tag:blogger.com,1999:blog-6933544261975483399.post5901822872539669067..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 818: Square, Triangle, AnglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-6933544261975483399.post-6862686494480590262017-06-06T03:42:41.535-07:002017-06-06T03:42:41.535-07:00Drop a perpendicular from D to AG to meet AG at X ...Drop a perpendicular from D to AG to meet AG at X and CF at Y. Let AG meet CD at Z.<br /> <br />DX is an altitude of right ∆ ADZ, hence < ZDY = DAZ = 2α.<br />But < DYC = 2α considering ∆GXY, so < DCY also = 90-α.<br />Therefore DY = DC<br /> <br />So DX = ½ DY = ½ DC = ½ AD<br />Hence ADX is a 30-60-90 ∆ and 2α = 30 and α = 15<br /> <br />Finally from ∆ AGF, x = 180 - (90 - 2α) - α =90 + α = 105<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34721339068656517772017-06-05T01:34:37.684-07:002017-06-05T01:34:37.684-07:00Excellent work PeterExcellent work PeterSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66061424869234772772013-10-10T06:53:42.512-07:002013-10-10T06:53:42.512-07:00http://www.mathematica.gr/forum/viewtopic.php?f=22...http://www.mathematica.gr/forum/viewtopic.php?f=22&t=32386&p=149846Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72198815005840770342013-01-04T14:51:55.125-08:002013-01-04T14:51:55.125-08:00correction:
The triangle CAE is isosceles And not ...correction:<br />The triangle CAE is isosceles And not "the triangle ΜΑΕ is isosceles"<br />(Michael TsourakakisAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91961743249182304672013-01-04T13:50:44.321-08:002013-01-04T13:50:44.321-08:00correct solution
(q) is Circumcircle of the squar...correct solution<br />(q) is Circumcircle of the square ABCD and (q) section AE=K<br />BD ,is mediator of AC and V is mediator of CE. BD section (v)=M.Then , CM is mediator of AE ,so CM is perpendicular of the AE.Beacause AC is ,diameter the circle (q) ,CK ,is perpendicular of AE and the points C,K,M , are collinear . The triangle ACE ,is isosceles.<br />So, angle CAE=α and angleCAD=3α=45 . Τhen,α=15<br />x=180-45-2α =135-30=105 degrees<br />http://img191.imageshack.us/img191/5597/geogebra3.png<br />(Michael Tsourakakis )<br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48516136297897316102012-12-31T18:25:25.616-08:002012-12-31T18:25:25.616-08:00http://img839.imageshack.us/img839/3518/problem818...http://img839.imageshack.us/img839/3518/problem818.png<br /><br />Draw circumcircle of triangle CDE<br />Note that angleDOG= 2 α and AE cut arc CD at midpoint G<br />Since angle (GAD)= angle (DOG)= 2 α and AD//GO => ADOG is a parallelogram<br />And OD=AD=CD => triangle DOC is a equilateral <br />So 2 α= 30<br />In triangle AFE , we have x=180- α-(90-2 α)= 90+ α= 105 degrees<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49637503560223017062012-12-31T17:29:30.403-08:002012-12-31T17:29:30.403-08:00you're right.Thanks for the correction.I'l...you're right.Thanks for the correction.I'll look again<br />HAPPY NEW YEAR<br />Michael Tsourakakis<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14776645437895417892012-12-31T15:07:11.210-08:002012-12-31T15:07:11.210-08:00Anonymous
