tag:blogger.com,1999:blog-6933544261975483399.post3655999183401480643..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 409: Quadrilateral, Diagonals, Perpendicular, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger16125tag:blogger.com,1999:blog-6933544261975483399.post-56837968628935644192020-05-02T20:23:54.590-07:002020-05-02T20:23:54.590-07:00It is special case of problem (t, 30-t, 30+t, 150-...It is special case of problem (t, 30-t, 30+t, 150-3t) where t = 20.<br /><br />In fact, it is related with cevian problem below:<br />https://output.jsbin.com/fofecum#20,10,30,50<br /><br />It can be generated like that:<br />Let P be a point where circumcircle BCD meets AC.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46932706095064425612016-06-21T00:19:24.389-07:002016-06-21T00:19:24.389-07:00Problem 409
I take the point A symmetry of A on t...Problem 409<br />I take the point A symmetry of A on the AB, then triangle AED is equilateral (AE=ED=AD)and <EBA=<ABD=70, but <EBA+<ABD+<DBC=70+70+40=180.So E,B and C are collinear.<br />Is <EAC=<ECA=50 then AE=EC .So the point E is the center of the circumcircle of the triangle CAD.Then <ACD=(<AED)/2=60/2=30. Therefore <BDC=90-30=60.<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-22325840417825958182016-06-17T12:01:03.155-07:002016-06-17T12:01:03.155-07:00Problem 409
Let the point O is the center of th...Problem 409<br /> <br />Let the point O is the center of the circle of des triangle ABD.Then triangle OBD is equilateral<br />And <OAB=<OBA=10,<OAD=<ODA=20,<OAC=10. I take the point K on the side AC such that <br /><DKC=40,so <KDO=10=<KAO , then AOKD is cyclic,then <KOD=<KAD.Therefore KO=KD (OB=BD) so BK is perpendicular bisector to OD.Then <DKB=<BKO=90+10=100,so<br /><CKB=100-40=60. But <DKC=40=<DBC so DKBC is cyclic.Therefore < CDB=<CKB=60.<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-72646094040311859552015-12-20T07:01:56.826-08:002015-12-20T07:01:56.826-08:00Construct the equilateral triangle AED, like Felip...Construct the equilateral triangle AED, like Felipe Hernandez, then <EAC=50=<ACE, so E is the circumcenter of tr. ADC and <ACD=<AED/2=30, i.e. <BDC=90-30=60.<br /><br />Best regardsStan Fulgerhttps://www.facebook.com/stan.fulger?fref=ufinoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52819917801757676032015-12-19T18:33:43.951-08:002015-12-19T18:33:43.951-08:00If u can draw it with the given data u can solve i...If u can draw it with the given data u can solve it at least using trigonometry Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5777088206150114072015-12-04T09:43:07.574-08:002015-12-04T09:43:07.574-08:00Excellent solution!
Slight modification.
Let DP...Excellent solution!<br /><br />Slight modification. <br /><br />Let DP be an altitude of Tr. ADB. Let DP and CB meet at E. Then Tr.s BEP and BDP are congruent ASA hence PE = PD and so Tr. ADE is equilateral. <br /><br />Now EA = EC = ED so < EDC = 80 and < EDB = 20 so x = 60Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17822097459954205662015-11-21T19:40:59.611-08:002015-11-21T19:40:59.611-08:00Referring to your comment " It's easy to...Referring to your comment " It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees. ", Can you explain further .<br /><br />Peter Tran<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40851677910543193732015-11-20T05:52:19.709-08:002015-11-20T05:52:19.709-08:00If O is the circumcentre of triangle ABD and AO ex...If O is the circumcentre of triangle ABD and AO extended meets BD at E, triangle OBD is equilateral. It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63376758781052887262015-07-28T12:16:16.943-07:002015-07-28T12:16:16.943-07:00I forget say : F is point of intersection of ED an...I forget say : F is point of intersection of ED and BA <br />And the point E is above ADAnonymoushttps://www.blogger.com/profile/07565862185002686893noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57471413774183679012015-07-27T16:23:04.617-07:002015-07-27T16:23:04.617-07:00Geometric solution :http://i.imgur.com/byc89Ao.jpg...Geometric solution :http://i.imgur.com/byc89Ao.jpg?1<br />Let E a point ,where triangle AED is equilateral<br />Triangle EBD is isosceles since ∠EFA=90 and EF=FD so ∠DEC=20<br />Points E,B ,C are collinear since ∠BCA +∠BAC=∠EBA<br />EA=EC since∠ EAC =∠ECA <br />So Triangle DEC is isosceles so 20+40+2x=180 finally x=60Anonymoushttps://www.blogger.com/profile/07565862185002686893noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89170067526214398902014-08-18T09:50:22.764-07:002014-08-18T09:50:22.764-07:00A synthetic Proof:
Let P be the reflection of C o...A synthetic Proof:<br /><br />Let P be the reflection of C over BD.<br />Then the problem turns into Trigonometric Form of Ceva.<br />The angles from A in clockwise are: 10, 20, 30, 40,x, 80-x.<br />By a synthetic proof or as a result of trigonometric form of Ceva, 10,30,x are interchangeable; and 20,40,80-x are interchangeable.<br />Swap 20 and 40.<br />Then problem goes into 10,40,30,20,x,80-x.<br />In that case A=10+40=50 and B=30+20=50 and C=x+80-x=80.<br />Let CQ be the symmetry axis of isosceles triangle where Q is on BP.<br />AQP=PQC=60 and QAP=PAC=10 => P is the incenter of triangle QAC.<br />So PCA=PCQ=20 and PCB=x=60.<br /><br />While solving this kind of problems, I always swap appropriate angles. It is a powerful tool also for synthetic solutions.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48703778558416301602010-01-13T14:29:55.302-08:002010-01-13T14:29:55.302-08:00Yes, this problem can be solved without trigonomet...Yes, this problem can be solved without trigonometry.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4682365620021072202010-01-13T06:56:46.509-08:002010-01-13T06:56:46.509-08:00Can this problem be solved without trig?
bjhvash4...Can this problem be solved without trig?<br /><br />bjhvash44@sbcglobal.netAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75156528028588947812010-01-05T06:10:26.212-08:002010-01-05T06:10:26.212-08:00OD=a
OA=atan80
OB=atan80tan20
OC=atan80tan20tan40
...OD=a<br />OA=atan80<br />OB=atan80tan20<br />OC=atan80tan20tan40<br />tanx=OC/OD=tan20tan40tan80=tan60<br />x=60<br />.-.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85866254362224351202010-01-05T05:55:06.762-08:002010-01-05T05:55:06.762-08:00Values of angles ACB and CAD are relevant.Values of angles ACB and CAD are relevant.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23203701266705041672010-01-05T05:02:32.215-08:002010-01-05T05:02:32.215-08:00there is not enough data to solve this^probleme. V...there is not enough data to solve this^probleme. Values of angles ACB et ACD are irrelevant.Anonymousnoreply@blogger.com