Angle chasing with ellipses [Part 2]

by p_square, Jun 13, 2021, 8:51 AM

As promised, this post will mainly feature in-ellipses, and isogonal conjugates in quadrilaterals.

This post is going to be slightly hi-tech, pre-requisites include miquel points, inversion, and my previous post.

Before we dive right in, let's prove a quick lemma about the configuration we had before.
Let the tangents at $A$ and $B$ to an ellipse intersect at $P$ . If $O$ is the centre of the ellipse, then $[APO] = [BPO]$ . (Over here, [] denotes the area)

When one looks at the figure, the point $O$ really stands out. We know absolutely nothing about this point so far. Thus, it is only natural to try and convert these areas into the areas involving $F_1$ and $F_2$ instead of $O$ . Then, remembering how nice reflections of focii across tangents were, we add in $G$ which is $F_1$ reflected across $PA$ , and $H$ which is $F_2$ reflected across $PB$ . These points just make the whole proof come together:
$2[APO] = [APF_1] + [APF_2] = [APG] + [APF_2] = [GPF_2]$ Similarly,
$2[BPO] = [BPF_1] + [BPF_2] = [BPF_1] + [BPH] = [F_1PH]$
$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 1.4, b=0.3; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); pair G = 2*foot(F2, P, A) - F2, H = 2*foot(F1, P, B) - F1; pair O = (F1 + F2)/2; draw(F1--P--G--F1, royalblue); draw(F2--P--H--F2, purple); draw(P--O--A--P--B--O, brown); dot("$O$",O,S); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$H$", H, dir(H)); dot("$G$", G, dir(G)); [/asy]$

All that is left to show, is that $[APF_2] = [BPF_1]$ , but that's clear, the two triangles are simply congruent.

Consider any quadrilateral $ABCD$ with an in-ellipse tangent to sides $AB, BC, CD, DA$ at $X, Y, Z, W$ respectively.
Let $F_1, F_2$ and $O$ , be the focii and centre of the ellipse respectively. The first interesting result is that $\angle AF_1B + \angle CF_1D = \pi$ and the second is that $O$ lies on the gauss line of quadrilateral $ABCD$ .

Slightly restated, we want to show that $\angle AF_1B + \angle CF_1D = \angle AF_1D + \angle BF_1C$ , and that $[AOB] + [COD] = [AOD] + [BOC]$ .

The proofs of both of these results are pretty similar. The idea in a nutshell is to split and rearrange. [Another eerily similar idea is used to prove that the sums of opposite sides of a quadrilateral which has an incircle are equal]

$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.15, min = 1, w = 3.0, x=1.9, y = 1.3, z = 0.3; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair W = point(el,w), X = point(el,x), Y = point(el, y), Z = point(el, z); pair A = extension(X, X+dir(el,x), W+dir(el,w),W); pair B = extension(dir(el,x)+X,X,Y,Y+dir(el,y)); pair C = extension(Z, Z+dir(el,z), Y+dir(el,y), Y); pair D = extension(dir(el,z) + Z, Z, W, W+dir(el, w)); draw(A--B--C--D--A, royalblue); draw(A--F2--B, purple); draw(C--F2--D, purple); draw(W--F2--Y, brown); draw(X--F2--Z, brown); draw(anglemark(B,F2,A), orange); draw(anglemark(D,F2,C), orange); draw(anglemark(C,F2,B)); draw(anglemark(A,F2,D)); dot("$W$", W, dir(W)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$F_1$", F2, S); [/asy]$

