The A-Humpty point; Inversion

by p_square, Sep 9, 2018, 3:29 PM

It isn't a well known saying, that "When in doubt, invert". And there is a reason for that. It is not recommended to try and invert around points randomly without regard for rhyme or reason.
However it is generally a good idea to invert around points through which a lot of circles pass.

Inversion possesses the power to turn all these cruel complicated circles into simple lines.

Looking at the complete diagram of the $A$ -HM point, it can be seen that many circles pass through the $A$ -HM point. In this blog post I will invert around the $A$ -HM point with arbitrary radius. This turns out even better than expected.

Also, since I am not comfortable inverting the large complicated diagram at once, I shall invert it in many stages.

Stage 1; $A,B,C,X,M$
Take $XBC$ as an arbitrary triangle. We see that $A$ is the intersection of tangents from $B,C$ to the circumcircle of $XBC$ .
$M$ is the second intersection point of $AM$ and the circumcircle of $XBC$ .
Things look quite good at this point. We have a harmonic quadrilateral, with a tangent intersection point marked.
$[asy] defaultpen(fontsize(9pt)); import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3; B=(-0.766,0.643); X=(-0.174,0.985); C=(0.766,0.643); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; draw(M--A--B--C--A); draw(circumcircle(X,C,B)); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$X$",X,dir(90)); dot("$M$",M,dir(270)); [/asy]$

The next stage will be dealing with $H$ . We definitely cannot define it in the usual manner as the orthocentre of triangle $ABC$ . All those perpendicular lines will become orthogonal circles, a not-so-nice constraint.

Stage 2; $E,F,H$
The thing to notice over here, is that $E,F$ are much better characterized as lying on circles $XCM, XBM$ in the original figure than foot of altitudes from $B,C$ to $AC, AB$ .

$E$ and $F$ will simply become the intersection points of $CM, BM$ with circle $XAC, XAB$ .

Now, instead of thinking of $H$ as $BE$ intersection $CF$ in the original figure, we can simply remember property 2.
$H$ goes to $BC$ intersection $AEF$ .

We now have enough points for interesting things to start happening in the figure and indeed they do

$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5; B=dir(150); X=dir(100); C=dir(30); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; H=extension(F,E,B,C); draw(M--A--B--C--A); draw(E--M--F--H--C); draw(C--A--B,heavygreen); draw(F--E--C--B--F,heavygreen); draw(circumcircle(X,C,B)); draw(circumcircle(A,X,B),blue); draw(circumcircle(A,X,C),blue); draw(circumcircle(E,F,B), heavygreen); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$X$",X,dir(90)); dot("$M$",M,dir(270)); dot("$E$",E); dot("$F$",F);dot("$H$",H); [/asy]$

Interesting Property wrote:

$A$ is the center of a circle passing through $B,C,E,F$ .

We can prove this by simple angle chasing.
$\angle AEC = 180 - \angle AXC = \angle MXC = \angle CBM = 180 - \angle ACM = \angle ACE$ .
We have $AF = AB = AC = AE$ .

Also, we can redefine $E,F$ as the points on $MC,MB$ such that $AE = AB = AC = AF.$

This diagram is basically USAMO 2007/2

Stage 3: $D$
$D$ clearly becomes the intersection of circle $XBC$ and line $MH$ . $D$ also lies on the circle $AXH$ .
Observe that $\angle ADM = \angle ADH = \angle AXH = 90$ .
$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5,x6,x7,D; B=dir(150); X=dir(100); C=dir(30); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; H=extension(F,E,B,C); x6 = foot(O,M,H); D = 2*x6-M; draw(M--A--B--C--A); draw(E--M--F--H--C); draw(M--H, red); draw(C--A--B,heavygreen); draw(F--E--C--B--F,heavygreen); draw(circumcircle(X,C,B)); draw(circumcircle(E,F,B),heavygreen); draw(circumcircle(A,X,H),red); draw(circumcircle(F,X,C),dashed+heavycyan); draw(circumcircle(E,X,B),dashed+heavycyan); draw(circumcircle(A,X,C),dashed+orange);draw(circumcircle(A,X,B),dashed+orange); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$X$",X,dir(90)); dot("$M$",M,dir(270)); dot("$E$",E); dot("$F$",F);dot("$H$",H);dot("$D$",D); [/asy]$

From the usual definitions of $E,F,H,D$ we see that circle $FXCH$ is orthogonal to circle $ABEX$ , circle $EXBH$ is orthogonal to circle $ACEX$ and circle $AHDX$ is orthogonal to $XBC$ .

