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.

## Monday, November 30, 2015

### Geometry Problem 1168: Construction of the Inscribed Circle of the Arbelos, Semicircles, Diameter, Circle, Triangle, Circumcircle, Tangent

Labels:
arbelos,
circle,
circumcircle,
construction,
inscribed,
semicircle,
triangle

Subscribe to:
Post Comments (Atom)

See below for minor correction and clarification from previous comment

ReplyDeletehttp://s21.postimg.org/aw9ih96yf/pro_1168.png

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

ReplyDelete