ELMO 2014 SL G6

by Wolstenholme, Aug 1, 2014, 9:47 PM

Let $ABCD$ be a cyclic quadrilateral with center $O$ .
Suppose the circumcircles of triangles $AOB$ and $COD$ meet again at $G$ , while the circumcircles of triangles $AOD$ and $BOC$ meet again at $H$ .
Let $\omega_1$ denote the circle passing through $G$ as well as the feet of the perpendiculars from $G$ to $AB$ and $CD$ .
Define $\omega_2$ analogously as the circle passing through $H$ and the feet of the perpendiculars from $H$ to $BC$ and $DA$ .
Show that the midpoint of $GH$ lies on the radical axis of $\omega_1$ and $\omega_2$ .

Proof:

First, let $E = AB \cap CD$ and $F = BC \cap DA$ and denote the circumcircle of $ABCD$ by $\omega$ . Let $O_1, O_2$ be the centers of $\omega_1, \omega_2$ respectively. Let $M$ be the midpoint of $GH$ .

Consider the inversion about $\omega$ . It is clear that line $AB$ goes to $\omega_1$ and that line $CD$ goes to $\omega_2$ so $E$ goes to $G$ . Similarly $F$ goes to $H$ . Moreover, $\omega_1$ and $\omega_2$ are the circles with diameters $EG$ and $FH$ respectively. Now since $M$ is the midpoint of $GH$ and since $O_1$ is the midpoint of $GE$ we have that $O_{1}M \parallel HE$ and similarly $O_{2}M \parallel GF$ .

But since $GF$ is the polar of $E$ with respect to $\omega$ we have that $GF \perp OE$ so $O_{2}M \perp OE$ which implies $O_{2}M \perp OO_{1}$ . Similarly $O_{1}M \perp OO_{2}$ and so $M$ is the orthocenter of triangle $OO_{1}O_{2}$ . This means that $OM \perp O_{1}O_{2}$ so to show that $M$ is on the radical axis of $\omega_1$ and $\omega_2$ it suffices to show that $O$ is on this radical axis.

However, this is clear since both $\omega_1$ and $\omega_2$ are orthogonal to $\omega$ , so we are done.

Interestingly after doing the inversion step, a complex number solution is somewhat doable as well.

Also, apparently we can replace $O$ with any point on line $OQ$ where $Q = AC \cap BD$ but I have not worked on proving this.

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
this blog is so hot
by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
by User360702, Jan 10, 2019, 6:03 PM
helloooooo
by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
You were just featured on AoPS's facebook page.
by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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ELMO 2014 SL G6

by Wolstenholme, Aug 1, 2014, 9:47 PM

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