Fun Geometry Problem with Solution #130
⨷Âì



¡Ó˹´ãËé BE = AC + CE
¨§¾ÔÊÙ¨¹ìÇèÒ x = 45°
¾ÔÊÙ¨¹ì



ãËé AC = a áÅÐ CE = b      BE = a + b

(1) ∵ ∆ABC à»ç¹ ∆ÁØÁ©Ò¡ áÅШش D à»ç¹¨Ø´¡Öè§¡ÅÒ§´éÒ¹µÃ§¢éÒÁÁØÁ©Ò¡ (AB)      AD = BD = CD
∵ AD = CD      ∆ACD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´
∵ BD = CD      ∆BCD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´

(2) ¡Ó˹´¨Ø´ P º¹ AC ·Õè·ÓãËé DP ⊥ AC      DP à»ç¹ÊèǹÊÙ§¢Í§ ∆ACD      CP (= AP) = AC/2 = a/2

(3) ¡Ó˹´¨Ø´ Q º¹ BC ·Õè·ÓãËé DQ ⊥ BC      DQ à»ç¹ÊèǹÊÙ§¢Í§ ∆BCD      CQ (= BQ) = BC/2 = a/2 + b      EQ = a/2

(4) ¾Ô¨ÒóҠ☐CPDQ ¨ÐàËç¹ÇèÒ ∠C = ∠P = ∠Q = 90°      ☐CPDQ à»ç¹ ☐ÁØÁ©Ò¡      DQ = CP      DQ = a/2      DQ = EQ      ∆DEQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q (= 90°) à»ç¹ÁØÁÂÍ´      ∠DEQ (= ∠EDQ) = x = 45°   Q.E.D.

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Create Date : 16 ¾ÄÉÀÒ¤Á 2558
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Fun Geometry Problem with Solution #129
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì (â´Â¤Ø³ Javi Master Jans)



(1) ∠ACB = 60° áÅР∠ADB = 40°
∵ ∠BAD = ∠ABD      ∆ABD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´      AD = BD

(2) ¡Ó˹´¨Ø´ P º¹ AB ·Õè·ÓãËé CP = BC      ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      ∠BPC = ∠CBP      ∠BPC = 80°      ∠ACP = 40° áÅР∠APC = 100°
∵ ∠CAP = ∠ACP      ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      AP = CP      AP = BC

(3) ¡Ó˹´¨Ø´ Q à˹×Í AP ·Õè·ÓãËé AQ = PQ = AP (= BC)      ∆APQ à»ç¹ ∆´éÒ¹à·èÒ      ∠PAQ = 60° áÅР∠APQ = 60°      ∠DAQ = 10° áÅР∠CPQ = 40°
∵ CP = PQ      ∆CPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P (= 40°) à»ç¹ÁØÁÂÍ´      ∠PCQ = ∠CQP = 70°

(4) ÊѧࡵÇèÒ ∆ADQ  ∆BCD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = BD, ∠DAQ = ∠CBD, AQ = BC)      ∠ADQ = ∠BDC      ∠ADQ = x      ∠BDQ = 40° - x
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ DQ = CD      ∆CDQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 40°) à»ç¹ÁØÁÂÍ´      ∠DCQ = ∠CQD = 70°

(5) ÊѧࡵÇèÒ ∆CDQ  ∆CPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠DCQ = ∠PCQ, CQ = CQ, ∠CQD = ∠CQP)      CD = CP      CD = BC      ∆BCD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      ∠BDC = ∠CBD      x = 10°   Q.E.D.

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Fun Geometry Problem with Solution #128
⨷Âì



¡Ó˹´ãËé AC = BD + CD
¨§¾ÔÊÙ¨¹ìÇèÒ x = 40°
¾ÔÊÙ¨¹ì



(1) ∠CAD = x = 60° - α

(2) µèÍ CD ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè DP = BD      ∠BDP = 120°
∵ BD = DP      ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 120°) à»ç¹ÁØÁÂÍ´      ∠BPD (= ∠DBP) = 30°
∵ CP = CD + DP      CP = CD + BD      CP = AC      ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= α) à»ç¹ÁØÁÂÍ´      ∠CAP (= ∠APC) = 90° - α/2      ∠BAP = 30° + α/2

(3) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ C ¼èÒ¹ BP      ∆BPQ  ∆BCP      BQ = BC, PQ = CP, ∠BPQ = ∠BPC = 30° áÅÐ ∠BQP = ∠BCP = α
∵ CP = PQ áÅР∠CPQ = 60°      ∆CPQ à»ç¹ ∆´éÒ¹à·èÒ      CP = CQ áÅР∠PCQ = 60°

(4) ÊѧࡵÇèÒ AC = CP = CQ      ¨Ø´ C à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆APQ Ṻ㹠     ∠PAQ = (∠PCQ)/2      ∠PAQ = 30°      ∠BAQ = α/2   
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠AQP = (∠ACP)/2      ∠AQP = α/2      ∠AQB = α/2      ∠AQB = ∠BAQ      ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      AB = BQ      AB = BC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      ∠BAC = ∠ACB      60° - α = 2α   ⇔   α = 20°      x = 40°   Q.E.D.

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Fun Geometry Problem with Solution #127
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 18°
¾ÔÊÙ¨¹ì



(1) ∠ACD = 30° - x áÅР∠BCD = x

(2) ¡Ó˹´¨Ø´ O à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆BCD Ṻ㹠     ...
     • BO = CO = DO
     • ∠BOC = 2(∠BDC)      ∠BOC = 60°
     • ∠BOD = 2(∠BCD)      ∠BOD = 2x
∵ BO = CO áÅР∠BOC = 60°      ∆BCO à»ç¹ ∆´éÒ¹à·èÒ      CO = BC      DO = AD
∵ CO = DO      ∆CDO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O (= 60° + 2x) à»ç¹ÁØÁÂÍ´      ∠DCO = 60° - x ( ∠ACO = 30°) áÅР∠CDO = 60° - x

(3) ÊѧࡵÇèÒ ∆ACO  ∆ABC ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AC = AC, ∠ACO = ∠ACB, CO = BC)      ∠CAO = ∠BAC      ∠CAO = x
∵ AD = DO      ∆ADO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´      ∠AOD = ∠DAO      ∠AOD = 2x
¾Ô¨ÒóҠ∆ADO ¨Ðä´éÇèÒ ∠DAO + ∠AOD = ∠BDO      2x + 2x = 90° - x      x = 18°   Q.E.D.

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Fun Geometry Problem with Solution #126
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 70°
¾ÔÊÙ¨¹ì



(1) µèÍ CB ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè ∠BAP = 30°
¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠ABP = 70°
¾Ô¨ÒóҠ☐ADCP ¨ÐàËç¹ÇèÒ ∠DAP + ∠DCP = 180°      ☐ADCP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠ADP = ∠ACP      ∠ADP = 60°

(2) ∵ ∠DAP = ∠ADP = 60°      ∆ADP à»ç¹ ∆´éÒ¹à·èÒ      AD = AP
ÊѧࡵÇèÒ ∆ABD  ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AB = AB, ∠BAD = ∠BAP, AD = AP)      ∠ABD = ∠ABP      x = 70°   Q.E.D.

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