Fun Geometry Problem with Solution #100
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 6°
¾ÔÊÙ¨¹ì



(1) ∠ACB = 108°
∵ ∠BAC = ∠ABC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      AC = BC

(2) µèÍ BP ÍÍ¡ä»Âѧ¨Ø´ D â´Â·Õè ∠DAP = 24°      ∠CAD = 6°, ∠ADP = 126° áÅР∠APD = 30°
ãËé α = 6°
ÊѧࡵÇèÒ ☐ACBD à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ AC = BC, ∠A = α, ∠B = 2α áÅР∠C = 120° - 2α      BC = BD (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 2)    ∆BCD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 12°) à»ç¹ÁØÁÂÍ´      ∠BCD (= ∠BDC) = 84°    ∠ACD = 24°

(3) µèÍ CD Í͡仾º AP ·Õè¨Ø´ E      ∠ADE = 30°      ∠EDP = 96° áÅР∠DEP = 54°

(4) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ADP Ṻ㹠     ...
     • AO = DO = OP
     • ∠AOD = 2(∠APD)      ∠AOD = 60°
     • ∠DOP = 2(∠DAP)      ∠DOP = 48°
∵ AO = DO áÅР∠AOD = 60°      ∆ADO à»ç¹ ∆´éÒ¹à·èÒ      ...
     • AD = AO      AD = OP
     • ∠DAO = 60°      ∠OAP = 36°
     • ∠ADO = 60°      ∠EDO = 30°
ÊѧࡵÇèÒ ∆DEO  ∆ADE ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DO = AD, ∠EDO = ∠ADE, DE = DE)      ∠DOE = ∠DAE      ∠DOE = 24°
∵ AO = OP      ∆AOP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O à»ç¹ÁØÁÂÍ´      ∠APO = ∠OAP      ∠APO = 36°
¾Ô¨ÒóҠ∆EOP ¨Ðä´éÇèÒ ∠OEP = 72°      ∠OEP = ∠EOP      ∆EOP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      EP = OP      EP = AD

(5) ãËé DE = k
¡Ó˹´¨Ø´ Q à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆DEP Ṻ㹠     ...
     • DQ = EQ = PQ
     • ∠DQE = 2(∠DPE)      ∠DQE = 60°
     • ∠DQP = 2(∠DEP)      ∠DQP = 108°
∵ DQ = EQ áÅР∠DQE = 60°      ∆DEQ à»ç¹ ∆´éÒ¹à·èÒ      EQ = DE      EQ = k      PQ = k
∵ EQ = PQ      ∆EPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q (= 168°) à»ç¹ÁØÁÂÍ´      ∠PEQ = ∠EPQ = 6°

(6) ¡Ó˹´¨Ø´ F º¹ AC ·Õè·ÓãËé AF = k
¨ÐàËç¹ÇèÒ ∆ADF  ∆EPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = EP, ∠DAF = ∠PEQ, AF = EQ)      DF = PQ      DF = k
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠AFD = ∠EQP      ∠AFD = 168°      ∠CFD = 12°

(7) µèÍ FC ÍÍ¡ä»Âѧ¨Ø´ G ·Õè·ÓãËé ∠DGF = 12°      ∠DGF = ∠DFG      ∆DFG à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´      DG = DF      DG = k
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠DCG = 156°      ∠CDG = 12° ( ∠GDP = 72°)      ∠CDG = ∠CGD      ∆CDG à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      CD = CG

(8) ÊѧࡵÇèÒ ∆DEG  ∆EPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DE = EQ, ∠EDG = ∠EQP, DG = PQ)      ∠DEG = ∠PEQ      ∠DEG = 6°      ∠GEP = 48°
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ EG = EP      ∆EGP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠E (= 48°) à»ç¹ÁØÁÂÍ´      ∠EPG (= ∠EGP) = 66°      ∠DPG = 36°
¾Ô¨ÒóҠ∆DGP ¨Ðä´éÇèÒ ∠DGP = 72°      ∠DGP = ∠GDP      ∆DGP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      DP = GP
ÊѧࡵÇèÒ ∆CDP  ∆CGP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (CD = CG, DP = GP, CP = CP)      ∠DCP = ∠GCP      ∠DCP = (∠DCG)/2 = 78°      ∠BCP = x = 6°   Q.E.D.

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Fun Geometry Problem with Solution #99
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1



(1) ¡Ó˹´¨Ø´ P º¹ AB ·Õè·ÓãËé CP = BC (= AD)      ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      ∠BPC = ∠CBP      ∠BPC = 48°      ∠ACP = 30° áÅР∠APC = 132°

(2) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ P ¼èÒ¹ AC      ∆ACQ  ∆ACP      CQ = CP, ∠CAQ = ∠CAP = 18° áÅР∠ACQ = ∠ACP = 30°

(3) ∵ CP = CQ áÅР∠PCQ = 60°      ∆CPQ à»ç¹ ∆´éÒ¹à·èÒ      PQ = CP (= AD), ∠CPQ = 60° ( ∠APQ = 72°) áÅР∠CQP = 60°
¾Ô¨ÒóҠ∆APQ ¨ÐàËç¹ÇèÒ ∠P = 2(∠A) áÅÐÁըش D º¹ AP «Öè§ AD = PQ      DQ = PQ (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì)

(4) ÊѧࡵÇèÒ CQ = DQ = PQ      ¨Ø´ Q à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆CDP Ṻ㹠     ∠CDP = (∠CQP)/2      x = 30°   Q.E.D.