For the first theorem:
$\begin{align*} \angle AF_1B + \angle CF_1D &= \angle AF_1X + \angle XF_1B + \angle CF_1Z + \angle ZF_1D \\ &= \angle AF_1W + \angle YF_1B + \angle CF_1Y + \angle WF_1Z \\ &= \angle AF_1D + \angle BF_1C \end{align*}$
$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.15, min = 1, w = 3.0, x=1.9, y = 1.3, z = 0.3; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair W = point(el,w), X = point(el,x), Y = point(el, y), Z = point(el, z); pair A = extension(X, X+dir(el,x), W+dir(el,w),W); pair B = extension(dir(el,x)+X,X,Y,Y+dir(el,y)); pair C = extension(Z, Z+dir(el,z), Y+dir(el,y), Y); pair D = extension(dir(el,z) + Z, Z, W, W+dir(el, w)); pair O = (0,0); draw(A--B--C--D--A, royalblue); draw(A--O--B, purple); draw(C--O--D, purple); draw(W--O--Y, brown); draw(X--O--Z, brown); dot("$W$", W, dir(W)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$",O, dir(D-C)); [/asy]$
For the second theorem:
$\begin{align*} [AOB] + [COD] &= [AOX] + [XOB] + [COZ] + [ZOD] \\ &= [AOW] + [YOB] + [COY] + [EOD] \\ &= [AOD] + [BOC] \end{align*}$
Also, something that follows from the triangle case, is that the $8$ points which are the feet of altitudes from $F_1$ and $F_2$ to $AB, BC, CD, DA$ are all concyclic, and are centred around the $O$ . The proof is also exactly the same. In fact, for any tangent, the foot of $F_1$ to that tangent has some fixed distance to $O$ .

We'll now discuss a few seemingly unrelated things.
Let $ABCD$ be a quadrilateral, and let point $E$ be arbitrary. Suppose $(AEB), (CED)$ meet at $F$ and $(AED), (CEB)$ meet at $G$ . Then, circles $(AGB), (CGD), (AFD), (CFB)$ concur. This concurrency point is called the clawson schmidt conjugate of $E$ with respect to $ABCD$ .

$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(110); pair A = dir(205), B = 0.7*dir(155), C = 0.8*dir(70), D = dir(-25); pair E = (-0.34, -0.2), F, G, H; F = 2*foot(E, circumcenter(E,A,B), circumcenter(E,C,D)) - E; G = 2*foot(E, circumcenter(E,A,D), circumcenter(E,B,C)) - E; H = 2*foot(F, circumcenter(B,F,C),circumcenter(A,F,D)) - F; draw(A--B--C--D--A, royalblue); draw(circumcircle(A,H,B), purple); draw(circumcircle(B,E,C), springgreen); draw(circumcircle(A,E,D), springgreen); draw(circumcircle(D,H,C), purple); draw(circumcircle(D,E,C), fuchsia); draw(circumcircle(A,F,D), brown); draw(circumcircle(A,E,B), fuchsia); draw(circumcircle(B,F,C), brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); path square = (-1.8, -1) -- (1.8, -1) -- (1.8, 1.2) -- (-1.8, 1.2) -- cycle; clip(currentpicture, square); [/asy]$

We first prove that the schmidt conjugate exists.
The main thing to notice in the diagram is that there are many circles. In fact there are wayy too many circles. Seeing so many circles, we realize we just _need_ to invert, and since we need to pick a point, let's just invert across $E$ with arbitrary radius. The result is surprising, let me let the diagram speak for itself:

$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(60); pair A = dir(205), B = 0.9*dir(155), C = 0.3*dir(70), D = 0.4*dir(-25); pair E = (-0.34, -0.2), F, G = (0,0), H = (0,0); F = extension(A,B,D,C); G = extension(A,D,B,C); H = 2*foot(F, circumcenter(B,F,C),circumcenter(A,F,D)) - F; draw(B--G--A,springgreen); draw(A--F--D, fuchsia); draw(circumcircle(H,B,C), brown); draw(circumcircle(H,A,D), brown); draw(circumcircle(H,D,C), purple); draw(circumcircle(H,A,B), purple); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); [/asy]$
(In case someone can't speak diagram-ese, $H$ just turns out to be the miquel point of $ABCD$ in the inverted figure! Also I've colour coded it so that it is hopefully easy to see which circle/line is the inverse of which circle)

Keeping in mind that the clawson-schmidt conjugate exists, we now turns towards _yet_ another seeming unrelated thing. Yes, I promise that we're getting somewhere and it's not just a wild goose chase.

