- Dear friends!

Is the following property of Gergonne's point well-known?

ABC is given triangle.

A1 is the point on ray BA, so that AA1=BC, and A2 is the point on ray CA, so

that AA2=BC. In the same way we can define B1,B2, so that BB1=BB2=AC and

C1C2, so that CC1=CC2=BA.

Triangle, formed by (A1A2),(B1B2),(C1C2) is perespective with ABC triangle

and Gergonne point is the center.

Also that triangle is homothetic to Gergonne triangle of ABC (with the

ratio...? - I don't know)

Can somebody give easy synthethical proof of it?

Best regards,

Yours sincerely,

Alex - Dear Alex,

My first remark is that A1, A2, B1, B2, C1, C2 lie on one circle,

called the Conway circle of triangle ABC. The center of this circle

is the incenter of ABC.

See http://mathforum.org/epigone/geometry-puzzles/wilyangdimp/

What concerns the question why the perspector is the Gergonne point -

this is a special case of theorems on cevian hexagons, and I can't

give an elementary argument right off.

Sincerely, Darij Grinberg

>> ABC is given triangle.

>> A1 is the point on ray BA, so that AA1=BC,

>> and A2 is the point on ray CA, so

>> that AA2=BC. In the same way we can define

>> B1,B2, so that BB1=BB2=AC and C1C2, so

>> that CC1=CC2=BA.

>> Triangle, formed by (A1A2),(B1B2),(C1C2)

>> is perespective with ABC triangle and

>> Gergonne point is the center.

>> Also that triangle is homothetic to

>> Gergonne triangle of ABC (with the

>> ratio...? - I don't know) - Dear Alex,

Let A1 and C1 be on AC, B' be the vertix of the intouch triangle of ABC on

AC. Then B' is the midpoint of A1C1. Let triangle, formed by

(A1A2),(B1B2),(C1C2), be A"B"C". Then A1C1 is antiparallel to A"C" wrt

angle A"B"C". Hence B"B' is the symmedian of A"B"C". Since BB' is the

symmedian of A'B'C' (intouch), then B, B' and B" are collinear. The

gomothetic ratio of A"B"C" and A'B'C' is

((a+b+c)*(ab+bc+ca)-a^3-b^3-c^3-abc)/(2*(a+b+c)*r).

Best regards,

Lev

>> ABC is given triangle.

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>> A1 is the point on ray BA, so that AA1=BC,

>> and A2 is the point on ray CA, so

>> that AA2=BC. In the same way we can define

>> B1,B2, so that BB1=BB2=AC and C1C2, so

>> that CC1=CC2=BA.

>> Triangle, formed by (A1A2),(B1B2),(C1C2)

>> is perespective with ABC triangle and

>> Gergonne point is the center.

>> Also that triangle is homothetic to

>> Gergonne triangle of ABC (with the

>> ratio...? - I don't know)