¾ÔÊÙ¨¹ì 2



(1) ¡Ó˹´¨Ø´ P º¹ AB ·Õè·ÓãËé CP = BC (= AD)      ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      ∠BPC = ∠CBP      ∠BPC = 48°   ⇔   ∠ACP = 30°

(2) ãËé AC = a áÅÐ CP = b ( AD = b)
¡Ó˹´¨Ø´ Q à˹×Í AD ·Õè·ÓãËé AQ = DQ = a   ⇒   ∆ADQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ°Ò¹ÂÒÇ b áÅдéÒ¹»ÃСͺÁØÁÂÍ´ÂÒÇ a      ∠AQD = 24° (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì)    ∠DAQ (= ∠ADQ) = 78°      ∠CAQ = 60°

(3) ∵ AC = AQ áÅР∠CAQ = 60°      ∆ACQ à»ç¹ ∆´éÒ¹à·èÒ      CQ = a
ÊѧࡵÇèÒ AQ = CQ = DQ      ¨Ø´ Q à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ACD Ṻ㹠     ∠ACD = (∠AQD)/2      ∠ACD = 12°   ⇔   ∠BDC = x = 30°   Q.E.D.

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Fun Geometry Problem with Solution #98
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1



(1) ¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠ACB = 40°
¾Ô¨ÒóҠ∆ABD ¨Ðä´éÇèÒ ∠ADB = 50°      ∆ABD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´      AB = AD

(2) ¡Ó˹´¨Ø´ P º¹ BC ·Õè·ÓãËé AP = AB (= AD)      ∆ABP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´      ∠APB = ∠ABP      ∠APB = 80°      ∠CAP = 40° áÅР∠APC = 100°

(3) ∵ AD = AP áÅР∠DAP = 60°      ∆ADP à»ç¹ ∆´éÒ¹à·èÒ      AP = DP áÅР∠APD = 60° ( ∠CPD = 40°)
∵ ∠CAP = ∠ACP      ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      AP = CP      DP = CP      ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P (= 40°) à»ç¹ÁØÁÂÍ´      ∠DCP (= ∠CDP) = 70°      ∠ACD = x = 30°   Q.E.D.

¾ÔÊÙ¨¹ì 2
ËÅѧ¨Ò¡·Õè·ÃÒºÇèÒ AB = AD áÅéÇ ÍÒ¨¾ÔÊÙ¨¹ìµèÍ´éÇÂÇÔ¸Õà´ÕÂǡѹ¡Ñº Problem 52

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Fun Geometry Problem with Solution #97
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1



(1) ¡Ó˹´¨Ø´ P ·Õè·ÓãËé ☐ABPD à»ç¹ ☐¨ÑµØÃÑÊ      AB = AD = BP = DP      BC = BP = DP
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠DBP = 45°      ∠CBP = 60°

(2) ∵ BC = BP áÅР∠CBP = 60°      ∆BCP à»ç¹ ∆´éÒ¹à·èÒ      BP = CP áÅР∠BPC = 60°
ÊѧࡵÇèÒ BP = CP = DP      ¨Ø´ P à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆BCD Ṻ㹠     ∠BDC = (∠BPC)/2      x = 30°   Q.E.D.

¾ÔÊÙ¨¹ì 2



ãËé AB (= AD = BC) = a

(1) ¡Ó˹´¨Ø´ P áÅШش Q ·Õè·ÓãËé ☐ADQP à»ç¹ ☐ÁØÁ©Ò¡ â´ÂÁըش C ÍÂÙ躹 PQ      ...
     • AP = DQ
     • PQ = AD      PQ = a
     • ∠ADQ = ∠APQ = ∠DQP = 90°

(2) ∵ AB = AD      ∆ABD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 90°) à»ç¹ÁØÁÂÍ´      ∠ABD = 45° ( ∠CBP = 30°) áÅР∠ADB = 45° ( ∠BDQ = 45°)
∵ AB = BC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 150°) à»ç¹ÁØÁÂÍ´      ∠BAC (= ∠ACB) = 15°

(3) ¨ÐàËç¹ÇèÒ ∆BCP à»ç¹ ∆ÁØÁ©Ò¡ ·ÕèÁÕ ∠P à»ç¹ÁØÁ©Ò¡ áÅР∠B = 30°      CP = BC/2      CP = a/2      CQ = a/2
ÊѧࡵÇèÒ ∆CDQ  ∆ACP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DQ = AP, ∠CQD = ∠APC, CQ = CP)      ∠CDQ = ∠CAP      ∠CDQ = 15°      ∠BDC = x = 30°   Q.E.D.

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Fun Geometry Problem with Solution #96
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 24°
¾ÔÊÙ¨¹ì



(1) ∠ACB = 108°
∵ ∠BAC = ∠ABC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      AC = BC

(2) ¡Ó˹´¨Ø´ Q à˹×Í AB ·Õè·ÓãËé AQ = BQ = AB      ∆ABQ à»ç¹ ∆´éÒ¹à·èÒ      ∠ABQ = 60° ( ∠CBQ = 24°) áÅР∠AQB = 60°
ÊѧࡵÇèÒ ∆BCQ  ∆ACQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (BC = AC, BQ = AQ, CQ = CQ)      ∠BQC = ∠AQC = (∠AQB)/2 = 30°

(3) µèÍ BP ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè BR = BQ      BR = AB      ∆ABR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 12°) à»ç¹ÁØÁÂÍ´      ∠ARB (= ∠BAR) = 84°
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∆BCR  ∆BCQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = BC, ∠CBR = ∠CBQ, BR = BQ)      ∠BRC = ∠BQC      ∠BRC = 30°      ∠CRP = ∠CAP      ☐APCR ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠ACP = ∠ARP   ⇔   ∠ACP = 84°      ∠BCP = x = 24°   Q.E.D.

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