## Tuesday, September 30, 2014

### Geometry Problem 1048: Circles, Tangent, Perpendicular, Diameter, Angle Bisector

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

Click the figure below to view the complete problem 1048.

1. Hi Antonio,

I would like to add the folllowings:
The point B is the center of the inscribed circle of the triangle DFH and the smallest circles passing through B and touching the three sides are Archimedean.

Hiroshi

2. As far as I know, the two Archimedean circles touching the sides DF and DH are new.

Hiroshi

3. Hi, Hiroshi
Please explain why B is the center of the inscribed circle of triangle DFH concretely.

4. http://s25.postimg.org/ag5sakzpb/pro_1048.png

Connect lines per attached sketch
We have GM=GB=GN => MBN is a right triangle
In right triangles AEC and O1GO2 we have GB^2=BO1.BO2 and BE^2=BA.BC
So BG= ½ BE and MBNE is a rectangle
We have ∠ (BNE)= ∠ (BNC)=90 =>E, N, C are collinear
Similarly A,M,E are collinear
Note that ∠ (BCE)= Arc(AF)+Arc(FE)
And ∠ (EMN)= Arc(AF)+Arc(EH)
But ∠ (EMN)= ∠ (BNM)= ∠ (BCE) => Arc( FE)=Arc(EH)=> DE is an angle bisector of angle FDH

5. By theorem from previous problem, AMNC is concyclic. Therefore, AM and NC must meet on radical axis DE of O1 and O2 at point P. <APC must be 90 because <MAB+<NCB=90, so P is also on big circle. P must coincide with E as result.
<ACN=<AMF=<ACF+<EDH, but also