Fun Geometry Problem with Solution #85
H A P PY N E W Y E A R 201 5




٨ x = 20°
٨ 1



(1)∠ABC = 30°Р∠ADC = 110°

(2) ˹ش O 繨شٹҧͧǧՠ∆ACD Ṻ AO = CO = DO
͡ҡ ѧҠ∠AOD = 2(∠ACD) ∠AOD = 60°
∵ AO = DO Р∠AOD = 60° ∆ADO 繠∆ҹ AD = AO = DO AD = CO
͡ҡ ѧҠ∠DAO = 60° ∠DAO =∠BAD ش O 躹 AB
∵ AO = CO ∆ACO 繠∆˹Ҩ ՠ∠O ʹ ∠ACO =∠CAO ∠ACO = 20° ∠BCO = 110°

(3) ѧࡵҠ∆BCO∆ACD ¤ѹẺ -- (∠CBO =∠ACD,∠BCO =∠ADC, CO = AD) BC = CD ∆BCD 繠∆˹Ҩ ՠ∠C (= 160°) ʹ ∠CBD (=∠BDC) = 10° ∠ABD = x = 20° Q.E.D.

٨ 2



(1) ԨóҠ∆ABC Ҡ∠ABC = 30°

(2) ˹ش P AB CP⊥AB
ԨóҠ∆ACP Ҡ∠ACP = 70°
Ҡ∆BCP 繠∆ҡ ՠ∠P ҡ Р∠B = 30° BC = 2‧CP

(3) AD ͡ѧش Q · AQ = AC ∆ACQ 繠∆˹Ҩ ՠ∠A (= 40°) ʹ ∠ACQ =∠AQC = 70°
ԨóҠ∆ACD Ҡ∠CDQ = 70° ∠CDQ =∠CQD ∆CDQ 繠∆˹Ҩ ՠ∠C ʹ CD = CQ

(4) ˹ش R CQ AR ⊥ CQ AR ǹ٧ͧ∆ACQ CR = QR
͡ҡ ѧ ∆ACR∆ACP ¤ѹẺ -- (∠ARC =∠APC,∠ACR =∠ACP, AC = AC) CR = CP QR = CP
∴ CQ = 2‧CP CD = BC ∆BCD 繠∆˹Ҩ ՠ∠C (= 160°) ʹ ∠CBD (=∠BDC) = 10° ∠ABD = x = 20° Q.E.D.




Create Date : 01 Ҥ 2558
Last Update : 1 Ҥ 2558 0:00:01 .
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Fun Geometry Problem with Solution #84



٨ x = 10°
٨



(1)∵AB = AC = BC ∆ABC 繠∆ҹ ∠BAC = 60°Р∠ABC = 60° ∠CAP = 60°- 3x Р∠ABP = 60°- x ∠APC = 120°- 2x Р∠APB = 120°- 2x
∵ 0°<∠ACP < 60° 0°< 5x < 60° 0°< x < 12° 96°< 120°- 2x < 120°
∴ ∠APC =∠APB ҹ

(2) ˹ش Q AB AQ = AP Р∠BAQ = 60°- 3x
Ҡ∆ABQ∆ACP ¤ѹẺ -- (AB = AC,∠BAQ =∠CAP, AQ = AP) BQ = CP
͡ҡ ѧ ∠AQB =∠APC ∠AQB =∠APB (ҹ)
∵ AP = AQ ∆APQ 繠∆˹Ҩ ՠ∠A ʹ ∠APQ =∠AQP ∠BPQ =∠BQP ∆BPQ 繠∆˹Ҩ ՠ∠B ʹ BP = BQ
ѧࡵҠ∆ABP∆ABQ ¤ѹẺ -- (AB = AB, AP = AQ, BP = BQ) ∠BAP =∠BAQ 3x = 60°- 3x x = 10° Q.E.D.

