Fun Geometry Problem with Solution #160
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ Ԡ❀❀❀
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣




٨ x= 30°
٨ 1



(1) ԨóҠ☐ABPC Ҡ∠BPC (˭) = 360°- 110° ∠BPC () = 110°

(2) ˹ش Q Ҿз͹ͧش P ҹ AB ∆ABQ∆ABP AQ = AP, BQ = BP,∠BAQ =∠BAP = 10° ∠ABQ =∠ABP = 30°
∵ BP = BQ Р∠PBQ = 60° ∆BPQ 繠∆ҹ BP = PQ

(3) ˹ش R AC AR = AP
Ҡ∆APR∆APQ ¤ѹẺ -- (AP = AQ,∠PAR =∠PAQ, AR = AP) PR = PQ PR = BP
∵ AP = AR ∆APR 繠∆˹Ҩ ՠ∠A (= 20°) ʹ ∠ARP (=∠APR) = 80°

(4) ˹ش S AR PS = PR (= BP) ∆PRS 繠∆˹Ҩ ՠ∠P ʹ ∠PSR =∠PRS ∠PSR = 80°
ԨóҠ∆CPS Ҡ∠CPS = 50° ∠CPS =∠PCS ∆CPS 繠∆˹Ҩ ՠ∠S ʹ CS = PS

(5) ˹ش T ҧҹҢͧ CP CT = PT = CP ∆CPT 繠∆ҹ ∠CPT =∠CTP = 60°
Ҡ∆PST∆CST ¤ѹẺ -- (PS = CS, PT = CT, ST = ST) ∠PTS (=∠CTS) = (∠CTP)/2 = 30°

(6) ѧࡵҠ∆BCP∆PST ¤ѹẺ -- (BP = PS,∠BPC =∠SPT, CP = PT) ∠BCP =∠PTS x = 30° Q.E.D.

٨ 2 (Angel Lazo HK)



(1) ԨóҠ☐ABPC Ҡ∠BPC (˭) = 360°- 110° ∠BPC () = 110°

BP = L

(2) ˹ش Q AB PQ = L PQ = BP ∆BPQ 繠∆˹Ҩ ՠ∠P ʹ ∠BQP =∠PBQ ∠BQP = 30° ∠APQ = 20°

(3) ˹ش R AP QR = L QR = PQ ∆PQR 繠∆˹Ҩ ՠ∠Q ʹ ∠PRQ =∠QPR ∠PRQ = 20° ∠AQR = 10° ∠AQR =∠QAR ∆AQR 繠∆˹Ҩ ՠ∠R ʹ AR = QR AR = L
ԨóҠ∆PQR Ҡ∠PQR = 140°

(4) ˹ش S AC RS = L RS = AR ∆ARS 繠∆˹Ҩ ՠ∠R ʹ ∠ASR =∠RAS ∠ASR = 20° ∠PRS = 40°

(5)∵QR = RS Р∠QRS = 60° ∆QRS 繠∆ҹ ∠RQS = 60° ∠PQS = 80°
͡ҡ ѧ QS = QR (= L) QS = PQ ∆PQS 繠∆˹Ҩ ՠ∠Q (= 80°) ʹ ∠QPS (= ∠PSQ) = 50° ∠APS = 30° ∠CSP = 50° ∠CSP =∠PCS ∆CPS 繠∆˹Ҩ ՠ∠P ʹ CP = PS
ԨóҠ∆PRS Ҡ∠PSR = 110°

(6) ѧࡵҠ∆BCP∆PRS ¤ѹẺ -- (BP = RS,∠BPC =∠PSR, CP = PS) ∠BCP =∠RPS x = 30° Q.E.D.




Create Date : 12 ԧҤ 2558
Last Update : 12 ԧҤ 2558 0:00:01 .
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Fun Geometry Problem with Solution #159
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ ❀❀❀
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣




٨ x= 10°
٨



(1)∠CAP = 90°- 6x Р∠APC = 90°+ x

(2) ˹ش Q AD DQ = BD ...
♦∆CDQ∆BCD ¤ѹẺ -- (CD = CD,∠CDQ =∠BDC, DQ = BD) ∠DCQ =∠BCD ∠DCQ = 2x ∠ACQ = 3x
♦∆DPQ∆BDP ¤ѹẺ -- (DP = DP,∠PDQ =∠BDP, DQ = BD) ∠DQP =∠DBP ∠DQP = 3x ∠APQ = 2x

PQ = L

(3) ˹ش R AP QR = L QR = PQ ∆PQR 繠∆˹Ҩ ՠ∠Q ʹ ∠PRQ =∠QPR ∠PRQ = 2x ∠AQR = x
∵ ∠QAR =∠AQR ∆AQR 繠∆˹Ҩ ՠ∠R ʹ AR = QR AR = L
∵ ∠PCQ =∠PRQ ☐CPQR öṺǧ ∠QCR =∠QPR ∠QCR = 2x ∠ACR = x
͡ҡ ѧҠ∠CQR =∠CPR ∠CQR = 90°+ x

(4) CQ ͡ѧش S · RS = QR ∠RQS = 90°- x
∵ QR = RS ∆QRS 繠∆˹Ҩ ՠ∠R ʹ ∠QSR =∠RQS ∠QSR = 90°- x
ԨóҠ∆CRS Ҡ∠CRS = 90°- x ∠CRS =∠CSR ∆CRS 繠∆˹Ҩ ՠ∠C ʹ CR = CS

(5) ˹ش T Ҿз͹ͧش R ҹ AC ∆ACT∆ACR AT = AR = L, CT = CR,∠CAT =∠CAR = 90° - 6x Р∠ACT =∠ACR = x
ѧࡵҠ∆CRT∆CRS ¤ѹẺ -- (CR = CS,∠RCT =∠RCS, CT = CR) RT = RS RT = L
∴ AR = AT = RT ∆ART 繠∆ҹ ∠RAT = 60° 180°- 12x = 60° x = 10° Q.E.D.




