tag:blogger.com,1999:blog-6933544261975483399.post752686250310478028..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 199: Triangle anglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger16125tag:blogger.com,1999:blog-6933544261975483399.post-20800919951083262142013-02-12T08:22:14.427-08:002013-02-12T08:22:14.427-08:00Problem 199 solution by Mixalis Tsourakakis at:
h...Problem 199 solution by Mixalis Tsourakakis at: <br /><a href="http://www.gogeometry.com/problem/p199-geometry-mixalis-greece.pdf" rel="nofollow">http://www.gogeometry.com/problem/p199-geometry-mixalis-greece.pdf</a><br />Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64812325169733457632012-08-15T21:32:53.038-07:002012-08-15T21:32:53.038-07:00Problem 199: Solution by Mixalis Tsourakakis from ...Problem 199: <a href="http://www.gogeometry.com/problem/p199-geometry-mixalis-greece.pdf" rel="nofollow">Solution</a> by Mixalis Tsourakakis from Greece.<br />Thanks Mixalis.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51826088025064395162010-11-12T20:01:00.778-08:002010-11-12T20:01:00.778-08:00To Speed 2001
Porque θ=8 Grados, x=3θ=24 Grados.
...To Speed 2001<br /><br />Porque θ=8 Grados, x=3θ=24 Grados.<br />Because θ=8 Degrees, x=3θ=24 Degrees.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62799194236026167432010-11-11T17:21:58.795-08:002010-11-11T17:21:58.795-08:00Another solution:
http://triangles-geometry.blogsp...Another solution:<br />http://triangles-geometry.blogspot.com/2010/11/problem-001-auxiliary-lines.htmlFelipehttps://www.blogger.com/profile/18073880153170044270noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26473264185145040882010-11-11T08:23:22.960-08:002010-11-11T08:23:22.960-08:00Sorry, I take it back, there's no paradox.Sorry, I take it back, there's no paradox.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82259231662917080182010-11-09T18:43:53.574-08:002010-11-09T18:43:53.574-08:00Before coming up with a geometric solution, you sh...Before coming up with a geometric solution, you should discover something:<br /> Want to see a paradox within an auxiliary construction? <br /> Extend CE, intersecting AB at F (which means Angle CFB=60 Degrees). Extend BD, make Angle ECG=38 Degrees; G is on the extension of BD. Connect FG and AG. (Thus FBCG is cyclic, Angle FGB=14 Degrees and Angle CFG=68 Degrees. Angle AFG=52 Degrees. Triangles AGD and CGD are both 30-60-90 Triangles.) Angle AGF=16 Degrees. Extend FG until AFCH is cyclic, that means Triangle AGH is isoceles; Angle GAH=Angle GHA=8 Degrees. Construct median GI of Rt Triangle AGC to side AC; then AG=GI=GH, which means G is the circumcenter of Triangle AHI. So Angle GHI=Angle GIH=22 Degrees. Angle GCH=Angle GIH=22 Degrees, subtended on segment GH, GICH should be cyclic, however, Angle GHI=22 Degrees, Angle GCI=30 Degrees, both angle subtended on segment GI; therefore, there's a paradox!<br /> Who can help me get solve this paradox? And post some geometric solutions up!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11764211228079854422010-11-08T12:38:44.789-08:002010-11-08T12:38:44.789-08:00Who has an imageshack.us image for solving the pro...Who has an imageshack.us image for solving the problem?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86909435229410415252010-11-07T16:53:10.887-08:002010-11-07T16:53:10.887-08:00The correct answer is 24 Degrees; this case trigon...The correct answer is 24 Degrees; this case trigonometry solution is a lot quicker than geometric solution. I found only one geometric solution for this problem, but it's in Arabic. If anyone has a geometric solution; it's best to copy a link to a website with that solution, like "Geometri Problemleri".Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87228916877698298342010-08-15T21:17:06.317-07:002010-08-15T21:17:06.317-07:0024°24°Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41450320388140893712010-08-06T07:03:05.045-07:002010-08-06T07:03:05.045-07:00What is the correct answer?What is the correct answer?Vitornoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77281725415742471572010-01-07T15:13:01.288-08:002010-01-07T15:13:01.288-08:00To c.t.e.o:
The answer is wrong.To c.t.e.o:<br />The answer is wrong.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52005008277348649902010-01-07T13:28:31.845-08:002010-01-07T13:28:31.845-08:00draw BF = AB
draw EG perpendicular to BC
ang EFD ...draw BF = AB<br />draw EG perpendicular to BC<br /><br />ang EFD = x = BEG ( perpendicular sides )<br /><br />=> x = 22 ( from tr BEG )c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50769345897605843292009-04-02T13:40:00.000-07:002009-04-02T13:40:00.000-07:00Tips: Think in auxiliary constructions with 30, 60...<B>Tips:</B> Think in auxiliary constructions with 30, 60 degrees, equilateral triangle, isosceles triangle, congruence, cyclic quadrilateral.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11689610159243413992009-04-02T06:56:00.001-07:002009-04-02T06:56:00.001-07:00Antonio, would you like to give me some tip to do ...Antonio, would you like to give me some tip to do a geometric solution? What the first step? Please...Unknownhttps://www.blogger.com/profile/17597446174103329489noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68052946980451725122008-11-02T19:10:00.000-08:002008-11-02T19:10:00.000-08:00You can also choose to use a geometric solution ra...You can also choose to use a geometric solution rather than a trigonometric solution.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66856384380046075102008-11-02T18:03:00.000-08:002008-11-02T18:03:00.000-08:00tan68=dc/bdtan38=ad/bdtanx=ed/adtan8=ed/dcso, tan6...tan68=dc/bd<BR/>tan38=ad/bd<BR/>tanx=ed/ad<BR/>tan8=ed/dc<BR/>so, tan68/tan38=tanx/tan8<BR/>x=24Anonymousnoreply@blogger.com