There is a common inversion/reflection that often appears in quadrilateral olympiad geometry :-
Take quadrilateral $ABCD$ , with miquel point $M$ . Then, the transformation is as follows: for a point $P$ , invert it with radius $\sqrt{MA \cdot MC}$ , and reflect it across the bisector of $\angle AMC$ . The nice thing about this, is that since $M$ is the spiral centre of $AB \to DC$ , that $\sqrt{MA*MC} = \sqrt{MB*MD}$ , and that $\angle AMC, \angle BMD$ share an angle bisector. Thus in this transformation, $A$ and $C$ swap, $B$ and $D$ swap, and the other vertices of the complete quadrilateral also swap. We say call this transformation as "miquel inverting"

$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(60); pair A = dir(205), B = 0.7*dir(155), C = 0.8*dir(70), D = dir(-25); pair E = (-0.34, -0.2), F, G = (0,0), H = (0,0); F = extension(A,B,D,C); G = extension(A,D,B,C); H = 2*foot(F, circumcenter(B,F,C),circumcenter(A,F,D)) - F; draw(C--G--D, royalblue); draw(A--F--D, royalblue); draw(circumcircle(H,B,C), brown+dotted); draw(circumcircle(H,A,D), brown+dotted); draw(circumcircle(H,D,C), brown+dotted); draw(circumcircle(H,A,B), brown+dotted); draw(A--H--C, fuchsia); draw(B--H--D, orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", H, dir(H)); [/asy]$

Now that we've introduced enough seemingly random things ~~and probably hopelessly confused whoever is reading this~~, let's actually relate some of them.
The first surprising result; Miquel inverting is the same as taking the clawson schmidt conjugate!!

Suppose after inverting $E$ went to the point $H'$ , we'll show that both $H$ and $H'$ satisfy some set of properties that uniquely determine the point, i.e. that will show that $H'$ is $H$ . Despite the short length of the proof, it is somewhat tricky, since one needs to choose the angle condition carefully.
For instance, if we simply suppose we pick $\angle HCB$ . From the inversion, we know that that $\angle H'CB = \angle H'CM - \angle BCM = \angle AEM - \angle BCM$ . However, if we try to show that $\angle HCB = \angle AEM - \angle BCM$ , we'll get hopelessly stuck. This is because, (as of now at least), we have absolutely no way of talking about $\angle AEM$ , or really any angle which involves line segment $EM$ .

$[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(110); pair A = dir(205), B = 0.7*dir(155), C = 0.8*dir(70), D = dir(-25); pair E = (-0.36, -0.2), F, G, H; F = 2*foot(E, circumcenter(E,A,B), circumcenter(E,C,D)) - E; G = 2*foot(E, circumcenter(E,A,D), circumcenter(E,B,C)) - E; H = 2*foot(F, circumcenter(B,F,C),circumcenter(A,F,D)) - F; draw(A--B--C--D--A, royalblue); pair P, Q = (0,0), M = (0,0); P = extension(A,B,D,C); Q = extension(A,D,B,C); M = 2*foot(P, circumcenter(B,P,C),circumcenter(A,P,D)) - P; draw(circumcircle(D,E,C), fuchsia); draw(circumcircle(A,F,D), brown); draw(circumcircle(A,E,B), fuchsia); draw(circumcircle(B,F,C), brown); draw(D--H--C, springgreen); draw(M--H, springgreen); draw(A--E--B, purple); draw(E--M, purple); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$M$", M, dir(M)); path square = (-1.8, -1) -- (1.8, -1) -- (1.8, 1.2) -- (-1.8, 1.2) -- cycle; clip(currentpicture, square); [/asy]$

Instead, the key idea is to try and evaluate $\angle H'CB + \angle H'DA$ and compare it to $\angle HCB + \angle HDA$ . (or try and get $\angle DH'C = \angle DHC$ ).
$\begin{align*} \angle H'CB + \angle H'DA &= \angle H'CM - \angle BCM + \angle H'DM + \angle MDA \\ &= \angle AEM - \angle BCM + \angle BEM + \angle MDA \\ &= \angle AEB + \angle MDA - \angle ADM \\ &= \angle AEB \end{align*}$
On the other hand,
$\begin{align*} \angle HCB + \angle HDA &= \pi - \angle HFB + \pi - \angle HFA \\ &= \angle AFB \\ &= \angle AEB \end{align*}$
Thus these two angles are equal. Since this must be true when any other side of the quadrilateral is picked, (Not just $CD$ ), we get that $H' = H$ . QED.
This means that clawson schmidt conjugation, and miquel inversion are the same transformation.

