Fun Geometry Problem with Solution #115
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٨ x = 24°
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(1)∵AP = CP ∆ACP 繠∆˹Ҩ ՠ∠P ʹ ∠CAP =∠ACP ∠CAP = 6° ∠APC = 168°

(2) ˹ش Q ҧҹ¢ͧ AP AQ = PQ = AP (= CP) ∆APQ 繠∆ҹ ∠PAQ = 60°Р∠APQ = 60°(∠CPQ = 108°)
∵ CP = PQ ∆CPQ 繠∆˹Ҩ ՠ∠P (= 108°) ʹ ∠PCQ = 36°(∠ACQ = 30°) Р∠CQP = 36°

(3) ˹ش R CQ CR = CP ∆CPR 繠∆˹Ҩ ՠ∠C (= 36°) ʹ ∠CPR (=∠CRP) = 72° ∠QPR = 36° ∠QPR =∠PQR ∆PQR 繠∆˹Ҩ ՠ∠R ʹ PR = QR

(4) ѧࡵҠ∆APR∆AQR ¤ѹẺ -- (AP = AQ, AR = AR, PR = QR) ∠PAR (=∠QAR) = (∠PAQ)/2 = 30° ∠CAR = 24°
ѧࡵҠ∆BCP∆ACR ¤ѹẺ -- (BC = AC,∠BCP =∠ACR, CP = CR) ∠CBP =∠CAR x = 24° Q.E.D.




Create Date : 01 ¹ 2558
Last Update : 1 ¹ 2558 0:00:01 .
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Fun Geometry Problem with Solution #114



٨ x = 18°
٨



(1)∠ACB = 60°Р∠APB = 78°
∵ ∠BAC =∠ABC =∠ACB ∆ABC 繠∆ҹ AB = BC

(2) ˹ش Q AP BQ 觤觠∠ABC ∠CBQ (=∠ABQ) = (∠ABC)/2 = 30° ∠PBQ = 24°
Ҡ∆ABQ∆BCQ ¤ѹẺ -- (AB = BC,∠ABQ =∠CBQ, BQ = BQ) ∠BCQ =∠BAQ ∠BCQ = 48° ∠ACQ = 12°
ԨóҠ∆ACQ Ҡ∠CQP = 24°
ԨóҠ∆BPQ Ҡ∠BQP = 78° ∠BQP =∠BPQ ∆BPQ 繠∆˹Ҩ ՠ∠B ʹ BP = BQ

(3) ˹ش O 繨شٹҧͧǧՠ∆BCQ Ṻ BO = CO = OQ
͡ҡ ѧҠ∠COQ = 2(∠CBQ) ∠COQ = 60°
∵ CO = OQ Р∠COQ = 60° ∆COQ 繠∆ҹ CO = CQ,∠OCQ = 60°(∠BCO = 12°) Р∠CQO = 60°(∠OQP = 36°)
∵ BO = CO ∆BCO 繠∆˹Ҩ ՠ∠O ʹ ∠CBO =∠BCO ∠CBO = 12°

(4) ˹ش R ˹ BQ BR = QR = BQ (= BP) ∆BQR 繠∆ҹ ∠QBR = 60°(∠OBR = 18°) Р∠BRQ = 60°
Ҡ∆BOR∆OQR ¤ѹẺ -- (BO = OQ, BR = QR, OR = OR) ∠BRO (=∠ORQ) = (∠BRQ)/2 = 30°
ѧࡵҠ∆BOP∆BOR ¤ѹẺ -- (BO = BO,∠OBP =∠OBR, BP = BR) ∠BPO =∠BRO ∠BPO = 30°

(5) ԨóҠ∆OPQ Ҡ∠POQ = 36° ∠POQ =∠OQP ∆OPQ 繠∆˹Ҩ ՠ∠P ʹ OP = PQ
ѧࡵҠ∆COP∆CPQ ¤ѹẺ -- (CO = CQ, CP = CP, OP = PQ) ∠OCP (=∠PCQ) = (∠OCQ)/2 = 30° ∠BCP = x = 18° Q.E.D.




Create Date : 29 չҤ 2558
Last Update : 29 չҤ 2558 0:00:03 .
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Fun Geometry Problem with Solution #113



٨ x = 15°
٨



(1)∠CAP = 90°- x
ԨóҠ∆ACP Ҡ∠APC = 90°- x ∠APC =∠CAP ∆ACP 繠∆˹Ҩ ՠ∠C ʹ CP = AC CP = AB
∵ AB = AC ∆ABC 繠∆˹Ҩ ՠ∠A (= 90°) ʹ ∠ABC =∠ACB = 45° ∠ABP =∠BCP = 45°- 2x

(2) BP ͡ѧش Q · AQ = AB ∆ABQ 繠∆˹Ҩ ՠ∠A ʹ ∠AQB =∠ABQ ∠AQB = 45°- 2x ∠BAQ = 90°+ 4x ∠CAQ = 4x
͡ҡ ѧҠ∠CPQ = 45°

(3)∵AQ = AB AQ = AC ∆ACQ 繠∆˹Ҩ ՠ∠A (= 4x) ʹ ∠AQC (=∠ACQ) = 90°- 2x ∠CQP = 45° ∠CQP =∠CPQ ∆CPQ 繠∆˹Ҩ ՠ∠C ʹ CQ = CP CQ = AB
ѧࡵ AC = AQ = CQ ∆ACQ 繠∆ҹ ∠CAQ = 60° 4x = 60° x = 15° Q.E.D.




Create Date : 26 չҤ 2558
Last Update : 26 չҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #112



٨ x = 20°
٨



(1)∠ABC = 40°

(2) ˹ش O 繨شٹҧͧǧՠ∆ABC Ṻ AO = BO = CO
͡ҡ ѧ ∠BOC = 2(∠BAC) ∠BOC = 60°
∵ BO = CO Р∠BOC = 60° ∆BCO 繠∆ҹ BC = CO, ∠CBO = 60° (∠ABO = 20°) ∠BCO = 60° (∠OCP = 30°)
∵ AO = BO ∆ABO 繠∆˹Ҩ ՠ∠O ʹ ∠BAO =∠ABO ∠BAO = 20°

(3) ѧࡵҠ∠OAP =∠OCP ☐ACPO öṺǧ ∠COP =∠CAP ∠COP = 20°
ѧࡵҠ∆BCP∆COP ¤ѹẺ -- (BC = CO,∠BCP =∠OCP, CP = CP) ∠CBP =∠COP ∠CBP = 20° ∠ABP = x = 20° Q.E.D.




Create Date : 23 չҤ 2558
Last Update : 23 չҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #111



٨ x = 30°
٨



(1)∠ACB = 40°Р∠APB = 70°
∵ ∠ABP =∠APB ∆ABP 繠∆˹Ҩ ՠ∠A ʹ AB = AP

(2) ˹ش Q BC AQ = AB ∆ABQ 繠∆˹Ҩ ՠ∠A ʹ ∠AQB =∠ABQ ∠AQB = 80° ∠CAQ = 40°(∠PAQ = 20°) ∠AQC = 100°
∵ ∠CAQ =∠ACQ ∆ACQ 繠∆˹Ҩ ՠ∠Q ʹ AQ = CQ

(3) α= 10°
ѧࡵҠ☐APCQ AP = AQ = CQ,∠A = 2αР∠Q = 120°- 2α ∠PCQ =α (Click ʹԸվ٨⨷ 3) ∠PCQ = 10° ∠ACP = x = 30° Q.E.D.




Create Date : 20 չҤ 2558
Last Update : 20 չҤ 2558 0:00:00 .
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TIYHz
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