Fun Geometry Problem with Solution #75
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 90°
¾ÔÊÙ¨¹ì



(1) ∵ ¨Ø´ I à»ç¹ incenter ¢Í§ ∆ABC      AI, BI áÅÐ CI à»ç¹àÊé¹áºè§¤ÃÖè§ ∠BAC, ∠ABC áÅР∠ACB µÒÁÅӴѺ   ⇒   ∠BAI = ∠CAI, ∠ABI = ∠CBI áÅР∠ACI = ∠BCI

ãËé ∠ABI = θ      ∠CBI = θ

(2) ¡Ó˹´¨Ø´ P º¹ AB ·Õè·ÓãËé AP = AC
∵ AB = AC + CI      AP + BP = AC + CI      BP = CI
¨ÐàËç¹ÇèÒ ∆AIP  ∆ACI ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = AC, ∠IAP = ∠CAI, AI = AI)      IP = CI      IP = BP      ∆BIP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠BIP = ∠IBP      ∠BIP = θ

(3) ÊѧࡵÇèÒ CI = IP áÅР∠CBI = ∠IBP      ☐BCIP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠BCP = ∠BIP áÅР∠ICP = ∠IBP      ∠BCP = θ áÅР∠ICP = θ

(4) ∵ AC = BC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      ∠BAC = ∠ABC      ∠BAC = 2θ
∵ ∠ACI = ∠BCI      ∠ACI = 2θ
¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠BAC + ∠ABC + ∠ACB = 180°      2θ + 2θ + 4θ = 180°      θ = 22.5°      4θ = x = 90°   Q.E.D.

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Fun Geometry Problem with Solution #74
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 60°
¾ÔÊÙ¨¹ì



(1) ∠APC = 150° áÅР∠BPC = 110°

(2) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆BCP Ṻ㹠     ...
     • ¨Ø´ O ÍÂÙè¹Í¡ ∆BCP
     • BO = CO = OP
     • ∠BOP = 2(∠BCP)      ∠BOP = 80°
     • ∠COP = 2(∠CBP)      ∠COP = 60°
∵ CO = OP áÅР∠COP = 60°      ∆COP à»ç¹ ∆´éÒ¹à·èÒ      CP = OP
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠CPO = 60°      ∠APO = 150°

(3) ÊѧࡵÇèÒ ∆AOP  ∆ACP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (OP = CP, ∠APO = ∠APC, AP = AP)      ∠OAP = ∠CAP      ∠OAP = 20°
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠AOP = ∠ACP      ∠AOP = 10°      ∠AOB = 70°

(4) ÊѧࡵÇèÒ ∆ABO  ∆ACO ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BO = CO, ∠AOB = ∠AOC, AO = AO)      ∠BAO = ∠CAO      ∠BAO = 40°      ∠BAP = x = 60°   Q.E.D.

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Fun Geometry Problem with Solution #73
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 100°
¾ÔÊÙ¨¹ì



(1) ¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠ACB = 20°      ∠ACB = ∠BAC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      AB = BC

(2) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ABD Ṻ㹠     ...
     • ¨Ø´ O ÍÂÙèà˹×Í AD
     • AO = BO = DO
     • ∠BOD = 2(∠BAD)      ∠BOD = 60°
∵ BO = DO áÅР∠BOD = 60°      ∆BDO à»ç¹ ∆´éÒ¹à·èÒ      BD = BO
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠DBO = 60°      ∠ABO = 40°
∵ AO = BO      ∆ABO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O à»ç¹ÁØÁÂÍ´      ∠BAO = ∠ABO      ∠BAO = 40°      ∠AOB = 100°

(3) ÊѧࡵÇèÒ ∆BCD  ∆ABO ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = AB, ∠CBD = ∠ABO, BD = BO)      ∠BDC = ∠AOB      x = 100°   Q.E.D.

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Create Date : 27 ¾ÄȨԡÒ¹ 2557
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Fun Geometry Problem with Solution #72
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¡Ó˹´ãËé ∠BDC à»ç¹ÁØÁ»éÒ¹ 
¨§¾ÔÊÙ¨¹ìÇèÒ x = 24°
¾ÔÊÙ¨¹ì



(1) ∠CAD = 120° - x áÅР∠BDC = 150° - x

(2) ¡Ó˹´¨Ø´ P º¹ BC ·Õè·ÓãËé BP = CD
¨ÐàËç¹ÇèÒ ∆BDP  ∆ACD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BD = AC, ∠DBP = ∠ACD, BP = CD)      ∠BDP = ∠CAD      ∠BDP = 120° - x      ∠CDP = 30°

(3) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆BCD Ṻ㹠     ...
     • ¨Ø´ O ÍÂÙèà˹×Í BC
     • BO = CO = DO
     • ∠BOD = 2(∠BCD)      ∠BOD = 2x
     • ∠COD = 2(∠CBD)      ∠COD = 60°
∵ CO = DO áÅР∠COD = 60°      ∆CDO à»ç¹ ∆´éÒ¹à·èÒ      CD = CO = DO   
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠DCO = 60° áÅР∠CDO = 60°      ∠BCO = 60° - x áÅР∠ODP = 30°
∵ BO = CO      ∆BCO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O à»ç¹ÁØÁÂÍ´      ∠CBO = ∠BCO      ∠CBO = 60° - x

