Proposed Problem

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Complete Problem 323

Level: High School, SAT Prep, College geometry

## Tuesday, July 14, 2009

### Problem 323: Triangle, Incenter, Circumcenter, Equal circles, Collinearity

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Let circles with centres at E, D & F meet BA, AC & CB in P,Q,R,S,T & U resply. Hence PE=QD=RD=EF= TF UE which, in effect, means that ED is parallel to BA, DF to AC & FE to CB. and we can say that Tr, EDF is similar to Tr BAC and that their corresponding sides are parallel to each other.

ReplyDeleteFurther, GD=GE=GF = radius of the congruent circles which means that G is the circumcentre of Tr. EDF. Moreover, A, D & I are collinear. Likewise C, F & I and B, E & I are collinear. Thus, I is the incentre of Tr. EDF as well.

For any triangle the line joining the incentre and the circumcentre is unique. OI and GI are thus lines joining the circumcentre and the incentre for the two triangles and since the two triangles are similar with the sides parallel, we can say that OI and GI are, in fact, the same line or that O, G and I are collinear.

As a matter of fact the orthocentres of the two triangles would also belong to the same line.

Would that be correct, Antonio?

Ajit: ajitathle@gmail.com

Excellent! I agree with your solution.

ReplyDeletehttp://s25.postimg.org/mkbjl9uf3/Pro_323.png

ReplyDeleteDraw lines per sketch

Note that BE, AD and CF will concurred at I

Triangles DEF and ABC have sides parallel and I is incenter of both triangles

G will be circumcenter of triangle DEF

ABC will be an image of DEF in the homothetic transformation center I and ratio= IA/ID=IB/IE=IC/IF

Point O will be the image of point G in this transformation so I, G and O are collinear