tag:blogger.com,1999:blog-6933544261975483399.post593839022535499128..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 936: Circle, Semicircle, Diameter, Tangent, Radius, Chord, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-35771808019135884542020-02-18T10:21:44.139-08:002020-02-18T10:21:44.139-08:00Trigonometry Solution
Draw the Common Tangent at ...Trigonometry Solution<br /><br />Draw the Common Tangent at E, XEY<br /><br />If < XEC = p, then < COE = 2p = < OQD, so < QED = p = < XEC whence < CED = 90<br /><br />So Cos(2p) = 3/4. From Tr. QED, cos (p) = 3/2x<br /><br />Now cos (2p) = 2cos^2 (p) - 1<br />Hence 9/(2x^2) - 1 = 3/4<br /><br />Solving, x = 3.sqrt(2/7)<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62481202040514498352019-01-30T10:28:29.418-08:002019-01-30T10:28:29.418-08:00Super!Super!Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18422545668960341702013-12-01T11:44:13.705-08:002013-12-01T11:44:13.705-08:00If <BDE=a then <OED=90-a and <EOD=2a-90. ...If <BDE=a then <OED=90-a and <EOD=2a-90. By circle theorem if <EOD=2a-90 then <BCE=a-45 and <ECO=(a-45)+45=a=<EDB making CEDO cyclic. That makes <DEC=90 and CE=sqrt7 by Pythagoras. If QD cuts CE at P then Q is cicrumcenter of PED so QD=1/2*DP. <br /><br />By angle bisector PE=3/sqrt7 so DP=6*sqrt(2/7) by Pythagoras. QD is half so QD=3*sqrt(2/7)Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51156389387469393102013-11-22T14:28:11.102-08:002013-11-22T14:28:11.102-08:00OC^2=DC*(DC+DE)/2
QE=OC*DE/(DC+DE)=(DC*(DC+DE)/2)^...OC^2=DC*(DC+DE)/2<br />QE=OC*DE/(DC+DE)=(DC*(DC+DE)/2)^(1/2)*DE/(DC+DE)<br />=(DC/2)^(1/2)*DE/(DC+DE)^(1/2)<br />=(2)^(1/2)*3/(7)^(1/2)<br />=(14)^(1/2)*(3/7)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17043206247173625992013-11-22T06:35:32.869-08:002013-11-22T06:35:32.869-08:00Enlarge circle Q with center E, such that Q→O, (Ho...Enlarge circle Q with center E, such that Q→O, (Homothetic transformation)<br />let D→F, then EQO, COF and EDF are straight lines, with EF=7. <br /><br />Since then CF is the diameter of circle O, thus ∠CED=∠CEF=90°. <br /><br />Using Pythagoras theorem, we have <br />CE² = 4²−3² = 7<br />CF² = CE² + EF² = 7 + 49 = 56<br />So OE = OF = √14<br /><br />Since ΔEQD~ΔEOF, <br />x/√14 = 3/7<br />x = (3/7)√14Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com