∠APC = 10°

(3) ˹ش Q ҧҹҢͧ AC ∠ACQ = 170°- x (∠DCQ = 80°) CQ = CD
Ҡ∆ACQ∆BCD ¤ѹẺ -- (AC = BD,∠ACQ =∠BDC, CQ = CD) ∠AQC =∠BCD ∠AQC = 10° ∠AQC =∠APC ☐ACQP öṺǧ ∠AQP =∠ACP ∠AQP = 10°
∵ CD = CQ ∆CDQ 繠∆˹Ҩ ՠ∠C (= 80°) ʹ ∠CDQ = 50°Р∠CQD = 50°(∠DQP = 30°)

(4) ˹ش R 繨شѴҧ CD PQ
ԨóҠ∆CQR Ҡ∠CRQ = 80° ∠CRQ =∠QCR ∆CQR 繠∆˹Ҩ ՠ∠Q ʹ QR = CQ QR = DP

(5) ԨóҠ∆DQR Ҡ∠D = 50°,∠Q = 30°ըش P ǹ͢¢ͧ QR DP = QR ∠DPQ = 40° (Click ʹԸվ٨) ∠PDR = 60° 20°+ 2x = 60° x = 20° Q.E.D.




Create Date : 31 Ҥ 2558
Last Update : 31 Ҥ 2558 0:00:00 .
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Fun Geometry Problem with Solution #134



٨ x = 20°
٨ 1



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

(2) ˹ش Q ˹ AB AQ = BQ = AB ∆ABQ 繠∆ҹ ∠BAQ = 60°(∠PAQ = 30°),∠ABQ = 60° (∠PBQ = 20°) Р∠AQB = 60°

♦ ѧࡵҠ∆BCQ∆ACQ ¤ѹẺ -- (BC = AC, BQ = AQ, CQ = CQ) ∠BQC (=∠AQC) = (∠AQB)/2 = 30°

♦ ѧࡵҠ∆APQ∆ABP ¤ѹẺ -- (AP = AP,∠PAQ =∠BAP, AQ = AB) ∠AQP =∠ABP ∠AQP = 40° ∠BQP = 20° ∠CQP = 10°

(3)∵∠CBP =∠CQP ☐BPCQ öṺǧ ∠BCP =∠BQP x = 20° Q.E.D.

٨ 2(ɮպǡѺǧ)


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

(2) ˹ش Q ˹ AB AQ = BQ = AB ∆ABQ 繠∆ҹ ∠BAQ = 60°(∠PAQ = 30°),∠ABQ = 60°(∠CBQ = 10°) Р∠AQB = 60°

♦ ѧࡵҠ∆BCQ∆ACQ ¤ѹẺ -- (BC = AC, BQ = AQ, CQ = CQ) ∠BQC (=∠AQC) = (∠AQB)/2 = 30°

♦ ѧࡵҠ∆APQ∆ABP ¤ѹẺ -- (AP = AP,∠PAQ =∠BAP, AQ = AB) PQ = BP ∆BPQ ∆˹Ҩ ՠ∠P ʹ ∠BQP =∠PBQ ∠BQP = 20° ∠CQP = 10°

(3) QC ͡ѧش R ·∠BRQ = 100° ∠BCR = 40°
ԨóҠ∆BQR Ҡ∠Q = 30°,∠R = 100°ըش C QR ∠BCR = 40° BC = QR (Click ʹԸվ٨⨷ 1-1)

(4) ѧࡵҠ∆PQR∆BCP ¤ѹẺ -- (PQ = BP,∠PQR =∠CBP, QR = BC) ∠PRQ =∠BCP ∠PRQ = x
͡ҡ ѧ PR = CP ∆CPR 繠∆˹Ҩ ՠ∠P ʹ ∠PCR =∠CRP ∠PCR = x

(5)∵∠BCP +∠PCR =∠BCR x + x = 40° x = 20° Q.E.D.

٨ 3



(1) BP ͡仾 AC ش Q
ԨóҠ∆ABQ Ҡ∠AQB = 90° BP⊥AC

(2) α= 10°
ԨóҠ∆ABC ըش P 繨ش㹷 BP⊥AC,∠BAP = 30°,∠CAP = 30°-αР∠CBP =α ∠BCP = 2α (Click ʹԸվ٨) x = 20° Q.E.D.




Create Date : 28 Ҥ 2558
Last Update : 14 Զع¹ 2558 4:00:00 .
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Fun Geometry Problem with Solution #133



٨ x = 30°
٨



BP = L

(1)∠APC = 120°Р∠BPC = 140° ∠APB = 100°

(2) ˹ش Q Ҿз͹ͧش P ҹ BC ∆BCQ∆BCP BQ = BP , CQ = CP,∠CBQ =∠CBP = 30° ∠BCQ = ∠BCP = 10°
∵ BP = BQ Р∠PBQ = 60° ∆BPQ 繠∆ҹ PQ = BP PQ = L

(3) ˹ش R AC CR = CP CR = CQ
Ҡ∆CPR∆CPQ ¤ѹẺ -- (CP = CP,∠PCR =∠PCQ, CR = CQ) PR = PQ PR = L
∵ CP = CR ∆CPR 繠∆˹Ҩ ՠ∠C (= 20°) ʹ ∠CPR = 80°Р∠CRP = 80° ∠APR = 40° ∠ARP = 100°
∵ ∠PAR =∠APR ∆APR 繠∆˹Ҩ ՠ∠R ʹ AR = PR AR = L

(4) ˹ش S ˹ PR PS = RS = PR (= L) ∆PRS 繠∆ҹ ∠RPS =∠PRS = 60°
∵ AR = RS ∆ARS 繠∆˹Ҩ ՠ∠R (= 160°) ʹ ∠RAS (=∠ASR) = 10° ∠PAS = 30°

(5) ѧࡵҠ∆ABP∆APS ¤ѹẺ -- (AP = AP,∠APB =∠APS, BP = PS) ∠BAP =∠PAS x = 30° Q.E.D.




