tag:blogger.com,1999:blog-6933544261975483399.post975463921189003255..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 284: Circular Sector 90 degrees, Semicircles, Tangent, RadiusAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-13829715787306824632021-01-04T12:27:48.555-08:002021-01-04T12:27:48.555-08:00See the drawing
Define h=OF
OA=OB=R
D, E and C ar...See the <a href="http://sciences.heptic.fr/2021/01/04/gogeometry-problem-284/" rel="nofollow"><b>drawing</b></a><br /><br />Define h=OF<br />OA=OB=R<br />D, E and C are aligned<br />ΔCOD is right in O => DC^2=OC^2+OD^2<br />OD=r+h<br />=> (1) : (r+R/2)^2=(R/2)^2+(r+h)^2<br />=> (1) : (r+R)^2=R^2+4(r+h)^2<br />OB=OA => R=2r+h => r+h=R-r<br />(1) : (r+R)^2=R^2+4(R-r)^2<br />=> r^2+2rR+R^2=R^2+4(R^2-2rR+r^2)<br />=> 6rR= 3r^2<br />Therefore <b>3R=r</b><br /><br />rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65177817428710231362009-04-29T22:26:00.000-07:002009-04-29T22:26:00.000-07:00We can use Pythagorean Theorem to solve it.
Since ...We can use Pythagorean Theorem to solve it.<br />Since D, E, C are collinear,OD=R-r, OR=R/2 and DC=r+R/2,<br />So solving (R-r)^2+(R/2)^2=(r+R/2)^2 will give r=R/3.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56649063221670455172009-04-26T20:38:00.000-07:002009-04-26T20:38:00.000-07:00The two inner circles are given by:
(x-R/2)^2+y^2=...The two inner circles are given by:<br />(x-R/2)^2+y^2=R^2/4<br />x^2+(y-R+r)^2=r^2<br />Solve these simultaneously and set the discriminant to zero for the circles to touch each other. This gives us:-R^6+4rR^5-3r^2R^4=0 or -R^2+4rR -3r^2 =0 which can be solved to obtain r = R/3 or r=R of which only the first solution is admissible.<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com