Fun Geometry Problem with Solution #150
⨷ 1



˹ x < 35°
٨ x= 30°
٨ 1



(1)∠BCD = 50°- x Р∠ADC = 70°- x

(2) AC ͡ѧش P · CP = DP ∠BCP = 20°+ x
∵ CP = DP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ ∠CDP =∠DCP ∠CDP = 70° ∠CPD = 40°
ѧࡵ CP = DP Р∠CPD = 2(∠CBD) ش P circumcenter ͧ∆BCD BP = CP = DP

BP = L CP = L DP = L

(3) ˹ش Q AD AQ = L Р∠DAQ = 20°+ x
Ҡ∆ADQ∆BCP ¤ѹẺ -- (AD = BC,∠DAQ =∠BCP, AQ = CP) DQ = BP (= L) DQ = AQ ∆ADQ 繠∆˹Ҩ ՠ∠Q ʹ ∠ADQ =∠DAQ ∠ADQ = 20°+ x ∠AQD = 140°- 2x

(4) ∵DP = DQ ∆DPQ 繠∆˹Ҩ ՠ∠D (= 160°) ʹ ∠DPQ = 10°Р∠DQP = 10° ∠APQ = 30°Р∠AQP = 130°- 2x (> 60° x < 35°)

(5) ˹ش R Ҿз͹ͧش Q ҹ AP ∆APR∆APQ PR = PQ, AR = AQ = L ∠APR =∠APQ = 30°
∵ PQ = PR Р∠QPR = 60° ∆PQR 繠∆ҹ PQ = QR
͡ҡ ѧҠ∠PQR = 60° ∠AQR = 70°- 2x

(6) ѧࡵҠ∆AQR∆DPQ ¤ѹẺ -- (AQ = DP, AR = DQ, QR = PQ) ∠AQR =∠DPQ 70°- 2x = 10° x = 30° Q.E.D.

٨ 2



(1)∠BCD = 50°- x

(2) AC ͡ѧش P · CP = DP ∠BCP = 20°+ x
∵ CP = DP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ ∠CDP =∠DCP ∠CDP = 70° ∠CPD = 40°
ѧࡵ CP = DP Р∠CPD = 2(∠CBD) ش P circumcenter ͧ∆BCD BP = CP = DP

BP = L CP = L DP = L

(3) ˹ش Q AD DQ = L Р∠ADQ = 20°+ x
Ҡ∆ADQ∆BCP ¤ѹẺ -- (AD = BC,∠ADQ =∠BCP, DQ = CP) AQ = BP AQ = L

(4) ˹ش R AP DR = L DR = DP ∆DPR 繠∆˹Ҩ ՠ∠D ʹ ∠DRP =∠DPR ∠DRP = 40° ∠ADR = 40°- x

(5)∵DQ = DR Р∠QDR = 60° ∆DQR 繠∆ҹ QR = L Р∠DQR = 60°
∵ AQ = DQ = QR ش Q circumcenter ͧ∆ADR ∠DAR = (∠DQR)/2 x = 30° Q.E.D.



⨷ 2



˹ x > 35°
٨ x= 40°
٨



(1)∠BCD = 50°- x Р∠ADC = 70°- x

(2) AC ͡ѧش P · CP = DP ∠BCP = 20°+ x
∵ CP = DP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ ∠CDP =∠DCP ∠CDP = 70° ∠CPD = 40°
ѧࡵ CP = DP Р∠CPD = 2(∠CBD) ش P circumcenter ͧ∆BCD BP = CP = DP

BP = L CP = L DP = L

(3) ˹ش Q AD AQ = L Р∠DAQ = 20°+ x
Ҡ∆ADQ∆BCP ¤ѹẺ -- (AD = BC,∠DAQ =∠BCP, AQ = CP) DQ = BP (= L) DQ = AQ ∆ADQ 繠∆˹Ҩ ՠ∠Q ʹ ∠ADQ =∠DAQ ∠ADQ = 20°+ x ∠AQD = 140°- 2x

(4) ∵DP = DQ ∆DPQ 繠∆˹Ҩ ՠ∠D (= 160°) ʹ ∠DPQ = 10°Р∠DQP = 10° ∠APQ = 30°Р∠AQP = 130°- 2x (< 60° x > 35°)

(5) ˹ش R Ҿз͹ͧش Q ҹ AP ∆APR∆APQ PR = PQ, AR = AQ = L ∠APR =∠APQ = 30°
∵ PQ = PR Р∠QPR = 60° ∆PQR 繠∆ҹ PQ = QR
͡ҡ ѧҠ∠PQR = 60° ∠AQR = 2x - 70°

(6) ѧࡵҠ∆AQR∆DPQ ¤ѹẺ -- (AQ = DP, AR = DQ, QR = PQ) ∠AQR =∠DPQ 2x - 70° = 10° x = 40° Q.E.D.




