Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, November 30, 2015
Geometry Problem 1168: Construction of the Inscribed Circle of the Arbelos, Semicircles, Diameter, Circle, Triangle, Circumcircle, Tangent
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See below for minor correction and clarification from previous comment
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Let 2R is the diameter of circle O , 2r1=diameter of circle O1 and 2r2=diameter of circle O2
Complete 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 C
We have tri. O1O2O4 similar to tri. F’CB …( case AA)
So O2O4/BC= O1O2/CF’= ½ => CF’=2.O1O2= AC= 2.R
Similarly we also have AD’=AC= 2.R
Perform 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
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