Fun Geometry Problem with Solution #105
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 40°
¾ÔÊÙ¨¹ì 1



(1) ∠BDC = 100°

(2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ D ¼èÒ¹ BC   ⇒   ∆BCP ≅ ∆BCD   ⇒   BP = BD, CP = CD, ∠CBP = ∠CBD = 30° áÅÐ ∠BCP = ∠BCD = 50°
∵ BD = BP áÅР∠DBP = 60°   ⇒   ∆BDP à»ç¹ ∆´éÒ¹à·èÒ   ⇒   DP = BD   ⇔   DP = AC

(3) ¡Ó˹´¨Ø´ Q º¹ BD ·Õè·ÓãËé DQ = CD (= CP)   ⇔   ∆CDQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 100°) à»ç¹ÁØÁÂÍ´   ⇔   ∠CQD (= ∠DCQ) = 40°
ÊѧࡵÇèÒ ∆CDQ ≅ ∆CDP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CD = CD, ∠CDQ = ∠DCP, DQ = CP)   ⇒   CQ = DP   ⇔   CQ = AC   ⇔   ∆ACQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´   ⇔   ∠CAQ = ∠AQC   ⇔   x = 40°   Q.E.D.

¾ÔÊÙ¨¹ì 2



(1) µèÍ CD ÍÍ¡ä»Âѧ¨Ø´ P ·Õè·ÓãËé ∠CBP = 50° (⇔ ∠DBP = 20°)   ⇔   ∠CBP = ∠BCP   ⇔   ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´   ⇔   BP = CP
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BDP = 80°
¾Ô¨ÒóҠ∆BDP ¨Ðä´éÇèÒ ∠BPD = 80°   ⇔   ∠BPD = ∠BDP   ⇔   ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´   ⇔   BP = BD   ⇔   BP = AC

(2) ¡Ó˹´¨Ø´ Q º¹ BC ·Õè·ÓãËé PQ ⊥ BC   ⇒   PQ à»ç¹ÊèǹÊÙ§¢Í§ ∆BCP   ⇒   BQ (= CQ) = BC/2
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BPQ (= ∠CPQ) = (∠BPC)/2 = 40°

(3) ¡Ó˹´¨Ø´ R º¹ AB ·Õè·ÓãËé CR ⊥ AB
¨ÐàËç¹ÇèÒ ∆BCR à»ç¹ ∆ÁØÁ©Ò¡ ·ÕèÁÕ ∠R à»ç¹ÁØÁ©Ò¡ áÅР∠B = 30°   ⇒   CR = BC/2
ÊѧࡵÇèÒ ∆ACR ≅ ∆BPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ©-´-´ (∠ARC = ∠BQP = 90°, CR = BQ, AC = BP)   ⇒   ∠CAR = ∠BPQ   ⇔   x = 40°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 02 ÁÕ¹Ò¤Á 2558
Last Update : 2 ÁÕ¹Ò¤Á 2558 0:00:00 ¹.
Counter : 734 Pageviews.
 
0 comment
Share
Fun Geometry Problem with Solution #104
⨷Âì·Õè¹ÓÁÒàʹÍã¹ blog ¹Õé à»ç¹â¨·Âì¨Ò¡¢éÍÊͺ PAT1 (µ.¤. 53) ¢éÍ·Õè 7 «Öè§à»ç¹â¨·ÂìàÃ×èͧµÃÕ⡳ÁÔµÔ 
ÍÂèÒ§äáçµÒÁ ¨¢º. ¨ÐáÊ´§ÇÔ¸Õ¾ÔÊÙ¨¹ì¤ÓµÍºâ´ÂãªéÇÔ¸Õ·Ò§àâҤ³Ôµ

⨷Âì





¨§¾ÔÊÙ¨¹ìÇèÒ EC/BC = 1/sqrt(3)
¾ÔÊÙ¨¹ì



(1) ∵ AD áÅÐ AE áºè§ ∠BAC ÍÍ¡à»ç¹ 3 Êèǹà·èÒæ ¡Ñ¹   ⇒   ∠BAD = ∠DAE = ∠CAE = 135°/3 = 45°

(2) µèÍ BA ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè ∠BCP = 90°
¾Ô¨ÒóҠ☐AECP ¨ÐàËç¹ÇèÒ ∠ECP = ∠BAE   ⇔   ☐AECP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é   ⇔   ∠CPE = ∠CAE   ⇔   ∠CPE = 45°

(3) ¾Ô¨ÒóҠ∆CEP ¨Ðä´éÇèÒ ∠CEP = 45°   ⇔   ∠CEP = ∠CPE   ⇔   ∆CEP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´   ⇔   EC = CP
¾Ô¨ÒóҠ∆BCP ¨Ðä´éÇèÒ CP/BC = tan30°   ⇔   EC/BC = 1/sqrt(3)   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 27 ¡ØÁÀҾѹ¸ì 2558
Last Update : 27 ¡ØÁÀҾѹ¸ì 2558 0:00:00 ¹.
Counter : 886 Pageviews.
 
