Proposed Problem

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Complete Problem 320

Level: High School, SAT Prep, College geometry

## Saturday, July 11, 2009

### Problem 320: Triangle, Circumcircle, Incenter, Excenter, Collinear points

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ReplyDelete1. Connect BI, EC and AI

Note that A,I,E are collinear.

We have angle(BCE)= ½ angle (A) + ½ angle (B)

And angle (BIF)= ½ angle (A)+1/2 angle (B) so angle (BCE)= angle (BIF) and quad(BICE) is cyclic

2. Since H is the midpoint of arc AG >> HI.HM=HN.HE= HA ^2

And quad(NIME) is cyclic.

3. Note that MN is radical line of circle(ABC) and circle(NIME)

In circle(BICE) , F is the intersection of chords BC and IE >> FI.FE=FB.FC

Since power of F to circle (NIME)=power of F to circle(ABC)

So F must lie on radical axis MN of circles (ABC) and circle (NIME)

Peter Tran