The A-Humpty point; Preliminaries
by p_square, Aug 18, 2018, 12:15 PM
In this blog post I will discuss the intricacies of the 
-HM point. Hopefully, by the end, many of its mysteries will be thrown into light.
The-HM point has several nice properties: We will randomly pick one as it's definition and derive the rest.
Define the-HM point as the foot of perpendicular from the Orthocentre 
of 
to the -Median.
We will also henceforth denote the-HM point by 
.
Anyway,
Well, not exactly. But that's close enough.
Property 2:
is a cyclic quadrilateral
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E); draw(C--F); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C));
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));
[/asy]](//latex.artofproblemsolving.com/7/9/9/79902d279973ec1bc3f1faa209767f5c9c71301f.png)
Let
, 
and 
be the altitudes of the triangle 
. Let 
be the midpoint of 
.
Observe that
is cyclic.
We will invert at A with radius
.
As a fortunate co-incedence, By Power of Point,
. Actually, this is the only reason we inverted.
What this tells us, is that during the inversion
.
Before inversion, Points
were cyclic (Nine-point circle), and as it hath been foretold, that circles shall (usually) remain circles after inversion, so are cyclic. 
Property 3:
is tangential to Circles 
.
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E--M); draw(C--F--M); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C)); draw(circumcircle(A,X,B)); draw(circumcircle(A,X,C));
dot("$A$",A,dir(90)); dot("$B$",B,dir(130)); dot("$C$",C,dir(60)); dot("$D$",D,dir(250)); dot("$E$",E,dir(60)); dot("$F$",F,dir(210)); dot("$H$",H,dir(130)); dot("$M$",M,dir(300)); dot("$X$",X,dir(330));
[/asy]](//latex.artofproblemsolving.com/9/2/9/92976c4045df9dd3ff0349afab471804da23905c.png)
If we continue to work in the inverted diagram, what I wrote translates to
, 
are tangent to Circle 
This is very well known and the proof is simple angle chasing.
Property 4: Let the tangent from to the circumcircle of meet at 
. The orthocentre of 
is the midpoint of 
.
At first this doesn't seem related to at all! But on re-reading the relationship hopefully becomes clear. Say 
is the midpoint of and 
is the foot of perpendicular from to 
.
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(T--Q); draw(H--X);
draw(circumcircle(A,B,C));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270));
[/asy]](//latex.artofproblemsolving.com/0/b/7/0b7cc6d209b1e41315d52648f52c812790b10b48.png)
Proof:
It isnot very easy to see that 
. It is also well known that 
.
So, is indeed the orthocentre of .
This further implies that
The foot of perpndicular from
will be the midpoint of the foot of perpendicular from 
. i.e. 
This tells us that is on the perpendicular bisector of 
or 
Many people might recognise this as ELMO 2018/4 (I also used the same notation). Quite a few geometry questions become much simpler if one knows about the-HM point.
Property 5: If
is a harmonic quadrilateral, then is the reflection of 
across
Proof:
It isn't hard to see that the reflection of across lies on circle 
. (
)
Also
. Ignoring configuration issues (which I vehemently hate generally dislike), we see that must be the reflection of across .
Property 6:
is tangent to circle 
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(X--T--Q); draw(H--X);
draw(circumcircle(A,B,C)); draw(circumcircle(B,H,C));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270));
[/asy]](//latex.artofproblemsolving.com/5/5/0/550509db41606c8898975537769e088cc2a52e10.png)
The power of with respect to circles and 
is the same (
)
Thus, The length of the tangent is same, and we have already proved that
.
This finishes the proof
Property 7:
are cyclic
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q,L,O;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
L=circumcenter(B,H,C);
O=circumcenter(A,B,C);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(T--Q); draw(H--X--T); draw(L--P); draw(L--M); draw(L--X);
draw(circumcircle(A,B,C)); draw(circumcircle(B,H,C)); draw(circumcircle(T,M,L));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270)); dot("$O_A$",L,dir(320));
[/asy]](//latex.artofproblemsolving.com/a/6/a/a6a601613e8f72bc4a97f27aa6150a2911d42787.png)
Like many a geometry problem, this actually becomes a lot easier if you just add in one extra point.
