Fun Geometry Problem with Solution #155
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 15°
¾ÔÊÙ¨¹ì



(1) ¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠CBD = 180° - 5x
¾Ô¨ÒóҠ∆BCD ¨Ðä´éÇèÒ ∠BDC = 3x

(2) ÊÃéҧǧ¡ÅÁÅéÍÁÃͺ ∆ABD áÅСÓ˹´ãËéàÊé¹ÃͺǧµÑ´¡Ñº CD ·Õè¨Ø´ P      ☐ABPD ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠BAP = ∠BDP, ∠APB = ∠ADB áÅР∠APD = ∠ABD      ∠BAP = 3x, ∠APB = 3x áÅР∠APD = 2x
∵ ∠BAP = ∠APB      ∆ABP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      AB = BP
∵ ∠CBP + ∠BCP = ∠BPD      ∠CBP + 2x = 5x      ∠CBP = 3x

(3) ¡Ó˹´¨Ø´ Q º¹ AC ·Õè·ÓãËé BQ = AB      ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      ∠AQB = ∠BAQ      ∠AQB = 2x      ∠CBQ = x      ∠PBQ = 2x   
∵ ∠CBQ = ∠BCQ      ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´      CQ = BQ      CQ = AB

(4) ÊѧࡵÇèÒ ☐BPCQ à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ BP = BQ = CQ, ∠B = 2x áÅР∠C = x      ∠BPC = 120° - x (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 1)

(5) ¾Ô¨ÒóҠ∆BCP ¨Ðä´éÇèÒ ∠CBP + ∠BCP + ∠BPC = 180°      3x + 2x + (120° - x) = 180°      x = 15°   Q.E.D.

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Create Date : 30 ¡Ã¡®Ò¤Á 2558
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Fun Geometry Problem with Solution #154
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì



(1) ∠ACP = 20° - x áÅР∠BPC = 50°

(2) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ P ¼èÒ¹ BC      ∆BCQ  ∆BCP      BQ = BP, ∠CBQ = ∠CBP = 30° áÅР∠BQC = ∠BPC = 50°
∵ BP = BQ áÅР∠PBQ = 60°      ∆BPQ à»ç¹ ∆´éÒ¹à·èÒ      BP = PQ
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BPQ = ∠BQP = 60°      ∠CPQ = ∠CQP = 10°

(3) ÊѧࡵÇèÒ ∠CAP = ∠CQP      ☐APCQ ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠CAQ = ∠CPQ áÅР∠AQP = ∠ACP      ∠CAQ = 10° áÅР∠AQP = 20° - x

(4) ¡Ó˹´¨Ø´ R º¹ AB ·Õè·ÓãËé PR = AP      ∆APR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠ARP = ∠PAR      ∠ARP = 20°      ∠BPR = 20° - x

(5) ¡Ó˹´¨Ø´ S º¹ AQ ·Õè·ÓãËé PS = AP      ∆APS à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠ASP = ∠PAS   ⇔   ∠ASP = 20°      ∠QPS = x

(6) ÊѧࡵÇèÒ ∆BPR  ∆PQS ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠PBR = ∠QPS, BP = PQ, ∠BPR = ∠PQS)      BR = PS (= AP)   ⇔   BR = PR      ∆BPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      ∠PBR = ∠BPR      x = 20° - x      x = 10°   Q.E.D.

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Create Date : 27 ¡Ã¡®Ò¤Á 2558
Last Update : 27 ¡Ã¡®Ò¤Á 2558 0:00:01 ¹.
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Fun Geometry Problem with Solution #153
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 20°
¾ÔÊÙ¨¹ì



(1) ∠ACB = 30° - x, ∠ADB = 60° - x áÅР∠BDC = 80° + x
∵ ∠CAD = ∠ACD      ∆ACD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´      AD = CD

(2) ¡Ó˹´¨Ø´ P à˹×Í CD ·Õè·ÓãËé CP = AB áÅР∠DCP = 20° + x
¨ÐàËç¹ÇèÒ ∆CDP  ∆ABD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CD = AD, ∠DCP = ∠BAD, CP = AB)      ∠CDP = ∠ADB áÅР∠CPD = ∠ABD      ∠CDP = 60° - x áÅР∠CPD = 100°
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ DP = BD      ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 140°) à»ç¹ÁØÁÂÍ´      ∠DBP = 20° áÅР∠BPD = 20°      ∠CBP = 30° áÅР∠BPC = 80°

