Re: Geometry Problem


OwlHoot wrote:
philippe 92 wrote:
OwlHoot wrote:
Bractals wrote:

Let P and Q be the circumcircle and incircle respectively of an arbitrary triangle.
If ABC is a triangle inscribed in P with AB and AC tangent to Q, then BC is
tangent to Q.

[...]


all six vertices of the two triangles are distinct and in
order round the circumcircle they form vertices of a hexagon,
which suggests that Pascal's Theorem may be relevant: [...]

But two distinct triangles don't make *one* single hexagon !

Not sure what you mean here. It's true that the union of the
interiors of two triangles don't make the interior of a hexagon,
even if they're coplanar.


nothing to do with interiors.

But for triangles ABC and DEF with no shared vertices,
ADBECF can be viewed as a hexagon, albeit non-planar
(if the triangles are not coplanar, which isn't the case in
this problem) and/or with self-intersecting sides.


The hexagon involved in Pascal (and Brianchon's) theorems are not made of just *vertices*.
An hexagon is made of {A-B-C-D-E-F(-A)} *one* cycle (in any order)
not of {A-B-C(-A)} + {D-E-F(-D)} *two* cycles

That is the *sides* of the set of two triangles are not *sides* of an hexagon.

However Brianchon's is related with Gergonne point.
That is *one* triangle ABC, with contact points TUV to incircle could be considered as an hexagon AUCTBVA
UC is tangent (in U) to incircle,
CT in T, TB in T, BV in V, VA in V and AU in U
hence from Brianchon :
AT (T considered as vertex opposite to A), BU and CV concur :
Gergonne point. Of course elementary proof of that with just Ceva is simpler, here just to force use of hexagons.

But as I said, Gergonne point is not directly related with incenter and circumcenter, contrary to what seems to be a visual proof in the animated Gifs of Mathworld (if their assumption of a fixed Gergonne point were true, the porism property would result "at once").

Regards.
--
Philippe C.
chephip+news@xxxxxxx

.

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