Fun Geometry Problem with Solution #146
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì 1



(1) ∠ACB = 80°

(2) ¡Ó˹´¨Ø´ O à»ç¹ circumcenter ¢Í§ ∆ABP      ...
     • AO = BO = OP
     • ∠AOP = 2(∠ABP)      ∠AOP = 80°
     • ∠BOP = 2(∠BAP)      ∠BOP = 20°
∵ AO = BO      ∆ABO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O (= 100°) à»ç¹ÁØÁÂÍ´      ∠BAO = ∠ABO = 40°

ãËé BO = L      OP = L

(3) µèÍ BP Í͡仾º AC ·Õè¨Ø´ Q      ∠BQC = 80°
ÊѧࡵÇèÒ ∆ABQ  ∆ABO ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠BAQ = ∠BAO, AB = AB, ∠ABQ = ∠ABO)      BQ = BO      BQ = L
∵ ∠BQC = ∠BCQ      ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      BC = BQ      BC = L

(4) ãËé α = 10°
ÊѧࡵÇèÒ ☐BCPO à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ BC = BO = OP, ∠CBO = 120° - 2α áÅР∠BOP = 2α      ∠BCP = α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 3)      x = 10°   Q.E.D.

¾ÔÊÙ¨¹ì 2



(1) µèÍ BP Í͡仾º AC ·Õè¨Ø´ Q   ⇒   ∠APQ = 50° áÅÐ ∠BQC = 80°
∵ ∠BAQ = ∠ABQ      ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´      AQ = BQ

ãËé BP = a áÅÐ PQ = b      BQ = a + b      AQ = a + b

(2) µèÍ BQ ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè AR = a + b      ∠AQR = 80°
∵ AQ = AR      ∆AQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´      ∠ARQ = ∠AQR      ∠ARQ = 80°      ∠QAR = 20°

(3) ∵ ∠PAR = ∠APR      ∆APR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      PR = AR      PR = a + b      QR = a
¨ÐàËç¹ÇèÒ ∆BCQ  ∆AQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠CBQ = ∠QAR, BQ = AQ, ∠BQC = ∠AQR)      CQ = QR      CQ = a

(4) ¾Ô¨ÒóҠ∆BCQ ¨ÐàËç¹ÇèÒ ∠B = 20°, ∠Q = 80° áÅÐÁըش P º¹ BQ ·Õè·ÓãËé BP = CQ      ∠BCP = 10° (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì)      x = 10°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 03 ¡Ã¡®Ò¤Á 2558
Last Update : 29 ¡Ã¡®Ò¤Á 2558 23:46:00 ¹.
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