Stage 4: $T,P,O_A$ .
$O_A$ being the circumcentre of $XBC$ clearly becomes $X$ reflected across $BC$ ~~or the M-Humpty point of MBC~~.
After inversion, $BCXT$ is an isosceles trapezium.

$P$ was the circumcenter of $AEFHX$ , so it becomes $X$ reflected across $AEFH$ .

$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5,x6,x7,D,T,P,Oa,x8,x9; B=dir(150); X=dir(100); C=dir(30); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; H=extension(F,E,B,C); x6 = foot(O,M,H); D = 2*x6-M; x7 = foot(X,B,C); Oa = 2*x7 - X; x8 = foot(X,E,F); P=2*x8-X; x9 = foot(O,M,P); T=2*x9-M; draw(M--A--B--C--A); draw(E--M--F--H--C); draw(M--H, red); draw(C--A--B,heavygreen); draw(F--E--C--B--F,heavygreen); draw(M--P,orange); draw(circumcircle(X,C,B)); draw(circumcircle(E,F,B),heavygreen); draw(circumcircle(A,X,H),red); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$X$",X,dir(90)); dot("$M$",M,dir(270)); dot("$E$",E); dot("$F$",F);dot("$H$",H);dot("$D$",D);dot("$T$",T);dot("$P$",P);dot("$O_a$",Oa); [/asy]$

An interesting thing to note is that the $A$ -HM point configuration is partially self-inversive.

Since $A$ is the midpoint of $EF$ and $AB=AC=AE=AF$ , we see that $EB$ and $FC$ are perpendiculars to $ME, MF$ .
This further implies that $X$ is the $M$ -HM point of triangle $MEF$ .

For now, let us hop back to original diagram

We will prove a quick result or $2$ that were not touched upon in the previous entry.

Result $1$ : $HX, BC$ and $EF$ concur (Say at $Y$ ).
The simplest way to prove this is to note that $EFHX, EFDM, MDHX$ are all circles. Since we know that radical axis of $3$ circles taken pairwise concur at a point, we are done.

$[asy] defaultpen(fontsize(9pt)); import olympiad; unitsize(30); pair A,B,C,D,E,F,H,M,X,Y; B=(0,0); A=(1,4); C=(5,0); D=foot(A,B,C); E=foot(B,A,C); F=foot(C,A,B); H=orthocenter(A,B,C); M=(B+C)/2; X = foot(H,A,M); Y=extension(E,F,X,H); draw(D--A--B--C--A--M); draw(E--Y--X, blue); draw(C--F); draw(B--E); draw(Y--C,blue); draw(circumcircle(A,E,F),heavygreen); draw(circumcircle(D,E,F),heavygreen); draw(circumcircle(M,H,X),heavygreen); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$D$",D,dir(250)); dot("$E$",E,dir(60)); dot("$F$",F,dir(210)); dot("$H$",H,dir(150)); dot("$M$",M,dir(300)); dot("$X$",X,dir(330)); dot("$Y$",Y); [/asy]$

Result $2$ : $X$ is the Miquel point of quadrilateral $BEFC$ (Yes, that order).
The proof of this is extremely short.
Simply note that since $BE, CF$ intersect at $H$ , and as we know, $EFHX$ and $BCHX$ are cyclic.
This further implies that $YEXC$ and $YFXB$ are cyclic.

$[asy] defaultpen(fontsize(9pt)); import olympiad; unitsize(30); pair A,B,C,D,E,F,H,M,X,Y; B=(0,0); A=(1,4); C=(5,0); D=foot(A,B,C); E=foot(B,A,C); F=foot(C,A,B); H=orthocenter(A,B,C); M=(B+C)/2; X = foot(H,A,M); Y=extension(E,F,X,H); draw(D--A--B--C--A--M); draw(E--Y--X, orange); draw(Y--C,orange); draw(B--E--F--C--B, heavygreen); draw(circumcircle(A,E,F),blue); draw(circumcircle(B,H,C),blue); draw(circumcircle(Y,B,X),heavycyan); draw(circumcircle(Y,C,X),heavycyan); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$D$",D,dir(250)); dot("$E$",E,dir(60)); dot("$F$",F,dir(210)); dot("$H$",H,dir(150)); dot("$M$",M,dir(300)); dot("$X$",X,dir(330)); dot("$Y$",Y); [/asy]$

These two are actually well worth adding to our original figure.