Create Date : 25 Ҥ 2558
Last Update : 25 Ҥ 2558 0:03:00 .
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Fun Geometry Problem with Solution #132



٨ x = 3α
٨



(1)∠ACB = 60°+α

(2) ˹ش Q ˹ AB AQ = BQ = AB ∆ABQ 繠∆ҹ ...
∠BAQ = 60° ∠CAQ = 30°
∠ABQ = 60° ∠CBQ = 30°-αР∠PBQ = 30°
∠AQB = 60°

♦ ѧࡵҠ∆ACQ∆ABC ¤ѹẺ -- (AC = AC,∠CAQ =∠BAC, AQ = AB) CQ = BC ∆BCQ 繠∆˹Ҩ ՠ∠C ʹ ∠BQC =∠CBQ ∠BQC = 30°-α

♦ ѧࡵҠ∆BPQ∆ABP ¤ѹẺ -- (BP = BP,∠PBQ =∠ABP, BQ = AB) PQ = AP
͡ҡ ѧҠ∠BQP =∠BAP ∠BQP =α ∠AQP = 60°-α
ԨóҠ☐ABQP Ҡ∠APQ (˭) = 360°- (60°+ 2α) ∠APQ () = 60°+ 2α

(3) ˹ش R AQ ∠QAR = 30°-αР∠AQR = 30°-α(∠PQR = 30°)
Ҡ∆AQR∆BCQ ¤ѹẺ -- (∠QAR =∠CBQ, AQ = BQ,∠AQR =∠BQC) QR = CQ
∵ ∠QAR = ∠AQR ∆AQR 繠∆˹Ҩ ՠ∠R ʹ AR = QR

♦ ѧࡵҠ∆PQR∆APR ¤ѹẺ -- (PQ = AP, PR = PR, QR = AR) ∠QPR (=∠APR) = (∠APQ)/2 = 30°+α

♦ ѧࡵҠ∆CPQ∆PQR ¤ѹẺ -- (CQ = QR,∠CQP =∠PQR, PQ = PQ) ∠CPQ =∠QPR ∠CPQ = 30°+α

(4) ԨóҠ∆ACP Ҡ∠ACP = 60°- 2α ∠BCP = x = 3α Q.E.D.




Create Date : 22 Ҥ 2558
Last Update : 27 Ҥ 2558 16:30:00 .
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Fun Geometry Problem with Solution #131



٨ x = 48°
٨



(1)∠ACB = 78°

(2) BC ͡ѧش Q ·∠AQB = 72° ∠CAQ = 6°
∠ABQ =∠AQB ∆ABQ 繠∆˹Ҩ ՠ∠A ʹ AB = AQ
ѧࡵҠ∆APQ∆ABP ¤ѹẺ -- (AP = AP,∠PAQ =∠BAP, AQ = AB) PQ = BP ∆BPQ 繠∆˹Ҩ ՠ∠P ʹ ∠BQP =∠PBQ ∠BQP = 36° ∠BPQ = 108°

(3) ˹ش R Ҿз͹ͧش C ҹ AB ∆ABR∆ABC AR = AC, BR = BC,∠BAR =∠BAC = 30° ∠ARB = ∠ACB = 78°
∵ AC = AR Р∠CAR = 60° ∆ACR 繠∆ҹ AC = CR Р∠ACR =∠ARC = 60°(∠BCR =∠BRC = 18°)

(4) ˹ش S ˹ AC ∠CAS =∠ACS = 18°(∠ASC = 144°)
Ҡ∆ACS∆BCR ¤ѹẺ -- (∠CAS =∠BCR, AC = CR,∠ACS =∠BRC) AS = BC CS = BR AS = BC = CS

(5)∵AS = CS Р∠ASC = 2(∠AQC) ش S circumcenter ͧ ∆ACQ ∠CSQ = 2(∠CAQ) ∠CSQ = 12°
∵ BC = CS ∆BCS 繠∆˹Ҩ ՠ∠C (= 96°) ʹ ∠BSC (=∠CBS) = 42°
∵ BP = PQ Р∠BPQ = 2(∠BSQ) ش P circumcenter ͧ∆BQS BP = PS

(6) ѧࡵҠ∆BCP∆CPS ¤ѹẺ -- (BC = CS, BP = PS, CP = CP) ∠BCP (=∠PCS) = (∠BCS)/2 x = 48° Q.E.D.




Create Date : 19 Ҥ 2558
Last Update : 19 Ҥ 2558 0:55:55 .
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