tag:blogger.com,1999:blog-6933544261975483399.post7298829813393676385..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 336. Two equal circles, a Common Tangent and a SquareAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-60781284825545874202023-08-25T08:05:32.627-07:002023-08-25T08:05:32.627-07:00(r-x)^2+(r-x/2)^2=r^2
r^2-2rx+x^2+r^2-rx+x^2/4=r^2...(r-x)^2+(r-x/2)^2=r^2<br />r^2-2rx+x^2+r^2-rx+x^2/4=r^2<br />r^2-3rx+5x^2/4=0<br />4r^2-12rx+5x^2=0<br />(2r-5x)(2r-x)=0<br />x=2r (rej.) or x=2r/5Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26446489711159834872021-04-01T06:51:44.813-07:002021-04-01T06:51:44.813-07:00See the drawing
In ΔDMF : DM^2=DF^2+FM^2
(1) DM^2...See the <a href="http://sciences.heptic.fr/2021/04/01/gogeometry-problem-336/" rel="nofollow"><b>drawing</b></a><br /><br />In ΔDMF : DM^2=DF^2+FM^2<br />(1) DM^2= (r-x/2)^2+x^2<br />In ΔAMD : AM^2=AD^2+DM^2 -2xAD (Proposition 13 Euclide Book II)<br />r^2= r^2+ DM^2 -2xr<br />(2) DM^2=2xr <br />(1) and (2) 2xr=(r-x/2)^2+x^2<br />2xr=r^2-xr+x^2/4+x^2<br />3xr=r^2+5x^2/4<br />Dividing by x^2 (x not equal to 0) =>3r/x=5/4+(r/x)^2<br />Let y=r/x : y^2-3y+5/4=0<br />Delta = 4 and y=5/2 or y=1/2<br />Therefore x=2r/5 or x=2r<br /><br />Since the only definition of M and H are<br />M belongs to the square and circle A, <br />and H belongs to the square and Circle B<br />Therefore, <b>there are 2 different solutions:</b><br /> x=2r/5 (blue square)<br />or x=2r (green square)<br /><br /><br />harvey.littleman@sciences.heptic.frnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20424935038013312262021-03-21T07:12:06.558-07:002021-03-21T07:12:06.558-07:00See diagram here.
Let O middle of FG and extend EB...<a href="https://drive.google.com/file/d/1c2RS9qTkQAYlaOaCAluuzT6wCqypDZ_n/view?usp=sharing" rel="nofollow"> See diagram here.</a><br />Let O middle of FG and extend EB to intersect circle B in K<br />∆OGH and ∆OEK are SAS similar and DE tangent to B in E => ∠OHG = ∠HKE = ∠OEH<br />Hence ∆OGH and ∆HGE are similar => EG = 2x and OE = r = 5x/2 QEDGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31647481133904866212010-12-24T16:14:23.278-08:002010-12-24T16:14:23.278-08:00Extend MH so that it meets the radii AD and BE at ...Extend MH so that it meets the radii AD and BE at points P and Q<br />Let PM = HQ = y<br /><br />Then we have two equations:<br /><br />(r-x)^2 + y^2 = r^2 ..... (1)<br /><br />and<br /><br /> 2y + x = 2r ..........(2)<br /><br />Eliminate y between the two equations to get a quadratic which resolves to<br />x = (2/5)*r or x = 2r.<br />x = 2r is absurd.<br />So x = (2/5)*r is the answer<br />which willUnknownhttps://www.blogger.com/profile/07363309698823587597noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86055328066457236882009-08-08T22:29:46.490-07:002009-08-08T22:29:46.490-07:00Extend FM to meet AC in N. Also extend DA & EB...Extend FM to meet AC in N. Also extend DA & EB to meet the two circles in D' & E' respectively. <br />Now, AN = r - x/2 & NM = r - x. From triangle MAN, (r-x/2)^2+(r-x)^2=r^2 from which we get x=2r/5 or x=2r. The former is the side of square FGHM while the latter is the side of DEE'D'<br />vihaan: vihaanup@gmail.comVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.com