These $2$ facts shall prove to be neat in the inverted diagram. In fact $Y$ plays a very important role.
Circles $XBC$ , $XEF$ , meet at the points on $XH$ . i.e. $XH$ is the radical axis of these two circles.
$Y$ is the orthocentre of $MEF$ .

$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5,x6,x7,D,x8,x9,Y,x10,T,P,Oa,x11,Y; B=dir(150); X=dir(100); C=dir(30); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; H=extension(F,E,B,C); x6 = foot(O,M,H); D = 2*x6-M; x7 = foot(O,X,H); Y = 2*x7 - X; x10 = foot(X,B,C); Oa = 2*x10 - X; x8 = foot(X,E,F); P=2*x8-X; x9 = foot(O,M,P); T=2*x9-M; x11=foot(O,X,H); Y=2*x11-X; draw(M--A--B--C--A); draw(E--M--F--H--C); draw(M--H, red); draw(C--A--B,heavygreen); draw(F--E--C--B--F,heavygreen); draw(B--E,blue); draw(C--F,blue); draw(A--D,blue); draw(X--H,blue); draw(circumcircle(X,C,B)); draw(circumcircle(E,F,B),heavygreen); draw(circumcircle(A,X,H),red); dot("$A$",A,dir(90)); dot("$B$",B,dir(150)); dot("$C$",C,dir(60)); dot("$X$",X,dir(90)); dot("$M$",M,dir(270)); dot("$E$",E); dot("$F$",F);dot("$H$",H);dot("$D$",D);dot("$Y$",Y); [/asy]$

Now, we are ready for the grand finale, Presenting IMO 2015 P3.
We invert at an HM point configuration to arrive at the solution.

IMO 2015/3 wrote:

Let $ABC$ be an acute triangle with $AB > AC$ . Let $\Gamma$ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$ . Assume that the points $A$ , $B$ , $C$ , $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

The diagram looks like this:
$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(70); pair A,B,C,H,O,Q,K,F,M,x1; A = dir(70); B = dir(210); C = dir(330); O = circumcenter(A, B, C); H = orthocenter(A, B, C); M =(B+C)/2; F = foot(A, B, C); Q=foot(A,M,H); K=OP(CP((H+Q)/2,H),circumcircle(A,B,C)); draw(A--B--C--A); draw(H--A--Q--H--K--Q, red); draw(A--M--H--F,blue); draw(circumcircle(A,B,C),heavygreen); draw(circumcircle(K,F,M),heavycyan); draw(circumcircle(K,Q,H),heavycyan); dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$H$",H); dot("$M$",M); dot("$F$",F); dot("$Q$",Q);dot("$K$",K); [/asy]$
~~Since I drew the diagram, I leave the proof as an exercise to the reader~~

First of all, observe that $A$ and $Q$ are the $H$ -HM point of triangle $HBC$ respectively.
Let us invert the figure about $Q$ .
After inversion, the diagram appears like this:

$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5,x6,x7,D,x8,x9,Y,W,K,x10,x11; B=dir(215); X=dir(160); C=dir(325); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; x8 = -1*M; x9 = foot(O,x8,A); D = 2*x9-x8; x7 = (A-M)*dir(270)+A; K = extension(A,x7,B,C); x6 = foot(O,x8,K); W = 2*x6-x8; x10 = foot(O,D,K); Y = 2*x10-D; H = extension(D,M,C,B); draw(X--A--B--C--A); draw(H--D--A--K--H--A); draw(A--H--D, red); draw(X--A--K,blue); draw(circumcircle(X,C,B),heavygreen); dot("$H$",A); dot("$B$",B); dot("$C$",C); dot("$Q$",X); dot("$M$",M); dot("$F$",D); dot("$K$",K); dot("$A$",H); [/asy]$

Now, let us erase a few unimportant points that aren't doing much and add in our very own set of points

Define $W$ as the point diametrically opposite $M$ in circle $MBC$ .
We know that $\angle HFA = 90$ (Stage $3$ ). This means that $W,M,F$ are collinear.
We can get rid of $A$ by re-defining $F$ as the second point of intersection between $WH$ and circle $MBC$ .
Let $L$ be the second intersection of $WK$ with circle $MBC$ .
Let $Y$ be the second intersection of $FK$ with circle $MBC$ .
We can also erase $Q$ from the figure.

