Fun Geometry Problem with Solution #140
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1



(1) ∠ACB = 40°
∵ ∠BAC = ∠ABC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      AC = BC

(2) ÊÃéҧǧ¡ÅÁÅéÍÁÃͺ ∆ABC áÅéǵèÍ AP Í͡仾ºàÊé¹Ãͺǧ·Õè¨Ø´ Q      ☐ABQC ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠CBQ = ∠CAQ, ∠BCQ = ∠BAQ áÅР∠AQB = ∠ACB      ∠CBQ = 40°, ∠BCQ = 30° áÅР∠AQB = 40°
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BPQ = 80°

(3) ¾Ô¨ÒóҠ∆ABQ ¨ÐàËç¹ÇèÒ ∠A = 30°, ∠Q = 40° áÅÐÁըش P º¹ AQ ·Õè·ÓãËé ∠BPQ = 80°      AP = BQ (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 1-2)

(4) ÊѧࡵÇèÒ ∆ACP  ∆BCQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AC = BC, ∠CAP = ∠CBQ, AP = BQ)      ∠ACP = ∠BCQ      x = 30°   Q.E.D.

¾ÔÊÙ¨¹ì 2



(1) ∠ACB = 40°
∵ ∠BAC = ∠ABC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´      AC = BC

(2) ¡Ó˹´¨Ø´ Q º¹ AP ·Õè·ÓãËé CQ à»ç¹àÊé¹áºè§¤ÃÖè§ ∠ACB      ∠BCQ = ∠ACQ = (∠ACB)/2 = 20°
¨ÐàËç¹ÇèÒ ∆BCQ  ∆ACQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = AC, ∠BCQ = ∠ACQ, CQ = CQ)      ∠CBQ = ∠CAQ      ∠CBQ = 40°      ∠ABQ = 30° áÅР∠PBQ = 20°
¾Ô¨ÒóҠ∆ABQ áÅР∆ACQ ¨Ðä´éÇèÒ ∠BQP = 60° áÅР∠CQP = 60° µÒÁÅӴѺ

(3) ¾Ô¨ÒóҠ∆BCQ ¨ÐàËç¹ÇèÒ Áըش P à»ç¹¨Ø´ÀÒÂã¹·Õè·ÓãËé BP áÅÐ PQ à»ç¹àÊé¹áºè§¤ÃÖè§ ∠CBQ áÅР∠BQC µÒÁÅӴѺ      ¨Ø´ P à»ç¹ incenter ¢Í§ ∆BCQ      CP à»ç¹àÊé¹áºè§¤ÃÖè§ ∠BCQ      ∠BCP (= ∠PCQ) = (∠BCQ)/2 = 10°      ∠ACP = x = 30°   Q.E.D.

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Create Date : 15 ÁԶعÒ¹ 2558
Last Update : 15 ÁԶعÒ¹ 2558 0:00:00 ¹.
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Fun Geometry Problem with Solution #139
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 18°
¾ÔÊÙ¨¹ì



ãËé AD = L      BC = L

(1) ∠ACB = 138°

(2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ C ¼èÒ¹ AB      ∆ABP  ∆ABC      AP = AC, BP = BC = L, ∠BAP = ∠BAC = 30° áÅÐ ∠ABP = ∠ABC = 12°
∵ AC = AP áÅР∠CAP = 60°      ∆ACP à»ç¹ ∆´éÒ¹à·èÒ      AC = CP

(3) ¡Ó˹´¨Ø´ Q à˹×Í AC ·Õè·ÓãËé AQ = CQ = L
¨ÐàËç¹ÇèÒ ∆ACQ  ∆BCP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (AC = CP, AQ = BC, CQ = BP)      ∠AQC = ∠CBP      ∠AQC = 24°
∵ AQ = CQ      ∆ACQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q (= 24°) à»ç¹ÁØÁÂÍ´      ∠CAQ = 78° áÅР∠ACQ = 78° ( ∠BCQ = 144°)

(4) ∵ BC = CQ      ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 144°) à»ç¹ÁØÁÂÍ´      ∠BQC (= ∠CBQ) = 18°
∵ AD = AQ      ∆ADQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 108°) à»ç¹ÁØÁÂÍ´      ∠AQD (= ∠ADQ) = 36°      ∠CQD = 12°      ∠CQD = ∠CBD      ☐BDCQ ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠ADC = ∠BQC      x = 18°   Q.E.D.

