tag:blogger.com,1999:blog-6933544261975483399.post6071647933134516858..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1341: Isosceles Triangle, 80-20-80 Degrees, Circumcenter, Angle BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-6933544261975483399.post-81758873627119267112018-12-27T23:11:32.058-08:002018-12-27T23:11:32.058-08:00OC bisects < BCX so BO/OX = BC/CX
But BC/CX = ...OC bisects < BCX so BO/OX = BC/CX<br /><br />But BC/CX = AB / AC<br /><br />Now AB/ AC = BD/DC since AD bisects < BAC<br /><br />So BO/OX = BD/DC<br />Hence OD//DCSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17805499536174910942018-12-23T01:16:41.146-08:002018-12-23T01:16:41.146-08:00Sumith: How does OD//XC follows from the angle bis...Sumith: How does OD//XC follows from the angle bisector theorem (BO/BX = BC/XC = AB/AC =BD/DC)?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59705497836673436962018-08-06T11:06:54.589-07:002018-08-06T11:06:54.589-07:00Sorry for the typo
Presume All’s Well nowSorry for the typo<br /><br />Presume All’s Well nowSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53361772107425335562018-07-02T22:50:53.996-07:002018-07-02T22:50:53.996-07:00@Sumith -- Typo correction reqd. BO/OX = BC/XC@Sumith -- Typo correction reqd. BO/OX = BC/XCAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65944830837069913712018-03-11T12:00:40.872-07:002018-03-11T12:00:40.872-07:00Sorry, I didn't follow the figure: I used A at...Sorry, I didn't follow the figure: I used A at the top and BC for the base of the triangle, the solution still valid.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1285220859497855332018-03-10T19:37:53.106-08:002018-03-10T19:37:53.106-08:00Solution by Trigonometry:
Draw AE the height of T(...Solution by Trigonometry:<br />Draw AE the height of T(ABC), then AE=EB=1 (valid for any dimension). Draw a perp. AF from O to AC, so AF=FC. Draw OC T(AOC) is isosc. and A(OCF)=10.<br />T(OCF) is sim. to T(AEC) now because FC=AC/2= 1/2sin(10) and and OF=AE=1/tan(10), the ratio of sim. is (1/2sin(10))/(1/tan10)=1/2cos10 and OF=1x1/2cos(10)= 1/2cos(10)<br />Per T(BDC) DC/sin(40)=2/sin(60) and DC=2sin(40)/sin(60)<br />Then, FD=FC-DC= 1/2sin(10)-2sin(40)/sin(60), finally tan(x)=OF/FD<br />tan(x)=(1/2cos(10))/(1/2sin(10)-2sin(40)/sin(60)) = 0.363970 and x=20Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41645497474002105562018-03-01T06:13:24.522-08:002018-03-01T06:13:24.522-08:00This is not proper geometric reasoning using accep...This is not proper geometric reasoning using accepted and time tested Ecludean principles and axioms<br /><br />Sorry not acceptable Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18063129997385306162018-02-06T13:20:48.670-08:002018-02-06T13:20:48.670-08:00Intuition I guess. As I was looking and drawing th...Intuition I guess. As I was looking and drawing the setup for the problem, I noticed the circle went through the apex of the large triangle and point D. And if you look closely at segment AD, that circle is NOT tangent to it. I don't know how else to explain it.Geek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32958134883623599872017-11-15T09:32:57.272-08:002017-11-15T09:32:57.272-08:00Problem 1341
Solution 2
From Problem 1274, circumr...Problem 1341<br />Solution 2<br />From Problem 1274, circumradius of ∆ABC, R2 = a2b/(a+b)…(1)<br /> <br />Now since AD bisects < BAC, CD = ab/(a+b), <br />so CD.CB = a2b/(a+b) = R2 from (1).<br />Hence CO is tangential at O to circle BDO.<br />So < COD = 10 and hence < BDO = 10+10 = 20<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61516921826528554922017-11-14T11:04:49.081-08:002017-11-14T11:04:49.081-08:00Sumith - Will you please explain how BO/BX = BC/XC...Sumith - Will you please explain how BO/BX = BC/XC<br />Per Angle bisector theorem shouldn't it be BO/OX = BC/XCAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77909986056932491102017-11-13T09:00:02.960-08:002017-11-13T09:00:02.960-08:00How did u assume that the circle u drew is tangent...How did u assume that the circle u drew is tangent to AD?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68481360034687428252017-11-13T08:55:49.956-08:002017-11-13T08:55:49.956-08:00Extend BO to X so that AXC is an equilateral trian...Extend BO to X so that AXC is an equilateral triangle.<br /><br />OC bisects <BCX<br />So BO/BX = BC/XC = AB/AC =BD/DC since AD bisects <A<br /><br />Hence OD//XC<br /><br />So < BDO = < BCX = 20<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1799850276076066222017-11-06T10:33:32.951-08:002017-11-06T10:33:32.951-08:00Here is my solution...