We are left with
$[asy] defaultpen(fontsize(9pt)); import math; import olympiad; unitsize(50); pair A,B,C,M,X,O,x1,x2,x3,E,F,H,x4,x5,x6,x7,D,x8,x9,Y,W,K,x10,x11; B=dir(215); X=dir(160); C=dir(325); O=circumcenter(X,B,C); A=(-1/((B+C)/2-O))+O; x1=orthocenter(X,B,C); x2=foot(x1,X,((B+C)/2)); x3 = foot(x2,B,C); M = 2*x3-x2; x4=foot(A,M,C); E=2*x4-C; x5=foot(A,M,B); F=2*x5-B; x8 = -1*M; x9 = foot(O,x8,A); D = 2*x9-x8; x7 = (A-M)*dir(270)+A; K = extension(A,x7,B,C); x6 = foot(O,x8,K); W = 2*x6-x8; x10 = foot(O,D,K); Y = 2*x10-D; draw(A--B--C--A--M--B--C--M); draw(A--W--M--A--K--A--x8, red); draw(x8--K--D--M, blue); draw(circumcircle(M,B,C),heavygreen); draw(circumcircle(A,K,M),heavycyan); dot("$H$",A); dot("$B$",B); dot("$C$",C); dot("$M$",M); dot("$F$",D); dot("$K$",K); dot("$W$",x8); dot("$L$",W); dot("$Y$",Y); [/asy]$
$(FW;BC) = (YL;BC) = -1$ implies that $H,Y,L$ are collinear.
Also, $\angle MLK = \angle MLW = 90 = \angle MHK$ . $\therefore M,L,K,H$ are cyclic
We end by noting that $\angle HKM = \angle HLM = \angle YLM = 180 - \angle YFM = 180 - \angle KFM$ .
This means that $HK$ is tangent to circle $MKF$ .

Note: IMO 2015 P3 is actually even easier to solve via inversion at $H$ which is the $Q-HM$ point of triangle $QBC$ . Readers are 'encouraged' to also try solving it that way.

Credits go to @anantmudgal09 for telling me that IMO 2015 P3 could be done via inversion.

This post has been edited 6 times. Last edited by p_square, Feb 25, 2019, 11:19 AM

HM-Point

geometry

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6 Comments

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I'll post a "complete" diagram of the inverted and normal $A$ -HM point diagrams tomorrow.

Also, if unclear, inverting at $A$ and $X$ will actually give precisely the same results if we plot in every point. However, given that we are treating $\triangle ABC$ as more important than $\triangle XBC$ in the choice of the points we plot, inversion at $X$ seems to be more interesting.

This post has been edited 1 time. Last edited by p_square, Sep 9, 2018, 3:36 PM

by p_square, Sep 9, 2018, 3:34 PM

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You never fail to impress!! Sadly I don't understand...

by enthusiast101, Sep 9, 2018, 4:00 PM

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Wow! Thanks for an awesome post.

Thanks!

This post has been edited 1 time. Last edited by p_square, Sep 12, 2018, 5:00 PM

by AnArtist, Sep 9, 2018, 4:54 PM

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Yummy,lots of new geo stuff

by Rmo, Sep 13, 2018, 2:46 AM

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Wat will next post be on?

Grid bipartition
Change of plan. That'll be the next to next post.

This post has been edited 2 times. Last edited by p_square, Oct 18, 2018, 11:29 AM

by Rmo, Sep 13, 2018, 2:49 AM

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Should add 2017 G3 also obvious by HM point

by Pluto1708, Sep 29, 2018, 8:22 AM

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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

The A-Humpty point; Inversion

by p_square, Sep 9, 2018, 3:29 PM

6 Comments