In cyclic quadrilateral ABCD

n.die · New Member Offline Joined: 23 Jan 2008 Posts: 4

There is cyclic quadrilateral ABCD(center O)

And intersection of AC,BD is E.

There is a point P is ABCD and X,Y,Z,W is circumcenter of ABP,BCP,CDP,DAP

Then,

XZ, YW, OE are concurrent!!

**Posted:** Thu Feb 07, 2008 5:42 am

Erken

This is the Generalization of the China TST 2006 geometry problem,though it is not hard,maybe even easy.But anyway this is nice application of the pole-polar theory.
Proof:
Let $\omega(O),\omega_1(O_1),\omega_2(O_2),\omega_3(O_3),\omega_4(O_4)$ be the circumcircles of the $ABCD,\triangle APB,\triangle BPC,\triangle CPD,\triangle DPA$ ,respectively.Denote $\omega_1\cap\omega_3 =P,N$ and $\omega_2\cap \omega_4 =P,M$ .Then
The point of intersection of $O_1O_3$ and $O_2O_4$ is the circumcenter of $\triangle PNM$ .
Consider circumcircle of $\triangle PNM$ ,denote it as $\zeta(I)$ .Now we must prove that $I,O,E$ are collinear.
Denote $AD\cap BC= F$ , $AB\cap CD=G$ .Then it is well known that $OE\perp FG$ ( $FG$ is the polar of $E$ w.r.t $\omega$ ).Hence
it is sufficiently to prove that $OI\perp EG$ .
It is well known that radical axes of the 3 circles passes through one point,hence
$PM,AD,BC$ are concurrent.
$PN,AB,CD$ are concurrent.
Since $PM$ is the radical axe of $\zeta$ and $\omega_2$ ,and $AD$ is radical axe of $\omega$ and $\omega_2$ ,thus radical axe
of $\zeta$ and $\omega$ passes through $F=PM\cap AD$ .Similarly the radical axe of $\zeta$ and $\omega$ passes through $G$ ,hence
$FG$ is radical axe of $\zeta$ and $\omega$ ,and we are done.

**Posted:** Thu Feb 07, 2008 6:16 am

n.die · New Member Offline Joined: 23 Jan 2008 Posts: 4

**Posted:** Thu Feb 07, 2008 8:54 am

Erken

You need to know some properties of harmonic fours and polar theory.If you know them,see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168866

**Posted:** Thu Feb 07, 2008 5:45 pm

Erken

See my blog for the proof of the fact above Smile

**Posted:** Sat Mar 01, 2008 5:29 am

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

PDatK40SP · Poincare Conjecture Offline Joined: 09 Oct 2006 Posts: 124

Very nice solution, Erken Smile

I have an another solution base on polar theory, but it is worse than yours Smile

Let $S \in XZ \cap YW$
Denote $u$ is the polar of the point $U$ wrt $(O)$ , and $(mn)$ is the pole of the line passes through $M,N$ wrt $(O)$
Let $A_1 \in WX \cap AA$ , and $B_1, C_1, D_1$ are defined similarly.
We have $A_1,B_1, C_1, D_1$ are collinear, because all of them lie on the radical axis of the circle $(O)$ and the point-circle $P$
Then $A(wx) = a_1, B(xy) = b_1, C(yz) = c_1, D(zw) = d_1$ are concurrent at a point, called $Q$
From this, and not that $(wx)(xy) = x \parallel AB, (xy)(yz) = y \parralel BC, (yz)(zw) = z \parallel CD, (zw)(wx) = w \parallel DA$ , we have there exists a homothety $f$ , centered at $Q$ such that: $f(A) = (wx), f(B) = (xy) , f(C) = (yz), f(D) = zw$
Now let $AB \cap CD = F, AD \cap BC = G$ , we have $f(F) = x \cap z = (xz), f(G) = (yw)$
So $FG \parallel (xz)(yw) = s$ , then $E$ - the pole of $FG$ , $S$ - the pole of $s$ and $O$ are collinear.

**Posted:** Tue Mar 11, 2008 8:20 pm