0 comment
Share
Fun Geometry Problem with Solution #103
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì 1



(1) ¾Ô¨ÒóҠ∆ABP ¨Ðä´éÇèÒ ∠APB = 180° - 4x
¾Ô¨ÒóҠ∆BCP ¨ÐàËç¹ÇèÒ ∠CBP = ∠BCP   ⇔   ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´   ⇔   BP = CP
¾Ô¨ÒóҠ☐ABCP ¨Ðä´éÇèÒ ∠APC (ÁØÁãË­è) = 360° - 8x   ⇔   ∠APC (ÁØÁàÅç¡) = 8x

(2) ¡Ó˹´¨Ø´ Q ãµé AP ·Õè·ÓãËé PQ = AP áÅР∠BPQ = 8x (⇔ ∠APQ = 180° - 12x)
¨ÐàËç¹ÇèÒ ∆BPQ ≅ ∆ACP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BP = CP, ∠BPQ = ∠APC, PQ = AP)   ⇒   ∠BQP = ∠CAP   ⇔   ∠BQP = 2x
∵ AP = PQ   ⇔   ∆APQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P (= 180° - 12x) à»ç¹ÁØÁÂÍ´   ⇔   ∠AQP (= ∠PAQ) = 6x

(3) ¾Ô¨ÒóҠ∆ABP áÅШش Q ¨ÐàËç¹ÇèÒ ∠AQP = 2(∠ABP) áÅР∠BQP = 2(∠BAP)   ⇒   ¨Ø´ Q à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ABP Ṻ㹠(Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇒   AQ = PQ
∴ AP = AQ = PQ   ⇔   ∆APQ à»ç¹ ∆´éÒ¹à·èÒ   ⇒   ∠AQP = 60°   ⇔   6x = 60°   ⇔   x = 10°   Q.E.D.

¾ÔÊÙ¨¹ì 2



(1) ¡Ó˹´¨Ø´ Q º¹ AP ·Õè·ÓãËé ∠ABQ = x   ⇔   ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´   ⇔   AQ = BQ
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠PBQ = 2x áÅР∠BQP = 2x

(2) ÊѧࡵÇèÒ ∆BCP ≅ ∆BPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠BCP = ∠BQP, ∠CBP = ∠PBQ, BP = BP)   ⇒   BC = BQ

(3) ãËé α = 2x
ÊѧࡵÇèÒ ☐ACBQ à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ AQ = BQ = BC, ∠A = α áÅР∠B = 2α   ⇒   ∠ACB = 120° - α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 1) ⇔   ∠ACB = 120° - 2x
¾Ô¨ÒóҠ∆ABC ¨Ðä´éÇèÒ ∠BAC + ∠ABC + ∠ACB = 180°   ⇔   3x + 5x + (120° - 2x) = 180°   ⇔   x = 10°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 24 ¡ØÁÀҾѹ¸ì 2558
Last Update : 24 ¡ØÁÀҾѹ¸ì 2558 0:00:00 ¹.
Counter : 810 Pageviews.
 
0 comment
Share
Fun Geometry Problem with Solution #102
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì



(1) ¡Ó˹´¨Ø´ P º¹ BC ·Õè·ÓãËé AP = AC (= BD)   ⇔   ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´   ⇔   ∠APC = ∠ACP   ⇔   ∠APC = 6x   ⇔   ∠BAP = 4x

(2) ¾Ô¨ÒÃ³Ò ∆ABP ¨ÐàËç¹ÇèÒ ∠A = 2(∠B) áÅÐÁըش D º¹ AB «Öè§ AP = BD   ⇒   DP = AP (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔   DP = BD   ⇔   ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´   ⇔   ∠BPD = ∠DBP   ⇔   ∠BPD = 2x   ⇔   ∠CDP = x   ⇔   ∠CDP = ∠DCP   ⇔   ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´   ⇔   CP = DP   ⇒   CP = AP

(3) ÊѧࡵÇèÒ AC = AP = CP   ⇔   ∆ACP à»ç¹ ∆´éÒ¹à·èÒ   ⇒   ∠ACP = 60°   ⇔   6x = 60°   ⇔   x = 10°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 21 ¡ØÁÀҾѹ¸ì 2558
Last Update : 23 ¡ØÁÀҾѹ¸ì 2558 16:35:00 ¹.
Counter : 698 Pageviews.
 
0 comment
Share
Fun Geometry Problem with Solution #101
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì



(1) ∠BPC = 100°

(2) µèÍ BP ÍÍ¡ä»Âѧ¨Ø´ Q â´Â·Õè ∠BQC = 20°   ⇔   ∠BQC = ∠CBQ   ⇔   ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´   ⇔   CQ = BC   ⇔   CQ = AP
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠CPQ = 80° áÅР∠APQ = 60°
¾Ô¨ÒóҠ∆CPQ ¨Ðä´éÇèÒ ∠PCQ = 80°   ⇔   ∠PCQ = ∠CPQ   ⇔   ∆CPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´   ⇔   PQ = CQ   ⇔   PQ = AP

(3) ∵ AP = PQ áÅР∠APQ = 60°   ⇒   ∆APQ à»ç¹ ∆´éÒ¹à·èÒ   ⇒   AQ = AP
ÊѧࡵÇèÒ AQ = CQ = PQ   ⇔   ¨Ø´ Q à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆ACP Ṻ㹠  ⇒   ∠CAP = (∠CQP)/2   ⇔   x = 10°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 18 ¡ØÁÀҾѹ¸ì 2558
Last Update : 18 ¡ØÁÀҾѹ¸ì 2558 0:00:00 ¹.
Counter : 717 Pageviews.
 
0 comment
Share
1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  
TIYHz
Location :
¡Ãا෾Ϡ Thailand

[´Ù Profile ·Ñé§ËÁ´]
 ½Ò¡¢éͤÇÒÁËÅѧäÁ¤ì
 Rss Feed
 Smember
 ¼ÙéµÔ´µÒÁºÅçÍ¡ : 20 ¤¹ [?]



¨Ø´»ÃÐʧ¤ì·Õè·ÓºÅçÍ¡¤³ÔµÈÒʵÃì¢Öé¹ÁÒ... ¡çäÁèÁÕÍÐäÃÁÒ¡¤ÃѺ á¤èÍÂÒ¡ãËé»ÃÐà·ÈàÃÒÁÕÍÐäÃẺ¹ÕéºéÒ§
All Blog