Create Date : 09 ԧҤ 2558
Last Update : 11 ԧҤ 2558 1:00:00 .
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Fun Geometry Problem with Solution #158
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ Ԡ❀❀❀
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣




٨ x= 30°
٨



(1) ˹ش P 繨شѴҧ AC BD ∠ADP = 80°Р∠BPC = 60°

(2) ˹ش Q AB PQ = BP ∆BPQ 繠∆˹Ҩ ՠ∠P ʹ ∠BQP =∠PBQ ∠BQP = 40° ∠APQ = 20°Р∠AQP = 140°
∵ ∠PAQ =∠APQ ∆APQ 繠∆˹Ҩ ՠ∠Q ʹ AQ = PQ

(3) ˹ش R AD PR = DP ∆DPR 繠∆˹Ҩ ՠ∠P ʹ ∠DRP =∠PDR ∠DRP = 80° ∠APR = 40° ∠APR =∠PAR ∆APR 繠∆˹Ҩ ՠ∠R ʹ AR = PR

(4) ѧࡵҠ∆PQR∆AQR ¤ѹẺ -- (PQ = AQ, PR = AR, QR = QR) ∠PQR (=∠AQR) = (∠AQP)/2 = 70°

(5) ѧࡵҠ∆BCP∆PQR ¤ѹẺ -- (∠CBP =∠PQR, BP = PQ,∠BPC =∠QPR) CP = PR CP = DP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ ∠CDP =∠DCP ∠CDP = x
∵ ∠DCP +∠CDP =∠BPC x + x = 60° x = 30° Q.E.D.




Create Date : 07 ԧҤ 2558
Last Update : 7 ԧҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #157
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ Ԡ❀❀❀
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣




٨ x= 20°
٨



(1)∠BPC = 100°
ԨóҠ☐ACBP Ҡ∠APB (˭) = 360°- 120° ∠APB () = 120°

(2) ˹ش O circumcenter ͧ∆BCP BO = CO = OP
͡ҡ ѧҠ∠COP = 2(∠CBP) ∠COP = 20°
∵ CO = OP ∆COP 繠∆˹Ҩ ՠ∠O (= 20°) ʹ ∠CPO (=∠OCP) = 80° ∠BPO = 20°
∵ BO = OP ∆BOP 繠∆˹Ҩ ՠ∠O ʹ ∠OBP =∠BPO ∠OBP = 20°

(3) α= 10°
ѧࡵҠ☐ACOP CO = OP,∠A =α,∠O = 2αР∠C = 120°-α AP = CO (Click ʹԸվ٨⨷ 2) AP = BO

(4) PO ͡ѧش Q · BQ = BP ∆BPQ 繠∆˹Ҩ ՠ∠B ʹ ∠BQP =∠BPQ ∠BQP = 20° ∠PBQ = 140° ∠OBQ = 120°

(5) ѧࡵҠ∆ABP∆BOQ ¤ѹẺ -- (AP = BO,∠APB =∠OBQ, BP = BQ) ∠ABP =∠BQO x = 20° Q.E.D.




Create Date : 05 ԧҤ 2558
Last Update : 5 ԧҤ 2558 0:03:00 .
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Fun Geometry Problem with Solution #156
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣
❀❀❀ ❀❀❀
╠╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╦╩╣




٨ x= 10°
٨



(1)∠ABC = 50°Р∠ADC = 50°

(2) ˹ش P AB CP = BC ∆BCP 繠∆˹Ҩ ՠ∠C ʹ ∠BPC =∠CBP ∠BPC = 50° ∠ACP = 30°

(3) ˹ش Q CD AQ = AD ∆ADQ 繠∆˹Ҩ ՠ∠A ʹ ∠AQD =∠ADQ ∠AQD = 50° ∠CAQ = 20°Р∠DAQ = 80°

(4) ѧࡵҠ∆ACQ∆ACP ¤ѹẺ -- (∠CAQ =∠CAP, AC = AC,∠ACQ =∠ACP) CQ = CP CQ = BC ∆BCQ 繠∆˹Ҩ ՠ∠C (= 140°) ʹ ∠CBQ (=∠BQC) = 20° ∠ABQ = 30°

(5) ˹ش R Ҿз͹ͧش Q ҹ AB ∆ABR∆ABQ AR = AQ (= AD), BR = BQ,∠BAR =∠BAQ = 40° ∠ABR =∠ABQ = 30°
∵ BQ = BR Р∠QBR = 60° ∆BQR 繠∆ҹ BQ = QR

(6) ѧࡵҠ∆ADQ∆AQR ¤ѹẺ -- (AD = AR,∠DAQ =∠QAR, AQ = AQ) DQ = QR DQ = BQ ∆BDQ 繠∆˹Ҩ ՠ∠Q ʹ ∠DBQ =∠BDQ ∠DBQ = x
∵ ∠DBQ +∠BDQ =∠BQC x + x = 20° x = 10° Q.E.D.




Create Date : 02 ԧҤ 2558
Last Update : 2 ԧҤ 2558 0:00:00 .
Counter : 598 Pageviews.

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