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Create Date : 12 ÁԶعÒ¹ 2558
Last Update : 12 ÁԶعÒ¹ 2558 0:00:00 ¹.
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Fun Geometry Problem with Solution #138
⨷Âì



¡Ó˹´ãËé BC = AB + AD + BD
¨§¾ÔÊÙ¨¹ìÇèÒ x = 12°
¾ÔÊÙ¨¹ì



ãËé AB = a, AD = b áÅÐ BD = c      BC = a + b + c

(1) ∠ADB = 84°

(2) µèÍ AB ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè BP = c      ∠DBP = 120°
∵ BD = BP      ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 120°) à»ç¹ÁØÁÂÍ´      ∠BPD (= ∠BDP) = 30°

(3) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ A ¼èÒ¹ DP      ∆DPQ  ∆ADP      PQ = AP = a + c, DQ = AD = b áÅР∠DPQ = ∠APD = 30°
∵ AP = PQ áÅР∠APQ = 60°      ∆APQ à»ç¹ ∆´éÒ¹à·èÒ      AQ = a + c
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠PAQ = 60°      ∠DAQ = 24°

(4) ¡Ó˹´¨Ø´ R ·Ò§´éÒ¹«éÒ¢ͧ AD ·Õè·ÓãËé DR = a + c áÅР∠ADR = 24°
¨ÐàËç¹ÇèÒ ∆ADR  ∆ADQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = AD, ∠ADR = ∠DAQ, DR = AQ)      AR = DQ      AR = b
∵ AD = AR      ∆ADR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´      ∠ARD = ∠ADR      ∠ARD = 24°      ∠DAR = 132°

(5) µèÍ BA Í͡仾ºÊèǹµèÍ¢ÂÒ¢ͧ DR ·Õè¨Ø´ S      ∠RAS = 12°
¾Ô¨ÒóҠ∆ARS ¨Ðä´éÇèÒ ∠ASR = 12°      ∠ASR = ∠RAS      ∆ARS à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      RS = AR      RS = b

(6) ÊѧࡵÇèÒ ∆BCD  ∆BDS ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = DS, ∠CBD = ∠BDS, BD = BD)      ∠BCD = ∠BSD      x = 12°   Q.E.D.

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Create Date : 09 ÁԶعÒ¹ 2558
Last Update : 9 ÁԶعÒ¹ 2558 0:00:00 ¹.
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Fun Geometry Problem with Solution #137
⨷Âì (Kadir Altintas)



¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì (Kadir Altintas)



(1) ∠APB = 140°
∵ ∠BAP = ∠ABP      ∆ABP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      AP = BP

(2) ¡Ó˹´¨Ø´ Q à˹×Í AP ·Õè·ÓãËé AQ = PQ = AP (= BP)      ∆APQ à»ç¹ ∆´éÒ¹à·èÒ      ∠APQ = 60° ( ∠BPQ = 160°) áÅР∠AQP = 60°
∵ BP = PQ      ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P (= 160°) à»ç¹ÁØÁÂÍ´      ∠PBQ = ∠BQP = 10°

(3) ¡Ó˹´¨Ø´ R à»ç¹¨Ø´µÑ´ÃÐËÇèÒ§ AC áÅÐ BQ
∵ AQ = PQ áÅР∠PQR = 2(∠PAR)      ¨Ø´ Q à»ç¹ circumcenter ¢Í§ ∆APR      ∠ARP = (∠AQP)/2      ∠ARP = 30°
∵ ∠PBR = ∠PCR      ☐BCRP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠CBP = ∠ARP      x = 30°   Q.E.D.

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Create Date : 06 ÁԶعÒ¹ 2558
Last Update : 6 ÁԶعÒ¹ 2558 0:00:00 ¹.
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Fun Geometry Problem with Solution #136
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 48°
¾ÔÊÙ¨¹ì



(1) ∠ACB = 96°

(2) µèÍ CB ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè ∠APC = 30°      ∠BAP = 24°
¾Ô¨ÒóҠ∆ACP ¨ÐàËç¹ÇèÒ ∠C = 96°, ∠P = 30° áÅÐÁըش B º¹ CP ·Õè·ÓãËé ∠ABC = 54°      BP = AC (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 2)      BP = b

(3) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ B ¼èÒ¹ AP      ∆APQ  ∆ABP      AQ = AB = a, PQ = BP = b, ∠PAQ = ∠BAP = 24° áÅР∠APQ = ∠APB = 30°
∵ BP = PQ áÅР∠BPQ = 60°      ∆BPQ à»ç¹ ∆´éÒ¹à·èÒ      BQ = b

(4) ÊѧࡵÇèÒ ∆DEF  ∆ABQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (DE = BQ, DF = AB, EF = AQ)      ∠DFE = ∠BAQ      x = 48°   Q.E.D.

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Create Date : 03 ÁԶعÒ¹ 2558
Last Update : 3 ÁԶعÒ¹ 2558 0:00:00 ¹.
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