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
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¨§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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