(3) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ P ¼èÒ¹ BC      ∆BCQ  ∆BCP      BQ = BP, ∠CBQ = ∠CBP = 30° áÅР∠BQC = ∠BPC = 80°
∵ BP = BQ áÅР∠PBQ = 60°      ∆BPQ à»ç¹ ∆´éÒ¹à·èÒ      BP = PQ
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BPQ = ∠BQP = 60°      ∠CPQ = ∠CQP = 20°

(4) ÊѧࡵÇèÒ ∆BDP  ∆CPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠DBP = ∠CQP, BP = PQ, ∠BPD = ∠CPQ)      DP = CP      ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠DCP = ∠CDP      20° + x = 60° - x      x = 20°   Q.E.D.

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Fun Geometry Problem with Solution #152
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì



(1) ∠APB = 120°

(2) ¡Ó˹´¨Ø´ Q º¹ AB ·Õè·ÓãËé PQ = BP      ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠BQP = ∠PBQ      ∠BQP = 40°      ∠APQ = 20°   ⇔   ∠APQ = ∠PAQ      ∆APQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´      AQ = PQ

(3) ¡Ó˹´¨Ø´ R ãµé AP ·Õè·ÓãËé ∠PAR = 30° áÅР∠APR = 40°      ∠QAR = 10°, ∠BPR = 80°, ∠QPR = 20° áÅР∠ARP = 110°

♦ ãËé α = 10°
ÊѧࡵÇèÒ ☐AQPR à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ AQ = PQ, ∠QAR = α, ∠QPR = 2α áÅР∠ARP = 120° - α      PR = PQ (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 3)      PR = BP

♦ ∵ ∠ACP + ∠ARP = 180°      ☐ACPR ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠PCR = ∠PAR      ∠PCR = 30°

(4) ÊѧࡵÇèÒ BP = PR áÅР∠BPR = 2(∠BCR)      ¨Ø´ P à»ç¹ circumcenter ¢Í§ ∆BCR      BP = CP      ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠CBP = ∠BCP      x = 10°   Q.E.D.

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Fun Geometry Problem with Solution #151
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¡Ó˹´ãËé AE = BE
¨§¾ÔÊÙ¨¹ìÇèÒ x = 22.5°
¾ÔÊÙ¨¹ì 1



(1) ∠ABC = 90° - 3x

(2) ¡Ó˹´¨Ø´ P º¹ BC ·Õè·ÓãËé PE ⊥ AB
¨ÐàËç¹ÇèÒ ∆AEP  ∆BEP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AE = BE, ∠AEP = ∠BEP, EP = EP)      ∠EAP = ∠EBP      ∠EAP = 90° - 3x      ∠APE = 3x

(3) ∵ ∠ACE = ∠APE      ☐ACPE ÊÒÁÒöṺã¹Ç§¡ÅÁä´é      ∠EAP = ∠ECP      90° - 3x = x      x = 22.5°   Q.E.D.

¾ÔÊÙ¨¹ì 2 (äÁèãªé·Äɮպ·à¡ÕèÂǡѺǧ¡ÅÁ)



(1) ∠ABC = 90° - 3x, ∠AEC = 90° - 2x áÅР∠BEC = 90° + 2x

(2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ B ¼èÒ¹ CE      ∆CEP  ∆BCE      ...
     • CP = BC
     • ∠ECP = ∠BCE      ∠ECP = x      ∠DCP = x
     • ∠CEP = ∠BEC      ∠CEP = 90° + 2x      ∠AEP = 4x
     • ∠CPE = ∠CBE      ∠CPE = 90° - 3x
     • EP = BE      EP = AE      ∆AEP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠E (= 4x) à»ç¹ÁØÁÂÍ´      ∠APE (= ∠EAP) = 90° - 2x      ∠APC = x

(3) ¡Ó˹´¨Ø´ Q à»ç¹¨Ø´µÑ´ÃÐËÇèÒ§ AB áÅÐ CP
ÊѧࡵÇèÒ ∆ACD  ∆CDQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠ACD = ∠DCQ, CD = CD, ∠ADC = ∠CDQ)      AC = CQ

(4) ÊѧࡵÇèÒ ∆ACP  ∆BCQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AC = CQ, ∠ACP = ∠BCQ, CP = BC)      ∠APC = ∠CBQ      x = 90° - 3x      x = 22.5°   Q.E.D.

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