Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Problem 1301 Let K medium the arc AB and L medium the arc BC, then K is medium M1O1 and L is medium M2O2.Draw IH perpendicular in AC , intersecting the circle (Ι,ΙΤ) in P.So from problem 1298 is IP=PH.But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear.APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
To ApostolisReferring to last line of your solution " But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear"In my opinion , we need to show that points C, P and K are collinear before conclude that M1, I and C are collinear.
Yes to peter. The points C,P and T1 are collinear from problem 1300 , <BT1C+<AT1B+<KT2A=45+90+45=180so the points K,T1,P and C are collinear. Thenks
Relation between three inradius (for all three circles inscribed in arbelos)Ri=[R1R2(R1+R2)]/(R1²+R1R2+R2²)
PN perpendicular to AC, P intersection CM1 and AM2, tr PNA ~ tr M2O2A and tr PNC ~ tr M1O1C + Theorem pythagore for tr O1IN and O2IN then just to prove PN = 2Ri