Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1168.
See below for minor correction and clarification from previous commenthttp://s21.postimg.org/aw9ih96yf/pro_1168.pngLet 2R is the diameter of circle O , 2r1=diameter of circle O1 and 2r2=diameter of circle O2Complete full circle O and let E’ is the midpoint of arc AC ( see sketch)Let Au and Cv are the tangents or circle O at A and CWe have tri. O1O2O4 similar to tri. F’CB …( case AA)So O2O4/BC= O1O2/CF’= ½ => CF’=2.O1O2= AC= 2.RSimilarly we also have AD’=AC= 2.RPerform geometry inversion with inversion center at B and inversion power= -BA. BC=- BE.BE’=-BF.BF’=-BD.BD’In this transformation Circle O → circle O circle O2 → line Au Circle O1 →Line Cv, Circle O4→ Line AE’ Circle O3 →line CE’Circumcircle of tri. DEF → Circumcircle of tri. D’E’F’But circumcircle of tri. D’E’F’ tangent to circle O, Lines Au and Cv ( images of circles O, O2 and O1)So circumcircle of tri. DEF will tangent to circles O, O1 and O2
just change ´´ given ´´ to ´´ to prove ´´ at P 638