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Problem related to Tangential Quadrilateral

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#1 h Nov 21, 2014, 8:44 AM • 8 Y

Y by langkhach11112, baopbc, GoJensenOrGoHome, toto1234567890, jty123, Adventure10, Mango247, and 1 other user

Let $ABCD$ be a tangential quadrilateral with incircle $\odot (I)$ .
Let $A', B', D'$ be the intersection of $\odot (I)$ with segment $AI, BI, DI$ , respectively .
Let $C'$ be the intersection of $\odot (I)$ with ray $CI$ and $X=AC \cap A'C' , Y=BD \cap B'D'$ .

Prove that $XY$ is tangent to $\odot (I)$

This post has been edited 1 time. Last edited by TelvCohl, Aug 20, 2015, 7:08 PM

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#2 h Jun 4, 2016, 3:44 AM • 1 Y

Y by Adventure10

Sorry for dig this topic out after 2 years but this problem is really nice. Can someone give a solution to it?

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SalaF

#3 h Jun 4, 2016, 3:42 PM • 8 Y

Y by BuiBaAnh, langkhach11112, GoJensenOrGoHome, TelvCohl, baopbc, jty123, Adventure10, Mango247

If you are interested I can sketch a proof (writing down every detail would make it too long).

Let $P,Q,R,S$ be the contact points of $\odot(I)$ with $AB,BC,CD,DA$ . Then $PQ\cap RS=M$ is the pole of $BD$ wrt $\odot(I)$ and $QR\cap SP=N$ is the pole of $AC$ wrt $\odot(I)$ . Let the tangent lines to $\odot(I)$ from $A',C'$ meet at $U$ and the tangent lines to $\odot(I)$ from $B',D'$ meet at $V$ . Then the problem is projectively equivalent to the following: $MV$ and $NU$ meet on $\odot(I)$ .

Now we can reformulate the exercize with respect to the cyclic quadrilateral $PQRS$ ; for clarity's sake I'll rename the points:

Let $ABCD$ be a quadrilateral inscribed into a circle $\omega$ ; let $L,M,N$ be the midpoints of the minor arcs $\widehat{AB},\widehat{BC},\widehat{AD}$ and let $K$ be the midpoint of the major arc $\widehat{CD}$ (the one containing the other two points). Let the tangents to $\omega$ to $K,L$ meet at $I$ and let the tangents to $\omega$ from $M,N$ meet at $J$ . Let $AB\cap CD=E, BC\cap AD=F, AC\cap BD=G$ . Then $EI,FJ$ meet on $\omega$ .

Proof: Let $\gamma$ be the circle tangent to the segments $AD,BC$ and to the minor arc $\widehat{AB}$ at a point $U$ : we will show that $U=EI\cap FJ$ . Let $\Gamma$ be the circle externally tangent to the minor arc $\widehat{AB}$ at a point $W$ and tangent to $AD,BC$ . I will not prove the following well known

Lemma 1: The line through the tangency points of $\gamma$ with $AD,BC$ passes through the incenters of $\triangle ABD,\triangle ABC$ ; similarly the line through the tangency points of $\Gamma$ with $BC,AD$ passes through the $D,C-$ excenters of the same triangles.
Lemma 2: There is a circle $\Gamma_1$ externally tangent to the minor arc $\widehat{AB}$ at $W$ and to the lines $AC,BD$ .

Clearly the homothety centered at $U$ which sends $\gamma$ to $\omega$ sends $AD$ to the line through $N$ which is tangent to $\omega$ ; following this reasoning we easily see that $F$ is mapped to $J$ , so $U\in FJ$ .
Now let $\gamma_1$ be a circle tangent to the segments $AD,BC$ and internally tangent to the minor arc $\widehat{CD}$ at $V$ ; by Monge's theorem we get that $F,U,V$ are collinear. Moreover by an analogous version of Lemma 2 we know that there is a circle tangent to $AC,BD$ and internally tangent to $\omega$ at $V$ , so again by Monge's theorem (wrt the latter circle and $\Gamma_1,\omega$ ) we also have that $V,G,W$ are collinear. By polar arguments this immediately implies that $E,U,W$ are collinear.
What we are left to show is that $U,I,W$ are collinear, or equivalently that the quadrilateral $ULWK$ is harmonic. So it is sufficient to prove that the tangent lines to $\omega$ through $U,W$ meet on $KL$ . But $KL$ is parallel to the line through the centers of $\gamma,\Gamma$ (easy angle-chasing) and $L$ belongs to the radical axis of $\gamma,\Gamma$ (for $L$ has the same distance to the incenter and excenter of $ABD,ACD$ and by Lemma 1 this implies that $L$ belongs to the bisector of the parallel lines through the tangency points, which is in fact the radical axis), so $KL$ is itself the radical axis. Then the concurrency point that we were seeking is the radical center of $\gamma,\Gamma,\omega$ . This ends the proof.

This post has been edited 1 time. Last edited by SalaF, Jun 4, 2016, 3:53 PM

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#4 h Jun 6, 2016, 10:27 AM • 5 Y

Y by baopbc, BuiBaAnh, langkhach11112, Adventure10, Mango247

This problem is true for any point on line $CI$ as following

Let $ABCD$ be a tangential quadrilateral with incircle $\odot (I)$ . $P$ is an any point on line $IC$ . Let $A', B', D'$ be the intersection of $\odot (I)$ with segment $PA, PB, PD$ , respectively. Let $C'$ be the intersection of $\odot (I)$ with ray $CI$ and $X=AC \cap A'C' , Y=BD \cap B'D'$ . Prove that $XY$ is tangent to $\odot (I)$ .

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