tag:blogger.com,1999:blog-6933544261975483399.post4657461952917280101..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 423: Triangle, 30, 40 degree, Angle, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-6933544261975483399.post-9515165995719486262016-08-18T03:38:24.950-07:002016-08-18T03:38:24.950-07:00Problem 423
Forming the equilateral triangle AEC...Problem 423<br />Forming the equilateral triangle AEC (E is above the point Β) then triangleABC=triangleABE<br />And BE=BC,<BEC=<BCE=20.Draw EK perpendicular to AC (K is on AC) and point F on EK <br />such that AF=FC=BC=AD=BE. So < ADF=80=<FBC then FBCD is cyclic.Therefore <BDC=<BFC=80 (<FCB=20 ).<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19613869454504072942015-12-30T02:59:38.399-08:002015-12-30T02:59:38.399-08:00Simpler geometric proof
Let E be the circumcentre...Simpler geometric proof<br /><br />Let E be the circumcentre of Tr. ABC. Then < BEC = 2 < BAC = 60 so Tr. BCE is equilateral. Hence BC = AD = AE and since < CAE = 20 < AED = 80. But < AEB = 80 so B,D,E are collinear implying that x = 80 (180 - 60 - 40 in Tr. DBC)<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4466339095103078832015-09-10T19:35:53.558-07:002015-09-10T19:35:53.558-07:00Should read AB = BE as radius and not as centre as...Should read AB = BE as radius and not as centre as line 4 states. The error is regretted. Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29257975515001455622015-09-10T03:17:49.715-07:002015-09-10T03:17:49.715-07:00Extend ADC to E such that AB = AE. Draw CM perpend...Extend ADC to E such that AB = AE. Draw CM perpendicular to AB to meet BE at M. Let the circle with B as centre and AB = BE as centre cut AM extended at N.<br /><br />Since B is the centre of circumcircle of Tr. AEN, < ANE = 1/2 < ABE = 60 implying that CENM is cyclic with < CME = 30 = < BAC so ACMB is cyclic from which we deduce that < BCM = 20 = < BAN = < BNA = < NBM. <br /><br />Now we see that Tr. ABD & CBN are congruent SAS so < ABD = < BNC = 50 and so x = 80. <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42871825243949172302013-01-13T17:11:09.079-08:002013-01-13T17:11:09.079-08:00AEC , is equilateral triangle. Because ,angle CAB=...AEC , is equilateral triangle. Because ,angle CAB=30, AB is mediator of EC.So , EB=BC and angleBCE=angleBEC=20<br />CZ=//BD (DBCZ , is parallelogram).Therefore,BC=DZ=DA . Because angleCDZ=40,then,angleDAZ=angleDZA=20<br />So, the triangles, EBC,ADZ , are equal.Therefore,AZ=EC. But EC=AC.So,AZ=AC .<br />Therefore,angle AZC=angleACZ=x=80 degrees <br />see the image: http://img842.imageshack.us/img842/3751/p423triangleangle3040.gif<br />Michael Tsourakakishttps://www.blogger.com/profile/01337344627083642570noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38904333684441486342013-01-13T15:04:32.093-08:002013-01-13T15:04:32.093-08:00Firstly, you must now a very importan triangle: 30...Firstly, you must now a very importan triangle: 30-40-110, wich has a crucial property, "Let tr ABC with <A=30 and <B=40, if we extend AB beyond B to X such that BC=BX, then CX=AB"<br />the prove for this in spanish si here:<br />http://www.youtube.com/watch?v=2_Kq1_QU-xQ<br /><br />Now for the problem, lets make the same construction (extend AC to X such that CX=BC), then AC=DX=BX, so <BDX=<DBX=80 because <X=20.Anonymoushttps://www.blogger.com/profile/05208522114695973019noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47990833389063872282013-01-10T18:11:07.023-08:002013-01-10T18:11:07.023-08:00Hmm, this one was tricky, I feel like there has to...Hmm, this one was tricky, I feel like there has to be a simpler solution.<br /><br />Let E be on the extension of AC such AB=BE, and let F be on the opposite side of BE as D such that triangle BCF is congruent to triangle ADB, with angle FBC=30. Let AF and BE meet at G. Then triangles ABE, ABF, and FBE are isoceles, from which we find angle GBC= angle GAC=10, so ABGC is cyclic, so angle GCB= angle GAB=20. We also find angle GFE=60=angle GCD, so FGCE is cyclic. Thus angle GCF= angle GEC=80. Since triangle BCF is congruent to triangle ADB, we find x=80.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25545914955179304482010-04-28T17:52:21.561-07:002010-04-28T17:52:21.561-07:00in tr(ABD), AD/BD=sin(x-30)/sin30
in tr(BDC), BC/B...in tr(ABD), AD/BD=sin(x-30)/sin30<br />in tr(BDC), BC/BD=AD/BD=sinx/sin40<br />thus, sinx=2sin40sin(x-30) <br />or<br />cotx=((3^0.5)sin40-1)/sin40<br />=(2(3^0.5)sin40cos40-2cos40)/2sin40cos40<br />=((3^0.5)sin80-2sin(80-30))/sin80<br />=cot80<br />therefore, x=80KIM YOUNG DONGnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17380611625340332492010-02-17T04:46:05.028-08:002010-02-17T04:46:05.028-08:00To anonymous:
The answer is not 85 degrees.To anonymous:<br />The answer is not 85 degrees.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53886023344249651142010-02-17T03:20:43.269-08:002010-02-17T03:20:43.269-08:00is the answer 85 degrees?is the answer 85 degrees?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35386387625145471162010-02-04T10:11:17.680-08:002010-02-04T10:11:17.680-08:00a more simpler solution added to blog, thank you f...a more simpler solution added to blog, thank you for your question. it made me think again.<br /><br /><br />here is my mail:<br />geometri-problemleri@hotmail.com<br /><br />i am some sick, maybe we can not talk for now. but you can send mail to meAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32304222403913137062010-02-04T06:34:43.814-08:002010-02-04T06:34:43.814-08:00it's definitely a great solution, but it's...it's definitely a great solution, but it's far-fetched i would've done what you just did.<br /><br />i'm wondering if you have another solution.<br /><br />can you give me any mail so that we can talk?Listsnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40564916340197259362010-02-03T20:26:02.686-08:002010-02-03T20:26:02.686-08:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.com