8
$\begingroup$

I am looking a proof for the problem as follows:

Let a convex hexagon, such that its principal diagonals are concurrent. For each side of the hexagon, extend the adjacent sides to their intersection, forming a triangle exterior to the given side. Then show that: Three external (or internal) homothetic centers of three pair circumcircle of opposite triangles are collinear.

Please see the applet in Geogebra

enter image description here

PS: The line through the three external homothetic centers are perpendicular to the line through three internal homothetic centers.

Another applet: Three homothetic centers are collinear associated a circumscribed conic hexagon

Improve this question
$\endgroup$
7
  • $\begingroup$ Out of curiosity, how are you coming up with all these facts? Are you just playing around with GeoGebra (or a similar tool) and noticing that such-and-such relation always holds? $\endgroup$
    – Wojowu
    Jul 8, 2016 at 12:34
  • $\begingroup$ @Wojowu When I construct Dao's theorem on six circumcenters, I found this property true with a cyclic hexagon with principal diagonals are concurrent . Later I generalization for arbitrary hexagon with principal diagonals are concurrent. See Geogebra: $\endgroup$
    – Oai Thanh Đào
    Jul 8, 2016 at 12:53
  • $\begingroup$ While you do specify it ... are you sure this property is solely for a convex hexagon ? and not valid for a general case. $\endgroup$
    – ARi
    Jan 15, 2017 at 18:13
  • $\begingroup$ General case: Alway have three homothetic centers are collinear, and three other homothetic centers are collinear. But exactly three external homothetic center are collinear and three internal homothetic center are collinear we need convex hexagon $\endgroup$
    – Oai Thanh Đào
    Jan 16, 2017 at 7:26
  • $\begingroup$ @ARi maybe the result true for generalization case $\endgroup$
    – Oai Thanh Đào
    Jan 16, 2017 at 7:29

1 Answer 1

Reset to default
2
$\begingroup$

At least for the external homothetic centers, this follows from Monge's theorem, which actually tells you that "for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear".

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Which 3 circles do you apply that to? The three points in the question are shown as arising from 6 circles instead. $\endgroup$
    – user44143
    Jan 14, 2017 at 14:50
  • $\begingroup$ Yes, I think the proof of @ihm is mistake. Monge's theorem apply with three circles. But here is six circles $\endgroup$
    – Oai Thanh Đào
    Jan 15, 2017 at 4:55
  • $\begingroup$ The false proof $\endgroup$
    – Đào Thanh Oai
    Nov 4, 2018 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.