321. Harmonical quadrilateral.

by Virgil Nicula, Oct 6, 2011, 9:14 AM

Definition. A cyclical convex quadrilateral $ABCD$ is harmonically iff $AB\cdot CD=AD\cdot BC$ .

Properties. Denote $XX$ - the tangent to the circumcircle $w$ of $ABCD$ in the point $X\in w$ . If $T\in BB\cap DD$ , then

$ABCD$ is harmonically $\iff$ $T\in AC$ $\iff$ for any point $M\in w$ the pencil $M(A,B,C,D)$ is harmonicaly.


PP1. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ and the $A$-exincircle $w_a$ . Denote $\left\{\begin{array}{ccc}
D\in BC\cap w & ; & D^{\prime}\in BC\cap w_a\\\\
E\in CA\cap w & ; & F\in AB\cap w\end{array}\right|$ .

Consider the points $P\in DE$ and $Q\in DF$ so that $I\in PQ$ and $\overline {PIQ}\perp AD^{\prime}$ . Prove that $IP=IQ$ .


Proof. Denote $\{X,Y\}=AD'\cap w$ , where $PQ$ separates $A$ and $Y$ . Since $EXFY$ is harmonically, for the point $D\in w$ the pencil $D(F,X,E,Y)$

is harmonically. Therefore, the intersection $D(F,X,E,Y)\cap \overline {EIF}=\{Q,I,P,\infty\}$ is a harmonical division, i.e. $I$ is the midpoint of $PQ$ .


PP2. Let $ABC$ be a triangle with the circumcircle $w=C(O)$ . Denote the diameter $[AN]$ of $w$ , the intersection $M\in BB\cap CC$

and the points $P\in AC$ , $Q\in AB$ so that $O\in PQ$ and $\overline {POQ}\perp MN$ . Prove that $OP=OQ$ and $PQ\cap BC\cap NN\ne\emptyset$ .


Proof. Denote $\{N,D\}=MN\cap w$ . Since $BDCN$ is harmonically obtain that the pencil $A(B,D,C,N)$

is harmonically $\iff$ the division $\{Q,\infty , P,O\}=A(B,D,C,N)\cap\overline {POQ}$ is harmonically $\iff$ $OP=OQ$ .



PP3 Let $\triangle ABC$ with the incircle $w=(I)$ . Denote $\left\{\begin{array}{cc}
D\in BC\ ; & \widehat{DAB}\equiv\widehat {DAC}\\\\
E\in CA\ ; & \widehat{EBC}\equiv\widehat {EBA}\\\\
F\in AB\ ; & \widehat{FCA}\equiv\widehat {FCB}\end{array}\right|$ and $\left\{\begin{array}{cc}
X\in EF\ ; & IX\perp EF\\\\
Y\in FD\ ; & IY\perp FD\\\\
Z\in DE\ ; & IZ\perp DE\end{array}\right|$ . Prove that $AX\cap BY\cap CZ\ne\emptyset$ .

Proof Denote $P\in BC\cap w$ and $Q \in EF\cap BC$ , $R\in PI\cap AQ$ and $S\in CI\cap AQ$ . Therefore, the cross ratio $(C,F;I,S)$ is harmonically $\Longrightarrow$ the pencil

$Q(C,F;I,S)$ is harmonically $\Longrightarrow$ the cross ratio $(P,X;I,R)$ is harmonically. Since $AR\perp AI$ obtain that $\widehat{IAX}\equiv\widehat{IAP}$ , i.e. $AX$ and $AP$ are isogonals w.r.t.

$\angle BAC$ $\Longrightarrow$ $AX$ passes through the isogonal conjugate $L$ of the Gergonne's point of $\triangle ABC$ . Similarly, $L\in BY$ and $L\in CZ$ , i.e. $AX\cap BY\cap CZ\ne\emptyset$ .
This post has been edited 19 times. Last edited by Virgil Nicula, Nov 19, 2015, 9:18 PM

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