Now, we're finally ready for the main result.
Suppose a point $E$ in quadrilateral $ABCD$ has an isogonal conjugate. Then, it's isogonal conjugate must be its clawson schmidt conjugate!
The proof is similar to the above one. Suppose $H'$ is the isogonal conjugate of $E$ , and $H$ is the clawson schmidt conjugate. As we've shown above, $\angle HCB + \angle HDA = \angle AEB$ . On the other hand, $\angle H'CB + \angle H'DA = \angle ECD + \angle EDC = \pi - \angle CED = \angle AEB$ . This, and the other symmetric equations prove the desired result!

It's pretty astounding, how isogonal conjugation, clawson schmidt conjugation, and miquel inverting appear very different at first sight, but they are actually identical.

_______________________________________________________________________________________________

A couple of misc things about the above setup; especially the section provides a new interesting way to think about miquel points. Suppose we have a set of pairs $(X_1, Y_1), (X_2, Y_2), \dots$ , of size at least $2$ . We say that $M$ is the miquel point of the above set, if there is some inversion across $M$ followed by reflection across the line through $M$ , which swaps $X_i$ and $Y_i$ . The nice thing about this definition is that $M$ is themiquel point of quadrilateral $X_jY_kY_jX_k$ as per the usual definition. This is because $M$ is the centre of the spiral similarity taking $X_jY_k$ to $X_kY_j$ . Also, $X_tY_t$ are clawson schmidt conjugates in quadrilateral $X_jY_kY_jX_k$ . Of course, the set needn't be finite or even countable.

One interesting corollary of this is 'miquel nesting'
Suppose quadrilateral $ABCD$ has pairs of clawson conjugates $(P, Q), (U, V)$ and $(X, Y)$ . Then $(X, Y)$ are clawson conjugates in $PUQV$ as well.

We can go along this thought process a little further, and talk about isogonal nesting.
Suppose we have a set of pairs $(X_1, Y_1), (X_2, Y_2), \dots$ , of size at least $2$ , for which a miquel point exists. We call this as isogonal pair set, if the midpoints of $X_iY_i$ all lie on a straight line. The reason for this will become clear shortly.
Consider any pairs $X_1, Y_1$ , $X_2, Y_2$ and $X_3, Y_3$ from this isogonal set. Note that $X_3, Y_3$ are clawson schmidt conjugates in $X_1X_2Y_1Y_2$ . Further since their midpoint lies on the gauss line, by the converse of the first theorem we showed (We haven't proved the converse, but it should not be too hard), $X_3, Y_3$ are isogonal conjugates in $X_1X_2Y_1Y_2$ .
Again, one interesting corollary of this is 'isogonal nesting'
Suppose quadrilateral $ABCD$ has pairs of isogonal conjugates $(P, Q), (U, V)$ and $(X, Y)$ . Then $(X, Y)$ are isogonal conjugates in $PUQV$ as well.

__________________________________________________________________________________________________

I'll also soon (after IOI tgh) post a part 3 of this, where I showcase some oly problems which fall to these methods.

This post has been edited 2 times. Last edited by p_square, Jun 13, 2021, 3:18 PM

Comment

5 Comments

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

All the best for IOI!!!!!!!

Thanks!!!

This post has been edited 1 time. Last edited by p_square, Jun 13, 2021, 3:06 PM

by Rg230403, Jun 13, 2021, 9:14 AM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

All the best for IOI. Also are you trying your hand at the Reimann Hypothesis ( I am asking this because I saw a comment related to this in your last blog post)

I'm obviously not trying the Riemann hypothesis, and the comment was of course just a joke. It doesn't even make sense for me to waste much time on any open problem, forget an important one.

