tag:blogger.com,1999:blog-6933544261975483399.post868763662400449928..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1028: Right Triangle, Double Angle, Midpoint, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-70423781918908280242021-07-12T14:31:01.630-07:002021-07-12T14:31:01.630-07:00E is circumcenter of DBC => m(ABE)=90-X => A...E is circumcenter of DBC => m(ABE)=90-X => ABE is isosceles triangle.<br />Let AP be the altitude of ABE , EQ be perpendicular from E onto AC.<br />APE and AQE are congruent <br />=> EQ=EP=BE/2=EC/2 <br />=> EQC is 30-60-90 right triangle<br />=> m(QCE)=90-4X=30<br />=> x=15Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87272496641898678582015-11-12T09:00:21.979-08:002015-11-12T09:00:21.979-08:00Let AE = AB = q, BD = r and BE = DE = CE = p
BE i...Let AE = AB = q, BD = r and BE = DE = CE = p<br /><br />BE is tangential to ADE so p^2 = rq ....(1)<br /><br />From similar Tr.s EF altitude of Tr. AEF = qr/2p = p/2 from (1)<br /><br />So < ECA = 30 and hence x = 15<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44746886649064908202015-06-23T01:33:55.802-07:002015-06-23T01:33:55.802-07:00Wonderful solution, Prof. Radu. I wish your soluti...Wonderful solution, Prof. Radu. I wish your solutions were in English, though, for Google translation is never adequate or accurate.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57876371183295914382014-07-16T20:10:02.817-07:002014-07-16T20:10:02.817-07:00Hi Erina,
Great. Please send your geometric soluti...Hi Erina,<br />Great. Please send your geometric solutions. ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76348789122544843362014-07-16T19:23:28.921-07:002014-07-16T19:23:28.921-07:00Hi Antonio
We have 2 other geometric solutions f...Hi Antonio <br /><br />We have 2 other geometric solutions for problems 1028 . <br />If you want them I can send to you. <br /><br />Have a great night .<br /><br />Erina Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48446259197216861362014-07-16T19:22:17.818-07:002014-07-16T19:22:17.818-07:00Trigonometric Solution 999 and 1028 ( Because ...Trigonometric Solution 999 and 1028 ( Because they have the same picture and same solution<br /><br />1. EF perpendicular AB ---> BC=2EF <br /><br />2. Triangle AEF EF= ABsin2X <br />3 Triangle ABC BC= ABtg3X <br /><br />Tg3X=2sin2X From geometric conditions 0 <3X < 90 ---> 0 <X<30 <br /><br />Equation has this solution X=15 <br /><br /><br />Erina New Jersey Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39997967112290741632014-07-16T03:53:21.730-07:002014-07-16T03:53:21.730-07:00construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,av...construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,avemAEB ~ DEB(EM=NE=EB=BE/2=CE/2=>ECA=30, 90-4X =30=> X =15 ion radunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76841830924999403952014-07-15T20:40:11.677-07:002014-07-15T20:40:11.677-07:00Prob 1027
We build AH perpendicular BC and DF per...Prob 1027 <br />We build AH perpendicular BC and DF perpendicular AB <br />CH=HD=1/2DC=1/2ED <br />DF=DH=1/2ED<br /> α =10 <br />X=90-4α <br />X=50 <br /><br />Prob 1028 <br /><br />BE=ED=EC EM=EN EN =1/2 EC <br />Triangle ECN < ECN =30 ECN = 90-4X -> X =15 <br /><br />Prob 1029 <br />We note <BDF = 90 <br />BD perpendicular AC <br /><br /><br />Erina New Jersey Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73801283203486959142014-07-13T17:51:52.570-07:002014-07-13T17:51:52.570-07:00Trigonometry solution:
Connect BE and let F is the...Trigonometry solution:<br />Connect BE and let F is the midpoint of BE<br />BDE and BAE are isosceles and similar<br />Let BC= 1we have EC=EB= 1/(2.cos(x)) and BH= 1/(4.cos(x))<br />And AB= BF/sin(x)= 1/(2.sin(2x))<br />AC= AB/cos(3x)=BC/sin(3x) => tan(3x)=2. sin(2x)<br />This equation has 2 solutions: x= 0 and x=15<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55961396335045822372014-07-13T04:09:04.667-07:002014-07-13T04:09:04.667-07:00Solution:
1)Connect B and E Notice that Triangle A...Solution:<br />1)Connect B and E Notice that Triangle ABE is isosceles.<br />2)Let AE = AB = a, let DE = BE = EC = b(right triangle).<br />3)AEB ~ DBE (They both isosceles and have the same angles). From here we get that AE/BE = BE/DB =<br />a/b = b/DB --> DB = b^2/a.<br />4)Draw EN_|_BC and EM _|_AC (notice that EN = DB/2 = b^2/2a).<br />5)AEM ~ CEN (Same angles) --> AE/EM = EC/ EN = a / EM = b/(b^2/2a).<br />6)From the proportion in step 5 we get that EM = b/2, in triangle ECM angle ECM = 30 = 90 - 4x.<br />Solution : The measure of x is 15.Anonymoushttps://www.blogger.com/profile/02970933367484538195noreply@blogger.com