tag:blogger.com,1999:blog-6933544261975483399.post5899610728182445985..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 109Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-78229032696047754592022-12-27T01:59:30.694-08:002022-12-27T01:59:30.694-08:00[For easy typing, I use "a" instead of a...[For easy typing, I use "a" instead of alpha]<br /><DBC=90-a & x=90-2a<br />Consider triangle BCD<br />sin(90-a)/CD=sinx/BD<br />CD/BD=sin(90-a)/sinx=cosa/cos2a---------(1)<br /><br />Consider triangle ABD<br />sin2a/BD=sin(180-3a)/AB<br />AB/BD=sin3a/sin2a-------(2)<br /><br />Since AB=CD, by equating (1) & (2)<br />cosa/cos2a=sin3a/sin2a<br />sin3acos2a=sin2acosa<br />sin5a+sina=sin3a+sina<br />sin5a=sin3a<br />5a=3a or 5a=180-3a<br />a=0 (rej) or 22.5<br />x=90-2a=45<br />Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4720515827422848412020-07-11T23:38:27.003-07:002020-07-11T23:38:27.003-07:00<DBC=90-a; take E onto BC so that <BDE=2a. C...<DBC=90-a; take E onto BC so that <BDE=2a. Clearly tr. BDE is isosceles and <CDE=<ABD=a, thus triangles BAD and DCE are congruent, s.a.s., wherefrom <DCE=<BAD, i.e. 2x=90, x=45. Stan FULGERhttp://facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23987965601875542142019-01-17T02:40:14.101-08:002019-01-17T02:40:14.101-08:00 By External Angle Theorem: m∠BDC=3α. Let E be a p... By External Angle Theorem: m∠BDC=3α. Let E be a point on BC such that DE∥AB, we have m∠ABC=m∠DEC=90°, m∠A=m∠EDC=2α, m∠ABD=m∠BDE=α, they are pairs of corresponding angles.<br /> Let F be a point on BC such that DF is the angle bisector ∠EDC, by Leg – Angle theorem: ∆BED≅∆FED, it follows that BD=DF.<br /> Since AB=DC, by SAS: ∆ABD≅∆CDF, therefore: ∠A≅∠C,x=2α, we have x+2α=x+x=90°,x=45°.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69087152267810582016-01-25T09:12:53.646-08:002016-01-25T09:12:53.646-08:00Let O be the midpoint of AC the centre of circle A...Let O be the midpoint of AC the centre of circle ABC<br /><br />OB = b/2. AD = b-c and DO = c-b/2<br /><br />Since BD bisects < ABO<br /><br />c/b/2 =(b-c)/(c-b/2) and so b^2 = 2c^2<br /><br />Hence a= c and x = 45<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74383836313491270682016-01-10T05:51:50.352-08:002016-01-10T05:51:50.352-08:00E is centre of right Tr. ABC, that's why.E is centre of right Tr. ABC, that's why. Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67848111740485280522012-03-18T04:06:21.505-07:002012-03-18T04:06:21.505-07:00The hint about requiring a construction was very h...The hint about requiring a construction was very helpful. I thought about bisecting angle A to get an isosceles triangle with angles 'alpha' but that seemed to make the figure very complex. I then thought of translating AB such that A maps to D and B to a point E, say, to make another isosceles triangle CDE. It then turns out that CDBE is cyclic and this leads to a proof...Sigismundhttps://twitter.com/#!/ProfSmudgenoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75732513372613581452010-07-15T19:20:22.450-07:002010-07-15T19:20:22.450-07:00The trigonometric solution here is interestingly s...The trigonometric solution here is interestingly simple for tan(2α) = BC/AB = BC/CD = sin(3α)/cos(α) applying the Sine Rule in Tr. BCD. Thus sin(2α)/cos(2α) = sin(3α)/cos(α) leading to sin(3α)=sin(5α) which is possible only when α=22.5 deg. ruling out α=67.5 since 2α<90 deg. and thus x = 2α = 45 deg.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3623499048780350082009-05-08T11:55:00.000-07:002009-05-08T11:55:00.000-07:00http://geometri-problemleri.blogspot.com/2009/05/p...http://geometri-problemleri.blogspot.com/2009/05/problem9-solution-on-10-may.htmlAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77758083505925184842009-04-07T03:39:00.000-07:002009-04-07T03:39:00.000-07:00If in the construction suggested by Anonymous AE =...If in the construction suggested by Anonymous AE = BE then Tr. AEB is isosceles and BD bisects angle ABE. That's easily understood but why should Tr. BCE be isosceles too?Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56668271382287340752008-05-27T03:06:00.000-07:002008-05-27T03:06:00.000-07:00Construct E on AC where AE=BE, then BCE is isoscel...Construct E on AC where AE=BE, then BCE is isosceles. Let AE=BE=CE=1 and AB=x, then by angle bisector theorem DE=1/(x+1), so 1/(x+1)+1=x, giving x=sqrt(2). Thus angle BAC is pi/4, therefore also angle ACB.Anonymousnoreply@blogger.com