Create Date : 15 áҤ 2558
Last Update : 15 áҤ 2558 0:00:01 .
Counter : 621 Pageviews.

0 comment
Fun Geometry Problem with Solution #149



٨ x= 10°
٨ 1



AD = L

(1)∠ACB = 100°Р∠ADB = 40°

(2) ˹ش P Ҿз͹ͧش D ҹ AB ∆ABP∆ABD BP = BD,∠ABP =∠ABD = 30°Р∠APB =∠ADB = 40°
∵ BD = BP Р∠DBP = 60° ∆BDP 繠∆ҹ BD = DP
͡ҡ ѧҠ∠BDP =∠BPD = 60° ∠ADP =∠APD = 20°

(3) ˹ش Q AC DQ = L DQ = AD ∆ADQ 繠∆˹Ҩ ՠ∠D ʹ ∠AQD =∠DAQ ∠AQD = 80° ∠ADQ = 20° ∠BDQ = 20°

(4) ѧࡵҠ∆BDQ∆ADP ¤ѹẺ -- (BD = DP,∠BDQ =∠ADP, DQ = AD) ∠DBQ =∠APD ∠DBQ = 20° ∠DBQ =∠BDQ ∆BDQ 繠∆˹Ҩ ՠ∠Q ʹ BQ = DQ BQ = L
ԨóҠ∆BCQ Ҡ∠BQC = 40°

(5) ԨóҠ∆ABC Ҡ∠A = 30°,∠C = 100°ըش Q AC ∠BQC = 40° AC = BQ (Click ʹԸվ٨⨷ 1-1) AC = L AC = AD ∆ACD 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠ADC (=∠ACD) = 50° ∠BDC = x = 10° Q.E.D.

٨ 2



(1)∠ACB = 100°Р∠ADB = 40°

(2) ˹ش P 繨شѴҧ AC BD
∵ ∠BAP =∠ABP ∆ABP 繠∆˹Ҩ ՠ∠P ʹ AP = BP

(3) ˹ش Q AD PQ = AP ∆APQ 繠∆˹Ҩ ՠ∠P ʹ ∠AQP =∠PAQ ∠AQP = 80° ∠APQ = 20°Р∠DPQ = 40°
∵ ∠PDQ =∠DPQ ∆DPQ 繠∆˹Ҩ ՠ∠Q ʹ DQ = PQ DQ = AP

(4) ˹ش R PQ AR = AQ ∆AQR 繠∆˹Ҩ ՠ∠A ʹ ∠ARQ =∠AQR ∠ARQ = 80° ∠ARP = 100°
ѧࡵҠ∆BCP∆APR ¤ѹẺ -- (∠BCP =∠ARP,∠CBP =∠APR, BP = AP) CP = AR CP = AQ

(5)∵AP = DQ CP = AQ AP + CP = DQ + AQ AC = AD ∆ACD 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠ADC (=∠ACD) = 50° ∠BDC = x = 10° Q.E.D.




Create Date : 12 áҤ 2558
Last Update : 12 áҤ 2558 0:00:00 .
Counter : 504 Pageviews.

0 comment
Fun Geometry Problem with Solution #148



٨ x= 20°
٨ 1



(1)∠ACB = 80°Р∠ADB = 50°

(2) CA ͡ѧش P · CP = BC ∆BCP 繠∆˹Ҩ ՠ∠C (= 80°) ʹ ∠CBP (=∠BPC) = 50° ∠ABP = 10°

(3) ѧࡵҠ∆ABD∆ABP ¤ѹẺ -- (∠ADB =∠APB,∠ABD =∠ABP, AB = AB) BD = BP ∆BDP 繠∆˹Ҩ ՠ∠B (= 20°) ʹ ∠BPD (=∠BDP) = 80° ∠CPD = 30° ∠CPD =∠CBD ☐BCDP öṺǧ ∠DCP =∠DBP x = 20° Q.E.D.

٨ 2(ɮպǡѺǧ)



(1)∠ACB = 80°Р∠ADB = 50°

(2) CA ͡ѧش P · CP = BC ∆BCP 繠∆˹Ҩ ՠ∠C (= 80°) ʹ ∠CBP (=∠BPC) = 50° ∠ABP = 10°

(3) ѧࡵҠ∆ABD∆ABP ¤ѹẺ -- (∠ADB =∠APB,∠ABD =∠ABP, AB = AB) BD = BP

(4) PC ͡ѧش Q · BQ = BP (= BD) ∆BPQ 繠∆˹Ҩ ՠ∠B ʹ ∠BQP =∠BPQ ∠BQP = 50° ∠CBQ = 30°

(5) ѧࡵҠ∆BCD∆BCQ ¤ѹẺ -- (BC = BC,∠CBD =∠CBQ, BD = BQ) ∠BDC =∠BQC ∠BDC = 50° ∠BCD = 100° ∠ACD = x = 20° Q.E.D.