If we say the cut of BD with CF = P an...Anonymous <br /><br />If we say the cut of BD with CF = P and <br />the cut AE with CD =R you have accept without verification that angle CPR= 2 alpha or equivalent that angle APR =90 degree . <br />If you can not verify one of these two saying than the solution is wrong. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18021652871701230382012-12-31T15:00:44.655-08:002012-12-31T15:00:44.655-08:00To: Michael Tsourakakis---In your solution you hav...To: Michael Tsourakakis---In your solution you have implied that the ARC CKMD is split into 3 equal parts. This is not true based on your reasoning. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90954401063233066482012-12-21T03:36:07.595-08:002012-12-21T03:36:07.595-08:00Daca notam cu P intersectia dreptelor CF si AD vom...Daca notam cu P intersectia dreptelor CF si AD vom obtine:2α=m(m( in triunghiul AFP dreptunghic in A, m(<br />x=m(<AFE)=180-(90-α)=90+αProf Radu Ion,Sc.Gim.Bozioru,Buzaunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55620419044851172042012-12-09T09:10:24.802-08:002012-12-09T09:10:24.802-08:00http://1.bp.blogspot.com/-H0dtaRbb9bY/UMTFm5yQRyI/...http://1.bp.blogspot.com/-H0dtaRbb9bY/UMTFm5yQRyI/AAAAAAAAAAM/5Dco1bojssk/s1600/Untitled.jpgMJChenhttps://www.blogger.com/profile/02062293431521151890noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37907026775023930702012-10-31T05:40:59.918-07:002012-10-31T05:40:59.918-07:00Geometric solution by Michael Tsourakakis
EC sect...Geometric solution by Michael Tsourakakis<br />EC sectionAB=H<br />(q) is Circumcircle of the scare ABCD and (q) section AE=K <br />bisector of the angle EAD section (q) =M.<br />angleDCK=angleKAD=2α .So angleDCM=angleDKM=angleCDK=angleKDM=α.Therefore, angleCBK=α<br />But angleKBD=angleKAD=2α .So angleDBC=45=2α+α=3α. So,α=15<br />angleCAK=α=angleCEA.So, angleHKA=2α =30<br />From the triangle HCA : x=180-30-45=1050 <br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59216655874307267482012-10-28T19:29:40.980-07:002012-10-28T19:29:40.980-07:00Problem 818: Try to use elementary geometry (Eucli...Problem 818: Try to use elementary geometry (Euclid's Elements).Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48087604350764253362012-10-27T09:24:31.302-07:002012-10-27T09:24:31.302-07:00In the triangle AFE
AFE) = x = 180 - alpha -(90 -...In the triangle AFE <br />AFE) = x = 180 - alpha -(90 - 2 * alpha) = 90 + alpha.<br />Then (DCE) = (AFE) = 90 + alpha.<br /><br />In the triangle ADE : <br />AD/sin(alpha) = DE/sin(2 alpha) <br />Therefore: <br />DE = 2 AD cos(alpha) ----- (1)<br /><br />In the triangle DCE : <br />DC/sin(2 alpha) = DE/sin(90 + alpha) <br />Therefore: <br />DE = DC / ( 2 sen(alpha) ) ----- (2).<br /><br />DC = AD<br /><br />Therefore from (1) and (2) : <br />2 cos(alpha) = 1/( 2 sen(alpha) ) ------ (3)<br /><br />If we simplify the (3): <br />sin(2 alpha) = 1/2.<br />Therefore: alpha = 15<br /> x = 90 + 15 = 105Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39641915774511647392012-10-25T06:17:19.374-07:002012-10-25T06:17:19.374-07:00To Anonymous, problem 818: solution is not complet...To Anonymous, problem 818: solution is not complete.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1767957455313813192012-10-25T05:48:42.695-07:002012-10-25T05:48:42.695-07:00alpha=pi/12=15 G
x=pi/2+alpha= 105 G
I solve sys...alpha=pi/12=15 G <br />x=pi/2+alpha= 105 G<br /><br />I solve system of 3 equations<br /><br />c=tg(a)<br />c+1=b*tg(3a)<br />c+1=(b+1)*tg(2a)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63973987581160934272012-10-25T03:52:11.571-07:002012-10-25T03:52:11.571-07:00http://www.wolframalpha.com/input/?i=tg%283a%29%3D...http://www.wolframalpha.com/input/?i=tg%283a%29%3Dtg%28a%29%2B1<br /><br />alpha=0.30714520<br />x=pi/2+alphaAnonymousnoreply@blogger.com