Thursday, December 30, 2010

Problem 561: Triangle, Cevian, Incenter, Perpendicular, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 561.

Problem 561: Triangle, Cevian, Incenter, Perpendicular, 90 Degrees.
Zoom
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 6:40 AM
Labels: , , , , , , , ,

3 comments:

  1. PravinJanuary 1, 2011 at 5:25 AM

    Call the radii of circles (E) and (F) as r1 and r2
    Divide the segment EF in the ratio r1:r2 at K.
    Through K draw the other (transverse)common tangent LN.
    GE, GF bisect the supplementary angles MGN and NGH
    Hence angle EGF is a right angle

    ReplyDelete
  2. Peter TranJanuary 1, 2011 at 7:09 AM

    http://img84.imageshack.us/img84/9576/problem561.png

    Let I is the midpoint of EF and M is the projection of I over AC
    We have angle(EDF)=90 ( see problem 559) and HD=GK (See problem 560)
    M is the midpoint of HK . Since HD=GK so M is also the midpoint of DG and triangle DIG is isosceles.
    And IE=IF=ID=IG .
    Quadrilateral EDGF is cyclic with center of circumcircle at I and angle( EGF)= 90

    Peter Tran

    ReplyDelete
  3. Peter TranJanuary 1, 2011 at 5:11 PM

    To Pravin

    Your solution assume that L,N and G are collinear.
    It is not clear to me .Please explain .

    Peter

    ReplyDelete

Subscribe to: Post Comments (Atom)