Fun Geometry Problem with Solution #90



٨ x = 10°
٨



(1)∠APB = 120°

(2) ˹ش Q AB AQ = BP
ԨóҠ∆ABP Ҡ∠A = 40°,∠B = 20°ըش Q AB AQ = BP ∠AQP = 30° (Click ʹԸվ٨)

(3) ˹ش R BC BR = PQ
Ҡ∆BPR∆APQ ¤ѹẺ -- (BP = AQ,∠PBR =∠AQP, BR = PQ) ∠BPR =∠PAQ ∠BPR = 40°
͡ҡ ѧ PR = AP ∆APR 繠∆˹Ҩ ՠ∠P (= 160°) ʹ ∠PAR =∠ARP = 10°

(4) AP ͡仾 BC ش S
ԨóҠ∆ABS Ҡ∠ASB = 90° AS⊥BC
ԨóҠ∆ACR ǹ٧ (AS) 觤ʹ (∠A) ∆ACR 繠∆˹Ҩ ՠ∠A ʹ AC = AR
ѧࡵҠ∆ACP∆APR ¤ѹẺ -- (AC = AR,∠CAP =∠PAR, AP = AP) ∠ACP =∠ARP x = 10° Q.E.D.




Create Date : 16 Ҥ 2558
Last Update : 16 Ҥ 2558 0:00:00 .
Counter : 625 Pageviews.

0 comment
Fun Geometry Problem with Solution #89



٨ x = 70°
٨



˹ش P Ҿз͹ͧش B ҹ AD ∆ADP∆ABD ...
∠ADP =∠ADB ∠ADP = 70° ∠CDP = 180° ش C ش D Шش P 躹鹵çǡѹ
∠APD =∠ABD
AP = AB AP = AC ∆ACP 繠∆˹Ҩ ՠ∠A ʹ ∠ACP =∠APC ∠ACD =∠ABD ☐ABCD öṺǧ ∠ABC +∠ADC = 180° x + 110°= 180° x = 70° Q.E.D.




Create Date : 13 Ҥ 2558
Last Update : 13 Ҥ 2558 0:00:00 .
Counter : 553 Pageviews.

0 comment
Fun Geometry Problem with Solution #88



٨ x = 20°
٨(¤سΣΤΑΘΗΣΚΟΥΤΡΑΣ)



(1) AB ͡ѧش Q · AQ = AC ∠CBQ = 60°
Ҡ∆APQ∆ACP ¤ѹẺ -- (AQ = AC,∠PAQ =∠CAP, AP = AP) ∠AQP =∠ACP ∠AQP = 10° ∠AQP =∠PAQ ∆APQ 繠∆˹Ҩ ՠ∠P ʹ AP = PQ
ԨóҠ☐ACPQ Ҡ∠CPQ (˭) = 360°- 40° ∠CPQ () = 40°

(2) ˹ش R ˹ AQ AR = QR = AQ (= AC) ∆AQR 繠∆ҹ ∠QAR = 60°(∠CAR = 40°), ∠AQR = 60°Р∠ARQ = 60°
∵ AC = AR ∆ACR 繠∆˹Ҩ ՠ∠A (= 40°) ʹ ∠ARC (=∠ACR) = 70° ∠CRQ = 10°
Ҡ∆PQR∆APR ¤ѹẺ -- (PQ = AP, QR = AR, PR = PR) ∠PRQ (=∠ARP) = (∠ARQ)/2 = 30°

(3) ˹ش S 繨شѴҧ BC Ѻ QR
∵ ∠QBS =∠BQS = 60° ∆BQS 繠∆ҹ BQ = BS
∵ ∠PRS =∠PCS ☐CRPS öṺǧ ∠CPS =∠CRS ∠CPS = 10° ∠QPS = 30°
ѧࡵ BQ = BS Р∠QBS = 2(∠QPS) ش B 繨شٹҧͧǧՠ∆PQS Ṻ BP = BQ ∆BPQ 繠∆˹Ҩ ՠ∠B ʹ ∠BPQ =∠BQP ∠BPQ = 10° ∠ABP = x = 20° Q.E.D.




Create Date : 10 Ҥ 2558
Last Update : 10 Ҥ 2558 0:00:02 .
Counter : 574 Pageviews.

0 comment
Fun Geometry Problem with Solution #87



٨ x = 45°
٨



∠CBP =αР∠BCP = β

(1) BP ͡仾 AC ش Q BQ⊥AC BQ ǹ٧ͧ∆ABC
ԨóҠ∆BCP Ҡ∠BPC = 135°Рα+β= 45°

(2) ˹ش R AC PR = CP ∆CPR 繠∆˹Ҩ ՠ∠P ʹ ∠CRP =∠PCR ∠CRP = 45° ∠ARP = 135°
ѧࡵҠ∆APR∆BCP ¤ѹẺ -- (∠ARP =∠BPC ҹ, PR = CP, AP = BC) ∠PAR =∠CBP ∠PAR = α

(3) AP ͡仾 BC ش S
ԨóҠ∆ACS Ҡ∠CAS +∠ACS +∠ASC = 180° α+ (45°+β) +∠ASC = 180° ∠ASC = 90° AS⊥BC AS ǹ٧ͧ∆ABC

(4) CP ͡仾 AB ش T ∠BPT = 45°
∵ AS ѴѺ BQ ش P ش P orthocenter ͧ∆ABC CT ǹ٧ͧ∆ABC CT⊥AB
ԨóҠ∆BPT Ҡ∠PBT = x = 45° Q.E.D.




Create Date : 07 Ҥ 2558
Last Update : 7 Ҥ 2558 0:00:00 .
Counter : 552 Pageviews.

0 comment
Fun Geometry Problem with Solution #86
H A P PY N E W Y E A R 2 01 5




٨ x = 24°
٨



(1) ˹ش P Ҿз͹ͧش C ҹ AB ∆ABP∆ABC ⇒ ...
BP = BC BP = a
∠ABP =∠ABC ∠ABP = 18°
∠APB =∠ACB ∠APB = 30°

(2) ˹ش Q Ҿз͹ͧش B ҹ AP ∆APQ∆ABP ⇒ ...
AQ = AB
PQ = BP PQ = a
∠APQ =∠APB ∠APQ = 30°

(3) ∵ BP = PQ (= a) Р∠BPQ = 60° ∆BPQ 繠∆ҹ BQ = a
͡ҡ ѧҠ∠PBQ = 60° ∠CBQ = 24°
∵ BC = BQ ∆BCQ 繠∆˹Ҩ ՠ∠B (= 24°) ʹ ∠BCQ =∠BQC = 78°
∵ AB = AQ ∆ABQ 繠∆˹Ҩ ՠ∠A ʹ ∠AQB =∠ABQ ∠AQB = 42° ∠AQC = 36°
ԨóҠ∆ACQ Ҡ∠CAQ = 36° ∠CAQ =∠AQC ∆ACQ 繠∆˹Ҩ ՠ∠C ʹ CQ = AC CQ = b

(4) ѧࡵҠ∆DEF∆BCQ ¤ѹẺ -- (DE = BC, DF = BQ, EF = CQ) ∠EDF =∠CBQ x = 24° Q.E.D.




Create Date : 04 Ҥ 2558
Last Update : 4 Ҥ 2558 0:00:00 .
Counter : 596 Pageviews.

1 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