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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1048.
Hi Antonio, I would like to add the folllowings:The point B is the center of the inscribed circle of the triangle DFH and the smallest circles passing through B and touching the three sides are Archimedean.Hiroshi
As far as I know, the two Archimedean circles touching the sides DF and DH are new. Hiroshi
Hi, HiroshiPlease explain why B is the center of the inscribed circle of triangle DFH concretely.
http://s25.postimg.org/ag5sakzpb/pro_1048.pngConnect lines per attached sketchWe have GM=GB=GN => MBN is a right triangleIn right triangles AEC and O1GO2 we have GB^2=BO1.BO2 and BE^2=BA.BCSo BG= ½ BE and MBNE is a rectangleWe have ∠ (BNE)= ∠ (BNC)=90 =>E, N, C are collinearSimilarly A,M,E are collinearNote that ∠ (BCE)= Arc(AF)+Arc(FE)And ∠ (EMN)= Arc(AF)+Arc(EH)But ∠ (EMN)= ∠ (BNM)= ∠ (BCE) => Arc( FE)=Arc(EH)=> DE is an angle bisector of angle FDH
By theorem from previous problem, AMNC is concyclic. Therefore, AM and NC must meet on radical axis DE of O1 and O2 at point P. <APC must be 90 because <MAB+<NCB=90, so P is also on big circle. P must coincide with E as result.<ACN=<AMF=<ACF+<EDH, but also<ACN=(<ADF=<ACF)+<FDE.Therefore <ACF+<EDH=<ACF+<FDE, so <EDH=<FDE
Let radius of O1 = a and that of O2 = bLet MG = GB = GN = p and let GB = qSo 2a X 2b = (p+q)^2But MN^2 = 4p^2 = (a+b)^2 - (b-a)^2 = 4abSo p+q = 2p hence p=q hence BMEN is a rectangle, AME, CNE are collinear and ADCE is a kite. So < BGN = < MAD hence AMGD is cyclic and so < BMG = < ADG = < HFE + < FEM = = < HDE + < ADF ==>. < FDB = < HDB