Clearly 
lies on the perpendicular bisector of , so 
.
By property 6,
Also, Since
is a parallelogram.
This tells us that
(Property 4) i.e. 
= 90.
As a corollary, note that
is an isosceles trapezium.
Property 8: Duality; is the -HM point of 
Proof:
Property 3 (Use Power of Point from) finishes this off immediately.
However look at Property 2.
which is just another way of writing Property 2, with the roles of and switched.
orthocentre of 
is the orthocentre of 
.
Property 9: lies on the -apollonian circle
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E); draw(C--F); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C));
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));
[/asy]](//latex.artofproblemsolving.com/e/d/c/edc543faa8d6f096bcdf6cc30720f3053dbd8a3d.png)
It is well known that upon inversion at, the -apollonian circle becomes the perpendicular bisector of the new 
.
After performing
inversion 
whereas 
. Since 
, lies on the perpendicular bisector of 
.
Here is a diagram encompassing most of these properties.
![[asy]
import olympiad;
unitsize(30);
pair Z,A,B,C,D,E,F,H,T,P,Q,X,M,N,O,L,Y;
A=(6,4); B=(7,0); C=(11,0);
M=(B+C)/2;
D=foot(A,B,C);
E=foot(B,C,A);
F=foot(C,A,B);
H=orthocenter(A,B,C);
dot(H);
X=foot(H,A,M);
P=(A+H)/2;
T=orthocenter(A,P,M);
Q=foot(P,A,M);
O=circumcenter(A,B,C);
L=circumcenter(H,B,C);
N=orthocenter(B,C,X);
Z=foot(M,A,T);
draw(B--A--C--B); draw(C--F, heavygreen); draw(E--H, heavygreen); draw(A--H, heavygreen); draw(A--M, heavygreen); draw(C--H, heavygreen); draw(D--B--F, heavygreen); draw(A--T--Q, red); draw(M--Z, red); draw(A--D, dashed+red); draw(T--X--L--M,blue); draw(L--P, blue); draw(T--P, dashed+blue); draw(T--B, red); draw(F--M--E,dotted+purple);
draw(circumcircle(A,B,C),dashed+lightred); draw(circumcircle(H,B,C),blue); draw(circumcircle(L,M,T),blue);
draw(circumcircle(X, H, D),dotted+lightred); draw(circumcircle(X, B, F),dotted+lightred); draw(circumcircle(X, C, E),dotted+lightred); draw(circumcircle(A,X,C),heavycyan); draw(circumcircle(A,X,B),heavycyan); draw(circle(T,4.4),orange); draw(circumcircle(A,E,F),dotted+purple);
dot("$A$",A, dir(150)); dot("$B$",B,dir(215)); dot("$C$",C, dir(270)); dot("$X$",X,dir(30)); dot("$M$",M,dir(60)); dot("$T$",T,dir(150)); dot("$D$",D,dir(225)); dot("$E$",E,dir(0)); dot("$F$",F,dir(270)); dot("$P$",P,dir(45)); dot("$Q$",Q,dir(45)); dot("$O_A$",L,dir(310));
[/asy]](//latex.artofproblemsolving.com/5/0/7/507425db1454a70c6177233601d73589f5261343.png)
For those interested, @anantmudgal09's handout which can be found here also deals with the A-HM point. @Rmo was also helpful in property 7.
-HM point. Hopefully, by the end, many of its mysteries will be thrown into light.The
-HM point has several nice properties: We will randomly pick one as it's definition and derive the rest.Define the
-HM point as the foot of perpendicular from the Orthocentre 
of 
to the -Median.We will also henceforth denote the
-HM point by 
.Anyway,
p_square wrote:
Property 1: X is the foot of perpendicular from the Orthocentre of Triangle to the -Median.
of Triangle 
to the -Median.Property 2:
is a cyclic quadrilateralProof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E); draw(C--F); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C));
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));
[/asy]](http://latex.artofproblemsolving.com/7/9/9/79902d279973ec1bc3f1faa209767f5c9c71301f.png)
Let
, 
and 
be the altitudes of the triangle . Let 
be the midpoint of 
.Observe that
is cyclic.We will invert at A with radius
.As a fortunate co-incedence, By Power of Point,
. What this tells us, is that during the inversion
.Before inversion, Points
were cyclic (Nine-point circle), and as it hath been foretold, that circles shall (usually) remain circles after inversion, so are cyclic. 
