Fun Geometry Problem with Solution #105



٨ x = 40°
٨ 1



(1)∠BDC = 100°

(2) ˹ش P Ҿз͹ͧش D ҹ BC ∆BCP∆BCD BP = BD, CP = CD,∠CBP =∠CBD = 30° ∠BCP =∠BCD = 50°
∵ BD = BP Р∠DBP = 60° ∆BDP 繠∆ҹ DP = BD DP = AC

(3) ˹ش Q BD DQ = CD (= CP) ∆CDQ 繠∆˹Ҩ ՠ∠D (= 100°) ʹ ∠CQD (= ∠DCQ) = 40°
ѧࡵҠ∆CDQ∆CDP ¤ѹẺ -- (CD = CD,∠CDQ =∠DCP, DQ = CP) CQ = DP CQ = AC ∆ACQ 繠∆˹Ҩ ՠ∠C ʹ ∠CAQ =∠AQC x = 40° Q.E.D.

٨ 2



(1) CD ͡ѧش P ∠CBP = 50°(∠DBP = 20°) ∠CBP =∠BCP ∆BCP 繠∆˹Ҩ ՠ∠P ʹ BP = CP
͡ҡ ѧҠ∠BDP = 80°
ԨóҠ∆BDP Ҡ∠BPD = 80° ∠BPD =∠BDP ∆BDP 繠∆˹Ҩ ՠ∠B ʹ BP = BD BP = AC

(2) ˹ش Q BC PQ⊥BC PQ ǹ٧ͧ∆BCP BQ (= CQ) = BC/2
͡ҡ ѧҠ∠BPQ (=∠CPQ) = (∠BPC)/2 = 40°

(3) ˹ش R AB CR⊥AB
Ҡ∆BCR 繠∆ҡ ՠ∠R ҡ Р∠B = 30° CR = BC/2
ѧࡵҠ∆ACR∆BPQ ¤ѹẺ -- (∠ARC =∠BQP = 90°, CR = BQ, AC = BP) ∠CAR =∠BPQ x = 40° Q.E.D.




Create Date : 02 չҤ 2558
Last Update : 2 չҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #104
⨷ʹ blog ⨷ҡͺ PAT1 (.. 53) ͷ 7 ⨷ͧ⡳ԵԠ
ҧá . ʴԸվ٨ӵͺԸշҧâҤԵ






٨ EC/BC = 1/sqrt(3)
٨



(1)∵AD AE 觠∠BAC ͡ 3 ǹ ѹ ∠BAD =∠DAE =∠CAE = 135°/3 = 45°

(2) BA ͡ѧش P · ∠BCP = 90°
ԨóҠ☐AECP Ҡ∠ECP =∠BAE ☐AECP öṺǧ ∠CPE =∠CAE ∠CPE = 45°

(3) ԨóҠ∆CEP Ҡ∠CEP = 45° ∠CEP =∠CPE ∆CEP 繠∆˹Ҩ ՠ∠C ʹ EC = CP
ԨóҠ∆BCP CP/BC = tan30° EC/BC = 1/sqrt(3) Q.E.D.




Create Date : 27 Ҿѹ 2558
Last Update : 27 Ҿѹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #103



٨ x = 10°
٨ 1



(1) ԨóҠ∆ABP Ҡ∠APB = 180°- 4x
ԨóҠ∆BCP Ҡ∠CBP =∠BCP ∆BCP 繠∆˹Ҩ ՠ∠P ʹ BP = CP
ԨóҠ☐ABCP Ҡ∠APC (˭) = 360°- 8x ∠APC () = 8x

(2) ˹ش Q AP PQ = AP Р∠BPQ = 8x (∠APQ = 180°- 12x)
Ҡ∆BPQ∆ACP ¤ѹẺ -- (BP = CP,∠BPQ =∠APC, PQ = AP) ∠BQP =∠CAP ∠BQP = 2x
∵ AP = PQ ∆APQ 繠∆˹Ҩ ՠ∠P (= 180°- 12x) ʹ ∠AQP (=∠PAQ) = 6x

(3) ԨóҠ∆ABP Шش Q Ҡ∠AQP = 2(∠ABP) Р∠BQP = 2(∠BAP) ش Q 繨شٹҧͧǧՠ∆ABP Ṻ (Click ʹԸվ٨) AQ = PQ
∴ AP = AQ = PQ ∆APQ 繠∆ҹ ∠AQP = 60° 6x = 60° x = 10° Q.E.D.

٨ 2



(1) ˹ش Q AP ∠ABQ = x ∆ABQ 繠∆˹Ҩ ՠ∠Q ʹ AQ = BQ
͡ҡ ѧҠ∠PBQ = 2x Р∠BQP = 2x

(2) ѧࡵҠ∆BCP∆BPQ ¤ѹẺ -- (∠BCP =∠BQP,∠CBP =∠PBQ, BP = BP) BC = BQ

(3) α= 2x
ѧࡵҠ☐ACBQ AQ = BQ = BC,∠A =αР∠B = 2α ∠ACB = 120°-α (Click ʹԸվ٨⨷ 1) ∠ACB = 120°- 2x
ԨóҠ∆ABC Ҡ∠BAC +∠ABC +∠ACB = 180° 3x + 5x + (120°- 2x) = 180° x = 10° Q.E.D.




Create Date : 24 Ҿѹ 2558
Last Update : 24 Ҿѹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #102



٨ x = 10°
٨



(1) ˹ش P BC AP = AC (= BD) ∆ACP 繠∆˹Ҩ ՠ∠A ʹ ∠APC =∠ACP ∠APC = 6x ∠BAP = 4x

(2) Ԩó ∆ABP Ҡ∠A = 2(∠B) ըش D AB AP = BD DP = AP (Click ʹԸվ٨) DP = BD ∆BDP 繠∆˹Ҩ ՠ∠D ʹ ∠BPD =∠DBP ∠BPD = 2x ∠CDP = x ∠CDP = ∠DCP ∆CDP 繠∆˹Ҩ ՠ∠P ʹ CP = DP CP = AP

(3) ѧࡵ AC = AP = CP ∆ACP 繠∆ҹ ∠ACP = 60° 6x = 60° x = 10° Q.E.D.




Create Date : 21 Ҿѹ 2558
Last Update : 23 Ҿѹ 2558 16:35:00 .
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Fun Geometry Problem with Solution #101



٨ x = 10°
٨



(1)∠BPC = 100°

(2) BP ͡ѧش Q ·∠BQC = 20° ∠BQC =∠CBQ ∆BCQ 繠∆˹Ҩ ՠ∠C ʹ CQ = BC CQ = AP
͡ҡ ѧҠ∠CPQ = 80°Р∠APQ = 60°
ԨóҠ∆CPQ Ҡ∠PCQ = 80° ∠PCQ =∠CPQ ∆CPQ 繠∆˹Ҩ ՠ∠Q ʹ PQ = CQ PQ = AP

(3) ∵AP = PQ Р∠APQ = 60° ∆APQ 繠∆ҹ AQ = AP
ѧࡵ AQ = CQ = PQ ش Q 繨شٹҧͧǧՠ∆ACP Ṻ ∠CAP = (∠CQP)/2 x = 10° Q.E.D.




Create Date : 18 Ҿѹ 2558
Last Update : 18 Ҿѹ 2558 0:00:00 .
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