collinear excircle

Carlez Tevos · Poincare Conjecture Offline Joined: 13 May 2008 Posts: 116

In triangle $ABC$ , let $I,I_a$ be the incenter and the excenter with respect to side $BC$ . Let $A',M$ be the intersections of $II_a$ with $BC$ and the circumcircle of triangle $ABC$ respectively. Let $N$ be the midpoint of arc $MBA$ , and $S,T$ the intersection points of rays $NI,NI_a$ with the circumcircle of triangle $ABC$ . Prove that $S,T,A'$ are collinear.

**Posted:** Sun May 18, 2008 4:34 am

mr.danh

**Posted:** Mon May 19, 2008 11:21 pm

Erken

Solution 1:
-Let $\omega$ be a circumcircle of triangle $\triangle ABC$ .Since we will consider only circle $\omega$ ,I won't write "with respect to $\omega$ " everytime I use some pole and polar transformations.So let's start:

-Let $X$ be a pole of $ST$ .So it is enough to show that $X$ lies on a polar of $A'$ .

-Suppose that $NA'$ intersects $\omega$ at $R$ .It is well-known fact that division $(AA',II_A)$ is harmonic.Therefore,quadrilateral $ASRT$ is harmonic and it also follows that $AR$ passes through $X$ .

-Since $N$ is the midpoint of arc $MBA$ ,it follows that tangent line to $\omega$ and $N$ is parallel to $AA'$ .Combining it with fact that
$(II_A,M\infty) = - 1$ ,we conclude that quadrilateral $NSMT$ is harmonic,as well.Therefore,points $N,M,X$ are collinear.

-Since quadrilateral $NARM$ is cyclic and point $A'$ its diagonals intersection point and $AR$ intersects $NM$ at $X$ ,we finally conclude that $X$ lies on the polar of $A'$ .
$\blacksquare$ .

P.S:
"Solution 1" is actually my proof,which I found out during the exam.There exist a more elegant solution with using just some angle chasing and some other simple facts.
Best regards.

_________________
A mathematician is a scientist who can figure out anything except such simple things as squaring the circle and trisecting an angle

Erken

Solution 2:
This is the most beautiful and elegant proof to this problem and I would be very surprised if one would find a shorter one.

So let's start:

It is obviously that $\widehat{TI_AI}$ is the half of arc $NT$ .And $\widehat{IST}$ is equal to the half of arc $NT$ as well,therefore,quadrilateral $SITI_A$ is cyclic.

Now it is enough to use the well-known fact,which tell us that radical axes of three circles are concurrent,to three circumcircles of
$\triangle ABC,\triangle SII_A$ and $\triangle IBC$ .

$\blacksquare$ .

Awesome proof,isn't it? Rolling Eyes

**Posted:** Sat Jul 26, 2008 9:38 am

_________________
A mathematician is a scientist who can figure out anything except such simple things as squaring the circle and trisecting an angle

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 640

Dear Erken and Matlinkers,
to go in the shorter way of Erken without using angles, denote Tn the tangent to the circumcircle at N, applied a version of the Reim's theorem : S, I T Ia are coyclic. (http://perso.orange.fr/jl.ayme ,see A propos)
Then you have an enterely synthetic proof.
Sincerely
Jean-Louis

**Posted:** Sun Jul 27, 2008 3:13 am

Erken

To jayme:
Brilliant comment,thank you.
But in my opinion,using any theorem in a such easy stuff makes a solution to lose its beauty,furthermore,angle-chasing here is just a half of an one line.
Best regards.

**Posted:** Sun Jul 27, 2008 7:10 am

_________________
A mathematician is a scientist who can figure out anything except such simple things as squaring the circle and trisecting an angle

Virgil Nicula

**Posted:** Sun Jul 27, 2008 6:33 pm