Property 3:
is tangential to Circles 
.Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E--M); draw(C--F--M); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C)); draw(circumcircle(A,X,B)); draw(circumcircle(A,X,C));
dot("$A$",A,dir(90)); dot("$B$",B,dir(130)); dot("$C$",C,dir(60)); dot("$D$",D,dir(250)); dot("$E$",E,dir(60)); dot("$F$",F,dir(210)); dot("$H$",H,dir(130)); dot("$M$",M,dir(300)); dot("$X$",X,dir(330));
[/asy]](http://latex.artofproblemsolving.com/9/2/9/92976c4045df9dd3ff0349afab471804da23905c.png)
If we continue to work in the inverted diagram, what I wrote translates to
p_square in a complicated manner wrote:
, 
are tangent to Circle 
Property 4: Let the tangent from
to the circumcircle of meet at 
. The orthocentre of 
is the midpoint of 
.At first this doesn't seem related to
at all! But on re-reading the relationship hopefully becomes clear. Say 
is the midpoint of and 
is the foot of perpendicular from to 
.
p_square almost wrote:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(T--Q); draw(H--X);
draw(circumcircle(A,B,C));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270));
[/asy]](http://latex.artofproblemsolving.com/0/b/7/0b7cc6d209b1e41315d52648f52c812790b10b48.png)
Proof:
It is
. It is also well known that 
.So,
is indeed the orthocentre of .This further implies that
The foot of perpndicular from
will be the midpoint of the foot of perpendicular from 
. i.e. 
This tells us that
is on the perpendicular bisector of 
or 
Many people might recognise this as ELMO 2018/4 (I also used the same notation). Quite a few geometry questions become much simpler if one knows about the
-HM point.Property 5: If
is a harmonic quadrilateral, then is the reflection of 
across Proof:
It isn't hard to see that the reflection of
across lies on circle 
. (
)Also
. Ignoring configuration issues (which I must be the reflection of across . Property 6:
is tangent to circle 
Proof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(X--T--Q); draw(H--X);
draw(circumcircle(A,B,C)); draw(circumcircle(B,H,C));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270));
[/asy]](http://latex.artofproblemsolving.com/5/5/0/550509db41606c8898975537769e088cc2a52e10.png)
The power of
with respect to circles and 
is the same (
)Thus, The length of the tangent is same, and we have already proved that
.This finishes the proof
Property 7:
are cyclicProof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,T,M,P,X,Q,L,O;
A=(6,4);
B=(7,0);
C=(11,0);
M=(B+C)/2;
H=orthocenter(A,B,C);
X=foot(H,A,M);
P=(A+H)/2;
Q=(A+X)/2;
T=orthocenter(A,P,M);
L=circumcenter(B,H,C);
O=circumcenter(A,B,C);
draw(A--T--C--A); draw(B--A--M); draw(A--H); draw(T--Q); draw(H--X--T); draw(L--P); draw(L--M); draw(L--X);
draw(circumcircle(A,B,C)); draw(circumcircle(B,H,C)); draw(circumcircle(T,M,L));
dot("$A$",A,dir(150)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$M$",M,dir(60)); dot("$X$",X,dir(60)); dot("$H$",H,dir(180)); dot("$T$",T,dir(150)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(270)); dot("$O_A$",L,dir(320));
[/asy]](http://latex.artofproblemsolving.com/a/6/a/a6a601613e8f72bc4a97f27aa6150a2911d42787.png)
Like many a geometry problem, this actually becomes a lot easier if you just add in one extra point.
p_square almost wrote:
Let be the circumcentre of .
are cyclic with diametre 
be the circumcentre of .
are cyclic with diametre 
lies on the perpendicular bisector of , so 
.By property 6,
Also, Since
is a parallelogram.This tells us that
(Property 4) i.e. 
