## Thứ Hai, 25 tháng 7, 2016

### 1. Introduction

The following property of mixtilinear incircle is quite-known
Theorem 1.$[1]$ Let $(I)$ be incircle of triangle $ABC$. $P$ is an arbitrary point on the circumcircle of triangle $ABC$. Two tangent lines through $P$ of the incircle $(I)$ intersect $BC$ at $X$, $Y$.
Then the circle $PXY$ passes through the common point of $A-$mixtilinear incircle of triangle $ABC$ and the circumcircle of triangle $ABC$.

It was generalized by Vladimir Zajic(nick name yetti on AoPS) as follow
Theorem 2. $[2]$ Let two tangent lines of the incircle of triangle $ABC$ meet at $X$ and intersect $BC$ at $Y$, $Z$. $AX$ intersects the circumcircle of triangle $ABC$ at $K$.
Then the circle $KYZ$ passes through the common point of $A-$mixtilinear incircle of triangle $ABC$ and the circumcircle of triangle $ABC$.

### 2. Generalization and Proof

Recently, Nguyen Duc Bao(nick name baopbc on Vietnam Mathematics Forum) found a generalization of mixtilinear circle as follow
Theorem 3.$[3]$ $P$, $Q$ are isogonal conjugate points with respect to triangle $ABC$.
Tangent line at $P$ of the circle $PBC$ intersects $AB$, $AC$ at $P_{ab}$, $P_{ac}$, respectively.
Tangent line at $Q$ of the circle $QBC$ intersects $AB$, $AC$ at $Q_{ab}$, $Q_{ac}$, respectively.
Then $P_{ab}$, $P_{ac}$, $Q_{ab}$, $Q_{ac}$ lie on a circle, which is tangent to the circumcircle of triangle $ABC$.

Proof of theorem 3 can easily obtained by angle chasing and the fact: the tangency point is Miquel point of the quadrilateral $(PB,PC,QB,QC)$.
Independently, Telv Cohl has also found this generalization with various of interesting properties in $[4]$.
Let $T$ be the tangency point in theorem 3. Then we generalized theorem 2 as follow
Problem 4. Let $\mathcal{S}$ be the inconic of triangle $ABC$ of which foci are $P$, $Q$.
Two tangent lines of $\mathcal{S}$ meet at $X$ and intersect $BC$ at $Y$, $Z$. $AX$ intersects the circumcircle of triangle $ABC$ at $K$.
Prove that the circle $KYZ$ passes through $T$.

Proof. Thanks to the proof of Luis Gonzales for theorem 2 and Telv Cohl's results$[4]$,  I have completed a synthetic solution.
Follow Telv Cohl's proof for property 2$[4]$, if we let $A'$ be a point on the circumcircle of triangle $ABC$ such that $AA'$ is parallel to $BC$ then $T$, $D$, $A'$ are collinear.
Let $\mathcal{S}$ touches $BC$ at $D$ and $AK$ intersects $BC$ at $D'$.
According to Desargues's involution theorem, there exists a point $R$ on $BC$ such that
$RB\cdot RC=RD\cdot RD'=RY\cdot RZ$
Simply by angle chasing, $(TD,TK)=(TA',TK)=(AA',AK)=(BC,AK)=(D'D,D'K)$. Thus $T$, $K$, $D$, $D'$ are concyclic. Combine with $T$, $K$, $B$, $C$ are concylic, $TK$ passes through $R$.
$\Rightarrow RY\cdot RZ=RK\cdot RT$
Hence we conclude that $K$, $Y$, $Z$, $T$ are concyclic.

By the inversion centered at $T$, we obtain that
Corollary 5. $P'$, $Q'$ are reflections of $P$, $Q$ in the circumcircle of triangle $ABC$.
Then $T$, $A$, $P$, $Q'$ are concyclic and $T$, $A$, $Q$, $P'$ are concyclic.