Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 821.

## Wednesday, November 14, 2012

### Problem 821: Adams Circle Theorem, Incenter, Incircle, Gergonne Point, Contact triangle, Parallel, Six Concyclic points, Concentric Circles

Subscribe to:
Post Comments (Atom)

http://img651.imageshack.us/img651/1259/problem821.png

ReplyDeleteDraw lines per attached sketch

We have ∆CA’B’ and CA”’B” are isosceles => A’A””=B’B”

From B draw line C1A1//AC

This line cut B’A’, B’C’,GC”’ and GA”’ at A1, C1, C2 and A2 .

Note that ∆CA’B’, ∆ BA’A1, ∆AB’C’ and ∆BC’C1 are isosceles

So BC1=BC’=BA’=BA1 => BA1=BC1

Since GA2//B’A1 and GC2//B’C1 we have BA2/BA1=BG/BB’=BC2/BC1

B is the midpoint of A2C2 => B’ is the midpoint of B”B”’

Triangle IB”B”’ is isosceles

Similarly IC”C”’ and IA”A”’ are isosceles => IB”=IB”’=IC”=IC”’=IA”=IA”’

So I is the center of circle A”A”’B”B”’C”C

Denote points P1,P2,P3,...P6 as the points where the parallel lines cut the triangles (starting from the right hand side of A)

ReplyDeleteBy the properties of incircle, AC'=AB' and since P2P5 parrllel to B'C', AP2 = AP5 hence B'P5 = C'P2.

By symmetry, B'P5 = C'P2 = A'P3 = B'P6 = C'P1 = A'P4

Therefore P1P2P5P6 forms an isosceles trapezium and hence concylic. Again by symmetry, the 6 points concyclic.

Hi Antonio, here is my proof using geogebra as dynamic geometry visualization.

ReplyDeletehttp://www.youtube.com/watch?v=EaH7FwwGNwI