(4) ÊѧࡵÇèÒ ∆DOP  ∆CDP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DO = CD, ∠ODP = ∠CDP, DP = DP)      ∠DOP = ∠DCP      ∠DOP = x

(5) ∵ CD = CO      BP = BO      ∆BOP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      ∠BPO = ∠BOP      ∠BPO = 3x
¾Ô¨ÒóҠ∆BOP ¨Ðä´éÇèÒ ∠OBP + ∠BOP + ∠BPO = 180°      (60° - x) + 3x + 3x = 180°      x = 24°   Q.E.D.

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Fun Geometry Problem with Solution #71
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¡Ó˹´ãËé ∠ACD à»ç¹ÁØÁ»éÒ¹ 
¨§¾ÔÊÙ¨¹ìÇèÒ x = 6°
¾ÔÊÙ¨¹ì



(1) ¾Ô¨ÒóҠ∆BCD ¨Ðä´éÇèÒ ∠ADC = 3x

(2) ¡Ó˹´¨Ø´ P º¹ BC ·Õè·ÓãËé DP = BD   ⇔   BDP à»ç¹ Ë¹éÒ¨ÑèÇ ·ÕèÁÕ D à»ç¹ÁØÁÂÍ´   ⇔   BPD = DBP   ⇔   BPD = 2x   ⇔   CDP = x   ⇔   CDP = DCP   ⇔   CDP à»ç¹ Ë¹éÒ¨ÑèÇ ·ÕèÁÕ P à»ç¹ÁØÁÂÍ´   ⇔   CP = DP      CP = BD

(3) ¡Ó˹´¨Ø´ Q à˹×Í AB ·Õè·ÓãËé DAQ = x áÅРADQ = 2x   ⇔   CAQ = 30° - x áÅРCDQ = x
¨ÐàËç¹ÇèÒ ADQ  BCD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (DAQ = BCD, AD = BC, ADQ = CBD)   ⇒   AQ = CD áÅÐ DQ = BD
¨ÐàËç¹ÇèÒ CDQ  CDP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DQ = DP, CDQ = CDP, CD = CD)   ⇒   CQ = CP áÅРDCQ = DCP   ⇔   CQ = BD áÅРDCQ = x
¾Ô¨ÒóҠADCQ ¨Ðä´éÇèÒ AQC (ÁØÁãË­è) = 360° - 5x   ⇔   AQC (ÁØÁàÅç¡) = 5x

(4) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ACD Ṻ㹠  ⇒   ...
     • ¨Ø´ O ÍÂÙèãµé AD
     • AO = CO = DO
     • ∠AOC = 2(∠ADC)   ⇔   AOC = 6x
     • COD = 2(CAD)   ⇔   COD = 60°
 CO = DO áÅРCOD = 60°      ∆CDO à»ç¹ ´éÒ¹à·èÒ      CD = CO = DO   ⇔   AQ = AO
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ CDO = 60°   ⇔   ADO = 60° - 3x
 AO = DO   ⇔   ADO à»ç¹ Ë¹éÒ¨ÑèÇ ·ÕèÁÕ O à»ç¹ÁØÁÂÍ´   ⇔   DAO = ADO   ⇔   DAO = 60° - 3x   ⇔   OAQ = 60° - 2x

(5) ¡Ó˹´¨Ø´ R à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ C ¼èÒ¹ AQ   ⇒   AQR  ACQ   ⇒   ...
     • QR = CQ   ⇔   QR = BD
     • QAR = CAQ   ⇔   QAR = 30° - x   ⇔   OAR = 30° - x
     • AQR = AQC   ⇔   AQR = 5x
ÊѧࡵÇèÒ AOR  AQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AO = AQ, OAR = QAR, AR = AR)   ⇒   OR = QR   ⇔   OR = BD
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ AOR = AQR   ⇔   AOR = 5x   ⇔   COR = x

(6) ÊѧࡵÇèÒ COR  CDQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (OR = DQ, COR = CDQ, CO = CD)   ⇒   CR = CQ
 CQ = CR = QR   ⇔   CQR à»ç¹ ´éÒ¹à·èÒ   ⇒   CQR = 60°   ⇔   10x = 60°   ⇔   x = 6°   Q.E.D.

ËÁÒÂà˵ؠàÃÒÊÒÁÒöáÊ´§ÇèҨش Q ÍÂÙèã¹ ∆ACD ä´é´Ñ§¹Õé
∵ ∠ACD à»ç¹ÁØÁ»éÒ¹      150° - 3x > 90°      x < 20°      ∠DAQ < 20°
∵ ∠DAQ < ∠CAD áÅР∠ADQ < ∠ADC
∴ ¨Ø´ Q ÍÂÙèã¹ ∆ACD

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