https://www.youtube.com/wa...Here is my solution...<br /><br />https://www.youtube.com/watch?v=d4ZzPnoFogkGeek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4586614537949938542017-09-09T14:45:59.724-07:002017-09-09T14:45:59.724-07:00https://ibb.co/df8rta
Let AO intersects BC in E. ...https://ibb.co/df8rta <br />Let AO intersects BC in E. Since AB=BC and OA=OB we have DH is a perpendicular bisector of AE => FA=FE => EF=EB (5). From (3) and (4) and the AAS => triangle ADC=triangel ADF => AC=AF and DC=DF => AD is a perpendicular bisector of FC. Let the perpendicular bisector of BC intersects BC in M and AD in T => TC=TF=TB (6). From (5) and (6) => ET is a perpendicular bisector of FB, i.e. ET _|_ FB. Let TE intersects BF in K => triangle BEK~trianle TEM => D,O,E and T lie on a circle and from (7) => <ODE=<OTE=20.S.D.noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29435420427315050932017-09-09T13:22:13.755-07:002017-09-09T13:22:13.755-07:00In order to prove "triangle AOK=triangle COL&...In order to prove "triangle AOK=triangle COL" above, how do we know that "10=<LOC" since we don't know yet that L is on BC?S.D.noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68877699056120424182017-08-11T13:48:13.501-07:002017-08-11T13:48:13.501-07:00Problem 1341 (sulotion 2)
Let K point on AB such ...Problem 1341 (sulotion 2)<br />Let K point on AB such that KA=KO , then <KAO=10=<KOA. I form the equilateral triangle<br />ΟΚL (the point L is located on the right of BC).Then the point K is the circumcenter the <br />triangle AOL.So <LAO=<LKO/2=60/2=30=<DAO so the point L belongs to AD.<br />Is triangle AOK=triangle COL(OA=OC,OK=OL,<KOA=10=<LOC),then LC=LO=KO=KA and <br /><LCO=10,so the point L belongs to BC.<br />Therefore the points D,L coincide and <ODB=<OLB=<LOC+<LCO=10+10=20.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71463404754375413532017-08-10T16:17:14.343-07:002017-08-10T16:17:14.343-07:00Problem 1341
Suppose the AD crosses the circle at...Problem 1341<br />Suppose the AD crosses the circle at point E, then arcBE=arcEC and <ECB=40, <ECO=50=<OEC (<COE=80).I form the equilateral triangle OCF (F is arcEC ).<br />Is < FOE=20, if G is circumcenter of triangle FOE then GO=GE=GF and <GOE=10=<GEO.<br />Is triangle OGC=triangle GFC so <GCO=<GCF=60/2=30.But <FCE=<FOE/2=20/2=10<br />and <ECG=20.Now triangle DCE=triangle GEC so DC=GE.But triangle OCD=triangle OEG<br />(OC=OE,DC=GE,<OCD=10=<OEG) ,soOD=DC=GO=GE.Therefore <ODB=<DOC+<DCO=<br />=10+10=20.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com