This post has been edited 1 time. Last edited by p_square, Jun 14, 2021, 3:45 PM

by Bratin_Dasgupta, Jun 14, 2021, 5:20 AM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Really your performance in the TST's was amazing. Like its really sad you didnt get score 300 on the first day as well. You could have gotten a whopping 800 and crushed any cutoff that could have come for the team! But even then >650 is very amazing !

Haha thanks. That said, I'd say that all of the rest team is comfortably better than me.

This post has been edited 1 time. Last edited by p_square, Jun 14, 2021, 3:43 PM

by starchan, Jun 14, 2021, 9:41 AM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Your blogs are always awesome.

All the best for IOI and IMO.

by GeoKing, Jun 20, 2021, 12:26 AM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

hey, isn't this similar to the theory of perfect isogonal sets

by qwerty123456asdfgzxcvb, Sep 14, 2024, 9:45 PM

Report

Submit

thank you for teaching me the humpty point
by ihatemath123, Jun 27, 2024, 2:43 PM
ORZZZZZZZZZZ
by avisioner, Jun 19, 2024, 12:02 AM
He never had motivation to multiply out random polynomials
by the_mathmagician, Oct 25, 2023, 1:49 AM
Getting a part 3 is unlikely... anyways, if it ever gets released, it would be great if you listed some problems where this methods are useful!
by Alfombra12, Aug 13, 2023, 10:36 AM
still waiting for part 3
by kn07, Jun 14, 2023, 10:55 PM
Orz orz
by aansc1729, Mar 13, 2023, 1:13 PM
waiting for part 3 since 1 year
by anurag27826, Jul 18, 2022, 11:19 AM
pro kid moment
by the_mathmagician, Feb 15, 2022, 2:21 AM
Ded blog
by SPHS1234, Jan 31, 2022, 7:43 AM
orzity orz orz
by tigerzhang, Oct 15, 2021, 10:08 PM
congrats on IMO gold medal!
by mathlearner2357, Jul 25, 2021, 5:04 AM
orzorzorz
by PEKKA, Jul 21, 2021, 3:55 PM
Wow this is awesome!!

Waiting for part3
by lneis1, Jun 24, 2021, 5:42 PM
Part 3 hype
by NJOY, Jun 15, 2021, 9:29 AM
Ooh niceee! Waiting for part 3!
by L567, Jun 14, 2021, 4:43 AM
Something on FE's please
by Aryan-23, Feb 26, 2019, 5:33 PM
Post!!!!!!
by ELMOliveslong, Feb 22, 2019, 7:55 PM
Really amazing!! In fact you also have the skills of presenting things neatly!! waiting for more blog posts!!
by AlastorMoody, Jan 12, 2019, 7:40 AM
A good topic
by ahwawm, Dec 20, 2018, 3:22 PM
Nice post!
by Rmo, Dec 20, 2018, 2:17 PM
Great blog!!
by AhirGauss, Dec 19, 2018, 2:56 PM
@below He must have been busy these days. Have patience, boys. Legends are not made in ~~one day~~mere two months, after all.
by hansu, Dec 12, 2018, 6:52 PM
Nice blog! Please add NT
by Wizard_32, Nov 26, 2018, 12:47 PM
No.. this blog cannot die....
by enthusiast101, Nov 19, 2018, 1:14 PM
Nice blog!
by niyu, Nov 8, 2018, 5:02 AM
An "edge" was better
by anantmudgal09, Nov 1, 2018, 2:48 PM
@rajat1603_ Why do people want to know my "preparation story'??? In this blog most (all) of my posts will be mainly about math
by p_square, Aug 17, 2018, 5:10 AM
*PLEASE* let it be about your preparation story
by rajat1603_, Aug 17, 2018, 3:57 AM
I will post something during the weekend.
by p_square, Aug 16, 2018, 11:33 AM

63 shouts

A Vertex of Math

Angle chasing with ellipses [Part 2]

by p_square, Jun 13, 2021, 8:51 AM

5 Comments