Fun Geometry Problem with Solution #125



˹ AB = CP AC = AB + BP
٨ x = 20°
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AB = a BP = b CP = a AC = a + b

(1) ˹ش O 繨شٹҧͧǧՠ∆ABP Ṻ ⇒ ...
AO = BO = OP
∠AOP = 2(∠ABP) ∠AOP = 80°
∠BOP = 2(∠BAP) ∠BOP = 60°
∵ BO = OP Р∠BOP = 60° ∆BOP 繠∆ҹ OP = BP AO = BO = b

(2) AB ͡ѧش Q · BQ = b ∠PBQ = 140°
∵ BP = BQ ∆BPQ 繠∆˹Ҩ ՠ∠B (= 140°) ʹ ∠BQP (=∠BPQ) = 20°
ѧࡵҠ∆BPQ∆ABO ¤ѹẺ -- (BP = AO,∠PBQ =∠AOB, BQ = BO) PQ = AB PQ = a

(3) ѧࡵҠ∆ACP∆APQ ¤ѹẺ -- (AC = AQ, AP = AP, CP = PQ) ∠CAP =∠PAQ Р∠ACP =∠AQP ∠CAP = 30°Р∠ACP = 20°
∵ AC = AQ Р∠CAQ = 60° ∆ACQ 繠∆ҹ ∠ACQ = 60° ∠PCQ = 40° ∠PCQ +∠PBQ = 180° ☐BPCQ öṺǧ ∠BCP =∠BQP x = 20° Q.E.D.




Create Date : 01 Ҥ 2558
Last Update : 1 Ҥ 2558 0:00:00 .
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Fun Geometry Problem with Solution #124
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٨ AP + BP = AB + CP
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(1)∠ACB = 100°

(2) ˹ش Q BP ∠BAQ = 30° ∠PAQ = 10°Р∠AQP = 50°
ԨóҠ∆ABC ըش Q 繨ش㹷∠CAQ = 20°,∠BAQ = 30°,∠ABQ = 20°Р∠CBQ = 10° ∠BCQ = 40° (Click ʹԸվ٨) ∠ACQ = 60°

(3) ԨóҠ∆BCQ Ҡ∠CQP = 50°
ԨóҠ∆ACQ ըش P 繨ش㹷 AP PQ 觤觠∠CAQ Р∠AQC ӴѺ ⇒ ش P incenter ͧACQ CP 觤觠∠ACQ ∠PCQ (=∠ACP) = (∠ACQ)/2 = 30°

(4) AP ͡ѧش R · PR = BP ∠BPR = 60°
∵ BP = PR Р∠BPR = 60° ∆BPR 繠∆ҹ BP = BR ∠BRP = 60°

(5) AB ͡ѧش S · AS = AR ∆ARS 繠∆˹Ҩ ՠ∠A (= 40°) ʹ ∠ARS = 70°(∠BRS = 10°) Р∠ASR = 70°
ѧࡵҠ∆BRS∆BCP ¤ѹẺ -- (∠BSR =∠BCP,∠BRS =∠CBP, BR = BP) BS = CP
∵ AR = AS AP + PR = AB + BS AP + BP = AB + CP Q.E.D.




Create Date : 28 ¹ 2558
Last Update : 14 Զع¹ 2558 23:32:00 .
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Fun Geometry Problem with Solution #123
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٨ sin20°+ sin40°= sin80° ԸշҧâҤԵ

٨



(1) ҧ∆ABC · AB = BC =LР∠B = 80°
∴ ∆ABC 繠∆˹Ҩ ՠ∠B (= 80°) ʹ ∠BAC =∠ACB = 50°

(2) ҧ∆CDE · CD = DE = L,∠D = 40°Шش E 躹ǹ͢¢ͧ AC
∴ ∆CDE 繠∆˹Ҩ ՠ∠D (= 40°) ʹ ∠DCE =∠CED = 70°

(3) ԨóҨش C Ҡ∠BCD = 60°
∵ BC = CD (= L) Р∠BCD = 60° ∆BCD 繠∆ҹ BD = LР∠CBD =∠BDC = 60°

(4) ˹ش F BD ☐ABDF 繠☐¡ٹ AF = DF = L, ∠BAF = 40°(∠EAF = 10°) ∠BDF = 40°(∠CDF = 20°)

