## Wednesday, July 17, 2013

### Dynamic Geometry Problem 904: Second Fermat Point. Triangle, Equilateral triangles, Concurrent lines

Geometry Problem 904. GeoGebra, HTML5 Animation for iPad and Nexus.

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

1. Let CC2 meet BB2 at F2 and F2AA2 unknown interval.
A2AC=b
ACA2=B2CB=a, A2C=BC, B2C=AC => B2CB~=ACA2 => B2BC=AA2C=180-a-b
AB=BC2, A2B=BC, C2BC=ABA2=> C2BC~=ABA2 =>BA2A=a+b-120 =>BF2C=180-F2BC-F2CB=120=>A2+F2=180=>A2CF2B cyclic=> F2AA2 collinear [if we build Imaginary interval F2A2 F2A2C=AA2C=F2BC=180-a-b, and A must be on A2F2]

2. http://img607.imageshack.us/img607/9152/y2lu.png
Let A2A cut CC2 at F2 . We will prove that B, B2 and F2 are collinear
1. Note that ∆(BAA2) congruence to ∆(BC2C) ….( case SAS)
In the rotation center B, rotational angle=60, ∆(BAA2) become ∆ (BC2C) so ∠ (A2F2C)=60
So A, C, F2 and B2 are concyclic and ∠ (AF2B2)= 60 … ( angle face 120 arc)
2. Similarly ∆ (CBB2) congruence to ∆ (CA2A) in the rotation center C , angle of rotation =60
So angle form by AF2 and BB2 is 60.
Since both BB2 and B2F2 form the same angle to AF2 so B, B2 and F2 are collinear.