٨ 3

(1) ∠ADB = 50°

(2) α= 50°
ԨóҠ☐ABCD Ҡ∠BAC = 60°,∠CAD = 60°,∠CBD = 30°Р∠ADB =α ∠BDC =α (Click ʹԸվ٨) ∠BDC = 50°

(3) ԨóҠ∆ACD Ҡ∠ACD = x = 20° Q.E.D.




Create Date : 09 áҤ 2558
Last Update : 9 áҤ 2558 0:00:00 .
Counter : 548 Pageviews.

0 comment
Fun Geometry Problem with Solution #147



٨ x= 10°
٨ 1



(1) ԨóҠ☐ACBP Ҡ∠APB (˭) = 360°- 150° ∠APB () = 150°

(2) ˹ش Q BC CQ = AC
Ҡ∆CPQ∆ACP ¤ѹẺ -- (CP = CP,∠PCQ =∠ACP, CQ = AC) PQ = AP
͡ҡ ѧҠ∠CQP =∠CAP ∠CQP = 20° ∠BPQ = 10° ∠BPQ =∠PBQ ∆BPQ 繠∆˹Ҩ ∠Q ʹ BQ = PQ BQ = AP

(3) PQ ͡ѧش R · BR = BP ∆BPR 繠∆˹Ҩ ՠ∠B ʹ ∠BRP =∠BPR ∠BRP = 10° ∠PBR = 160° ∠QBR = 150°

(4) ѧࡵҠ∆ABP∆BQR ¤ѹẺ -- (AP = BQ,∠APB =∠QBR, BP = BR) ∠ABP =∠BRQ x = 10° Q.E.D.

٨ 2

α= 20°,β= 10°Рγ= 60° α+β+γ= 90°
ԨóҠ∆ABC ըش P 繨ش㹷∠CAP =α,∠CBP =β,∠ACP =γР∠BCP =γ ∠ABP =β (Click ʹԸվ٨) x = 10° Q.E.D.




Create Date : 06 áҤ 2558
Last Update : 6 áҤ 2558 0:00:00 .
Counter : 533 Pageviews.

0 comment
Fun Geometry Problem with Solution #146



٨ x= 10°
٨ 1



(1)∠ACB = 80°

(2) ˹ش O circumcenter ͧ∆ABP ...
AO = BO = OP
∠AOP = 2(∠ABP) ∠AOP = 80°
∠BOP = 2(∠BAP) ∠BOP = 20°
∵ AO = BO ∆ABO 繠∆˹Ҩ ՠ∠O (= 100°) ʹ ∠BAO =∠ABO = 40°

BO = L OP = L

(3) BP ͡仾 AC ش Q ∠BQC = 80°
ѧࡵҠ∆ABQ∆ABO ¤ѹẺ -- (∠BAQ =∠BAO, AB = AB,∠ABQ =∠ABO) BQ = BO BQ = L
∵ ∠BQC =∠BCQ ∆BCQ 繠∆˹Ҩ ՠ∠B ʹ BC = BQ BC = L

(4) α= 10°
ѧࡵ ☐BCPO ՠBC = BO = OP,∠CBO = 120°- 2αР∠BOP = 2α ∠BCP =α (Click ʹԸվ٨⨷ 3) x = 10° Q.E.D.

٨ 2



(1) BP ͡仾 AC ش Q ∠APQ = 50° ∠BQC = 80°
∵ ∠BAQ =∠ABQ ∆ABQ 繠∆˹Ҩ ՠ∠Q ʹ AQ = BQ

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

(2) BQ ͡ѧش R · AR = a + b ∠AQR = 80°
∵ AQ = AR ∆AQR 繠∆˹Ҩ ՠ∠A ʹ ∠ARQ =∠AQR ∠ARQ = 80° ∠QAR = 20°

(3)∵∠PAR =∠APR ∆APR 繠∆˹Ҩ ՠ∠R ʹ PR = AR PR = a + b QR = a
Ҡ∆BCQ∆AQR ¤ѹẺ -- (∠CBQ =∠QAR, BQ = AQ,∠BQC =∠AQR) CQ = QR CQ = a

(4) ԨóҠ∆BCQ Ҡ∠B = 20°,∠Q = 80°ըش P BQ BP = CQ ∠BCP = 10° (Click ʹԸվ٨) x = 10° Q.E.D.




Create Date : 03 áҤ 2558
Last Update : 29 áҤ 2558 23:46:00 .
Counter : 537 Pageviews.

0 comment
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 :
ا෾  Thailand

[ Profile ]
ԻҢͧ Blog [?]
 ҡͤѧ
 Rss Feed
 Smember
 Դ͡ : 20 [?]



شʧӺ͡Եʵ... ҡѺ ҡẺҧ
All Blog