(5) ѧࡵ DE = DF (= L) Р∠EDF = 60° ∆DEF 繠∆ҹ EF = L EF = AF ∆AEF 繠∆˹Ҩ ՠ∠F ʹ

ԨóҠ∆ABC 繠∆˹Ҩ ҡǹ٧ BH Ҡ∆ABH Р∆BCH 繠∆ҡ ҡѹءС
¹ͧ cosine 㹠∆ҡ AH = CH = Lcos50° AC = 2Lcos50°
㹷ӹͧǡѹ Ѻ ∆CDE Р∆AEF CE = 2Lcos70° AE = 2Lcos10°ӴѺ

∵ AC + CE = AE 2Lcos50°+ 2Lcos70°= 2Lcos10° cos70°+ cos50°= cos10° sin20°+ sin40°= sin80° Q.E.D.




Create Date : 25 ¹ 2558
Last Update : 25 ¹ 2558 15:55:51 .
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Fun Geometry Problem with Solution #122
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٨ x = 6°
٨ 1



(1)∵∠BAC =∠ABC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ BC = AC BC = BP

(2) AC ͡ѧش Q · AQ = AB
Ҡ∆APQ∆ABP ¤ѹẺ -- (AP = AP,∠PAQ =∠BAP, AQ = AB) PQ = BP
͡ҡ ѧҠ∠AQP =∠ABP ∠AQP = 2x

(3) ԨóҠ☐ABPQ Ҡ∠BPQ (˭) = 360°- 10x ∠BPQ () = 10x
ѧࡵ BC = BP Р∠CBP = 2(∠CQP) ش B 繨شٹҧͧǧՠ∆CPQ Ṻ BP = BQ
∴ BP = BQ = PQ ∆BPQ 繠∆ҹ ∠BPQ = 60° 10x = 60° x = 6° Q.E.D.


٨ 2(¤س Angel Lazo HK)



(1) ∵∠BAC =∠ABC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ BC = AC BC = BP ∆BCP 繠∆˹Ҩ ՠ∠B (= 4x) ʹ ∠BCP (=∠BPC) = 90°- 2x

(2) ˹ش Q AB PQ⊥AB С˹ش R AC PR⊥AC
Ҡ∆APR∆APQ ¤ѹẺ -- (∠ARP =∠AQP,∠PAR =∠PAQ, AP = AP) PR = PQ

(3) ˹ش S CP BS⊥CP BS ǹ٧ͧ∆BCP CS = PS Р∠PBS (=∠CBS) = (∠CBP)/2 = 2x
Ҡ∆BPS∆BPQ ¤ѹẺ -- (∠BSP =∠BQP,∠PBS =∠PBQ, BP = BP) PS = PQ CS = PQ

(4) ѧࡵҠ∆CPR 繠∆ҡ ՠ∠R ҡ CP = 2‧PR ∠PCR = 30°
ԨóҠ∆ABC Ҡ∠BAC +∠ABC +∠ACB = 180° 6x + 6x + (120°- 2x) = 180° x = 6° Q.E.D.




Create Date : 22 ¹ 2558
Last Update : 22 ¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #121
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˹ BD = 2‧AC
٨ x = 30°
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AC = BD = 2

(1) ˹ش P 繨ش觡ҧ BD BP = DP = BD/2 =
∆ABD 繠∆ҡ Шش P 繨ش觡ҧҹçҡ (BD) AP = BP = DP AP =
∆BCD 繠∆ҡ Шش P 繨ش觡ҧҹçҡ (BD) CP = BP = DP CP =

(2)∵AC = AP = CP ∆ACP 繠∆ҹ ∠APC = 60°
∵ AP = CP = DP ش P 繨شٹҧͧǧՠ∆ACD Ṻ ∠ADC = (∠APC)/2 x = 30° Q.E.D.




Create Date : 19 ¹ 2558
Last Update : 19 ¹ 2558 0:00:00 .
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TIYHz
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