= 90. As a corollary, note that
is an isosceles trapezium.Property 8: Duality;
is the -HM point of 
Proof:
Property 3 (Use Power of Point from
) finishes this off immediately.However look at Property 2.
p_square almost wrote:
Let N be the point where the perpendicular from to meets circle . Then is the orthocentre of 
to meets circle . Then is the orthocentre of 
and switched.
orthocentre of 
is the orthocentre of 
. Property 9:
lies on the -apollonian circleProof:
![[asy]
import olympiad;
unitsize(30);
pair A,B,C,D,E,F,H,M,X;
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);
draw(D--A--B--C--A--M); draw(B--E); draw(C--F); draw(H--X);
draw(circumcircle(A,E,F)); draw(circumcircle(D,E,F)); draw(circumcircle(B,H,C));
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));
[/asy]](http://latex.artofproblemsolving.com/e/d/c/edc543faa8d6f096bcdf6cc30720f3053dbd8a3d.png)
It is well known that upon inversion at
, the -apollonian circle becomes the perpendicular bisector of the new 
.After performing
inversion 
whereas 
. Since 
, lies on the perpendicular bisector of 
. Here is a diagram encompassing most of these properties.
![[asy]
import olympiad;
unitsize(30);
pair Z,A,B,C,D,E,F,H,T,P,Q,X,M,N,O,L,Y;
A=(6,4); B=(7,0); C=(11,0);
M=(B+C)/2;
D=foot(A,B,C);
E=foot(B,C,A);
F=foot(C,A,B);
H=orthocenter(A,B,C);
dot(H);
X=foot(H,A,M);
P=(A+H)/2;
T=orthocenter(A,P,M);
Q=foot(P,A,M);
O=circumcenter(A,B,C);
L=circumcenter(H,B,C);
N=orthocenter(B,C,X);
Z=foot(M,A,T);
draw(B--A--C--B); draw(C--F, heavygreen); draw(E--H, heavygreen); draw(A--H, heavygreen); draw(A--M, heavygreen); draw(C--H, heavygreen); draw(D--B--F, heavygreen); draw(A--T--Q, red); draw(M--Z, red); draw(A--D, dashed+red); draw(T--X--L--M,blue); draw(L--P, blue); draw(T--P, dashed+blue); draw(T--B, red); draw(F--M--E,dotted+purple);
draw(circumcircle(A,B,C),dashed+lightred); draw(circumcircle(H,B,C),blue); draw(circumcircle(L,M,T),blue);
draw(circumcircle(X, H, D),dotted+lightred); draw(circumcircle(X, B, F),dotted+lightred); draw(circumcircle(X, C, E),dotted+lightred); draw(circumcircle(A,X,C),heavycyan); draw(circumcircle(A,X,B),heavycyan); draw(circle(T,4.4),orange); draw(circumcircle(A,E,F),dotted+purple);
dot("$A$",A, dir(150)); dot("$B$",B,dir(215)); dot("$C$",C, dir(270)); dot("$X$",X,dir(30)); dot("$M$",M,dir(60)); dot("$T$",T,dir(150)); dot("$D$",D,dir(225)); dot("$E$",E,dir(0)); dot("$F$",F,dir(270)); dot("$P$",P,dir(45)); dot("$Q$",Q,dir(45)); dot("$O_A$",L,dir(310));
[/asy]](http://latex.artofproblemsolving.com/5/0/7/507425db1454a70c6177233601d73589f5261343.png)
For those interested, @anantmudgal09's handout which can be found here also deals with the A-HM point. @Rmo was also helpful in property 7.
This post has been edited 3 times. Last edited by p_square, Aug 19, 2018, 11:16 AM
.
be the pedal triangle of 
be a variable point in 
be the parallels to 
with 
respectively. Then the circumcircles of 
passes through a fixed point, which is 
.
one week.![$FDM[b]C[/b]$](http://latex.artofproblemsolving.com/3/4/e/34ef2ad09e31002a983b58855d095f9503f7181e.png)
were cyclic (Nine-point circle), and as it hath been foretold, that circles shall (usually) remain circles after inversion, so 
be 
?
concur at 
are concyclic.
June 2021
May 2021