OI passes through isogonal of I wrt triangle A'B'C', angle bisectors

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OI passes through isogonal of I wrt triangle A'B'C', angle bisectors

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#1 h Oct 29, 2021, 6:31 PM • 5 Y

Y by HWenslawski, MathLuis, Siddharth03, GeoKing, ei0natural

Let $ABC$ be a triangle with circumscribed and inscribed circles $(O)$ and $(I)$ respectively. $AA'$ , $BB'$ , $CC'$ are the bisectors of triangle $ABC$ . Prove that $OI$ passes through the the isogonal conjugate of point $I$ with respect to triangle $A'B'C'$ .

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#3 h Nov 1, 2021, 2:42 PM • 1 Y

Y by GeoKing

Do anybody have a solution for this tough problem ?

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#5 h Nov 7, 2021, 4:27 AM • 4 Y

Y by parmenides51, PRMOisTheHardestExam, ThisNameIsNotAvailable, GeoKing

Nice problem! Here's my solution :

Diagram

Let isogonal conjugate of point $I$ with respect to triangle $\triangle A'B'C'$ be $X$ .
Let the intouch points of $\triangle ABC$ be $D,E,F$ .
It's well-known (by inversion in $(I)$ ) that $OI$ is the Euler line of $\triangle DEF$ . $\quad\dots(1)$

We shall show a (sort of) similar argument to prove that $IX$ is also the Euler line of $\triangle DEF$ .

Let the projections of $I$ on the sides of $\triangle A'B'C'$ be $P,Q,R$ .
Note that as $I,X$ are isogonal conjugates w.r.t. $\triangle A'B'C'$ , we have that circumcenter of $PQR$ is the midpoint of $IX$ , say $M$

Now, with respect to $(I)$ we shall use the usual pole-polar notation i.e. capital letters for pole and small for its polar.
Let the sides of $\triangle A'B'C'$ be $u,v,w$ .
Observe that $U,V,W$ are the inverse points of $P,Q,R$ w.r.t. $(I)$ .
$\therefore \odot UVW,\odot PQR$ are mutually inverse w.r.t. $(I)$ .
$\therefore$ The circumcenter say $K$ of $\triangle UVW$ , the circumcenter $M$ of $\triangle PQR$ and $I$ are collinear.
So, we have that $K,I,X$ are collinear. $\quad\dots(2)$

Now, observe that the sides of $\triangle UVW$ are $a',b',c'$ .
By La Hire, $A'\in d\implies D\in a'$ .
$\therefore$ As $A'$ lies on the $A$ -angle bisector of $\triangle ABC$ , we have that $a'$ is the line passing through $D$ parallel to $EF$ and similar results.
$\therefore \triangle UVW$ , the triangle formed by the lines $a',b',c'$ is the antimedial triangle of $\triangle DEF$ .
$\therefore K$ , the circumcenter of $\triangle UVW$ is the orthocenter of $\triangle DEF$ .
$\therefore$ By $(2)$ , $IX$ is the Euler line of $\triangle DEF$ .

$\therefore$ By $(1)$ , $I,O,X$ are collinear and lie on the Euler line of $\triangle DEF$ .

@below, Motivation

This post has been edited 5 times. Last edited by Siddharth03, Nov 9, 2021, 3:27 PM
Reason: Motivation

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#6 h Nov 9, 2021, 11:43 AM • 1 Y

Y by GeoKing

@above Very nice solution!!! Why do you think of triangle UVW or in other words what is your idea when you solve this problem? Thanks!

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everythingpi3141592

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everythingpi3141592

#7 h Mar 10, 2024, 4:32 PM • 2 Y

Y by GeoKing, parmenides51

Solved with MathsCrazy and a few other for a while (he suggested the Pedal triangle idea) and also with help of Siddharth03 (he told me that $OI$ is the Euler line of the intouch triangle)

Note that $OI$ is the Euler line of the intouch triangle, say $\triangle DEF$ . Then, note that midpoint of $OI'$ where $I'$ is the isogonal conjugate of $I'$ in $\triangle A'B'C'$ , is the centre of the Pedal triangle of $I$ in $A'B'C'$ . We now invert about $I$ . Then, $A$ , $B$ , $C$ become midpoints of $EF$ , $FD$ , $DE$ respectively. Now, $I$ being circumcentre of $DEF$ becomes orthocentre of $ABC$ . Now, $I$ , $A$ and $A'$ are collinear and so are $B$ , $A'$ , $C$ so, $I$ , $A$ , $A'$ are still colinear and $A'$ lies on $(IBC)$ . infact, $(IBC)$ also passes through $D$ . So infact, $A'$ , $B'$ and $C'$ are reflections about $BC$ , $CA$ and $AB$ of $A$ , $B$ , and $C$ respectively. Now, feet from $I$ onto $A'B'$ becomes diametrically opposite point to $I$ in $(IA'B')$ . Further, the line connecting $I$ to the centre of the pedal triangle is same as that connecting $I$ to centre of the inverted pedal triangle. Take a homotethy of factor $\frac{1}{2}$ at $I$ , thus we get the centre of triangle formed by circumcentres of $(IA'B')$ , $(IB'C')$ and $(IC'A')$ . These are formed by the perpendicular bisectors of $IA'$ , $IB'$ and $IC'$ . Since $BC$ , $CA$ and $AB$ are perpendicular bisectors of $AA'$ , $BB'$ , $CC'$ , we have that the distances of the perpendicular bisectors of $IA'$ , $IB'$ and $IC'$ from $BC$ , $CA$ and $AB$ are $\frac{IA}{2}$ , $\frac{IB}{2}$ , $\frac{IC}{2}$ , which are also the distances of the circumcentre of $ABC$ from its sides. Thus the circumcentre of $(ABC)$ is the homotethy centre of $(ABC)$ and the inverted homotethic Pedal triangle, thus is centre of the homotethic Pedal triangle, and we are done.

This post has been edited 2 times. Last edited by everythingpi3141592, Mar 10, 2024, 4:38 PM

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#8 h Apr 1, 2024, 1:05 AM • 1 Y

Y by parmenides51

Here’s an overkill solution:
Consider the feurbach hyperbola, We’ll prove that $II’$ is tangent to the hyperbola which clearly finishes, it is well known that the excenters and incenters of cevial triangle of $I$ lie on feurbach hyperbola, now we finish by the following lemma:
Lemma: If $P$ is a point in $ABC$ and $P’$ is the isogonal conjugate of $P$ in orthic triangle then $PP’$ is tangent to the hyperbola passing through $A,B,C,P$

This post has been edited 2 times. Last edited by math_comb01, Apr 1, 2024, 8:18 AM

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