˵ҡ (1) öػҠ∆ABP∆ACP ¤ѹẺ -- (∠APB =∠APC ҹ, AP = AP, AB = AC) ∠BAP =∠CAP 3x = 60°- 3x x = 10°




Create Date : 30 ѹҤ 2557
Last Update : 3 Ҥ 2558 23:34:00 .
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Fun Geometry Problem with Solution #83



٨ x = 80°
٨



(1) ˹ش P AB ∠BPC = 40° ∠BPC =∠CBP ∆BCP 繠∆˹Ҩ ՠ∠C ʹ CP = BC CP = AD
ԨóҠ∆ACP Ҡ∠ACP = 10°

(2) ˹ش O 繨شٹҧͧǧՠ∆ACP Ṻ ...
AO = CO = OP
∠AOP = 2(∠ACP) ∠AOP = 20°
∠COP = 2(∠CAP) ∠COP = 60°
∵ AO = OP ∆AOP 繠∆˹Ҩ ՠ∠O (= 20°) ʹ ∠OAP (=∠APO) = 80°
∵ CO = OP Р∠COP = 60° ∆COP 繠∆ҹ OP = CP AO = AD ∆ADO 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠AOD (=∠ADO) = 50° ∠DOP = 30° ∠COD = 30°

(3) ѧࡵҠ∆CDO∆DOP ¤ѹẺ -- (CO = OP,∠COD =∠DOP, DO = DO) CD = DP ∆CDP 繠∆˹Ҩ ՠ∠D ʹ ∠DCP =∠CPD ∠DCP = 40° ∠BDC = x = 80° Q.E.D.




Create Date : 27 ѹҤ 2557
Last Update : 27 ѹҤ 2557 0:00:00 .
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Fun Geometry Problem with Solution #82



˹ [ABCDEFGHI] ٻҹ
٨ a + b = c
٨



(1)∵[ABCDEFGHI] ٻҹ AB = BC = CD = DE = a
͡ҡ ѧҠ∠ABC =∠BCD =∠CDE = [(9 - 2)× 180°]÷ 9 = 140°
Ҡ∆CDE∆ABC ¤ѹẺ -- (CD = AB,∠CDE =∠ABC, DE = BC) CE = AC CE = b
∵ AB = BC ∆ABC 繠∆˹Ҩ ՠ∠B (= 140°) ʹ ∠BAC =∠ACB = 20°
∵ CD = DE ∆CDE 繠∆˹Ҩ ՠ∠D (= 140°) ʹ ∠DCE (=∠CED) = 20° ∠ACE = 100°
∵ AC = CE ∆ACE 繠∆˹Ҩ ՠ∠C (= 100°) ʹ ∠CAE =∠AEC = 40°

(2) EC ͡ѧش P CP = a ∠BCP = 60°
∵ BC = CP Р∠BCP = 60° ∆BCP 繠∆ҹ BP = BCР∠CBP = 60°
ѧࡵ AB = BC = BP ش B 繨شٹҧͧǧՠ∆ACP Ṻ ∠CAP = (∠CBP)/2 ∠CAP = 30°
ԨóҠ∆AEP Ҡ∠APE = 70° ∠APE =∠EAP ∆AEP 繠∆˹Ҩ ՠ∠E ʹ EP = AE a + b = c Q.E.D.




Create Date : 24 ѹҤ 2557
Last Update : 24 ѹҤ 2557 0:00:00 .
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Fun Geometry Problem with Solution #81



˹ ∆ABC ∆BDE ∆ҹ
٨ x = α
٨



(1) AE = a BE = b
∵ ∆ABC 繠∆ҹ AC = a + b,∠BAC = 60°Р∠ACB = 60°(∠ACE = 60°- x)
∵ ∆BDE 繠∆ҹ DE = b Р∠BED = 60°(∠ADE = 60°-αР∠AED = 120°)

(2) ˹ش P AC AP = a CP = b
∵ AE = AP Р∠EAP = 60° ∆AEP 繠∆ҹ EP = a Р∠APE = 60°(∠CPE = 120°)

(3) ѧࡵҠ∆CEP∆ADE ¤ѹẺ -- (EP = AE,∠CPE =∠AED, CP = DE) ∠ECP =∠ADE 60°- x = 60°-α x =α Q.E.D.




Create Date : 21 ѹҤ 2557
Last Update : 21 ѹҤ 2557 0:00:00 .
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TIYHz
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