On the Complete Quadrilateral Configurations

Another interesting configuration, this time associated with a more tl;dr post. Firstly, a complete quadrilateral is the figure formed by four lines intersecting at 6 points, and the difference with a complete quadrangle is that this thing has three diagonals.

Before starting, I’d like to present a set of disclaimers.

Disclaimers.

1. This post is around 40% compilation and 60% observations. I have used geogebra for experimenting with the different configurations, and also searched a lot on the internet for different properties of this configuration. In the meanwhile, there are a lot of properties that I have independently figured out, but some I cannot prove myself.
2. I have mentioned “Awaiting proofs” to whichever property I have not been able to solve. For example, in Property 13, I was not able to prove all of the results thanks to the complexity. If I am able to come up with solutions, I will add them later. However, anyone eager to try is free to try and prove these results. Discussions regarding these are welcome as well.
3. Nothing else. Go ahead and enjoy reading. 🙂

Main Post.
Whenever we talk about complete quadrilaterals, we firstly think of Miquel’s theorem. A really well-known and useful theorem as it is, we will check it out along with some other basic results. I will refer to the following configuration always, ’cause it’s annoying to name each of the points every time I state a new theorem/result. We will just analyse the different properties of this configutation:

Property 1.(Miquel)
Let the lines \ell_1,\ell_2,\ell_3,\ell_4 meet in six points (named or unnamed 😛 ). Then the circumcircles of the four triangles that are formed out of these six vertices concur at a point, known as the Miquel Point, which I will designate as M.

Proof.
Surely this is well-known and simple angle-chasing. I won’t give a proof for this one, since the post is getting more and more tl;dr everyday!

\Box

Property 2.(Miquel Circle Theorem)
The circumcentres O_1,O_2,O_3,O_4 of these four triangles formed are concyclic and lie on a circle that passes through the Miquel Point.

Proof.
Refer to the diagram in Property 5. This Miquel circle doesn’t really look too hard, just note that \angle O_3MO_2=\pi-\angle MO_2O_3-\angle MO_3O_2=\pi-\angle MDE-\angle MFE=\angle DMF. Using this, working backwards of PP5, it is obvious that M,O_2,O_3,O_4 are concyclic. So, we are done.

\Box

Property 3.(Miquel-Steiner Line)
The orthocentres H_1, H_2, H_3, H_4 of these triangles lie on a line, known as the Miquel-Steiner Line.

Proof.
For the proof of this, we make use of PP12. So maybe you should first scroll down a little, read what it is all about, then come back. 🙂
Obviously the midpoints of MH_1, MH_2, MH_3, MH_4 lie on the pedal line mentioned in PP12, and therefore their image after homothety wrt (M,2) lie on a line. This line is known as the orthocentric or Miquel-Steiner line.

\Box

Property 4.
The orthocentre of O_2O_3O_4 lies on \ell _1.

Proof.
I don’t really care about the configurations much, so I’d like you to refer to the diagram in the next Property. Now, according to the notations of PP5, we need to show that H_1'\in DE.
Note that \angle DH_1'F=\angle DH_1'O_2+\angle O_3H_1'F+\angle O_2H_1'O_3=\angle O_2O_4O_3+\angle O_2H_1'O_3=180^{\circ}, because the reflection of H_1' in O_2O_3 lies on the Miquel Circle.

\Box

Property 5.
M is the centre of spiral similarity that maps O_2O_3O_4 to CAB.

Proof.
Since O_4O_2\perp DM and O_2O_3\perp ME, therefore \angle O_4O_2O_3=\angle DCE and similarly it can be shown that \angle O_2O_3O_4=\angle CBA. Therefore the triangles O_2O_3O_4 and CAB are similar. Also, note that \angle O_{3}MO_{4}=\angle O_{3}O_{2}O_{4}=\angle BCA=\angle BMA, so we see that M is the centre of spiral similarity that maps O_2O_3O_4 to CAB.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...
\Box

Property 6.(Newton-Gauss Line)
The midpoints of AC, BD, EF are collinear.

Proof.

Not really too eager to write it all down. The following stronger version of this theorem holds, which is known as Gauss-Bodenmiller theorem.

Extension of property 6:
The circles drawn on the diagonals of a complete quadrilateral taking them as diameters, are coaxal.
For a proof, see here.

\Box

Property 7.
Define the unique point K_1 on AD such that BC\parallel GK_1 and similarly define K_2\in BC, K_3\in DC, K_4\in AB. Then let M_1, M_2 be the midpoints of FG, DG. Then the points K_i and M_i lie on a straight line.

Proof.
Note that the sides DA, BC are harmonically separated by FG, FE. So, K_2 and K_1 are harmonically divided by FG\cap K_2K_1 and FG\cap K_2K_1. Since GK_1FK_2 is a parallelogram, so the point FG\cap K_2K_1 is the point at infinity of K_1K_1. Therefore FE\parallel K_1K_2. So, we are done.


\Box

Property 8.
The projections of Miquel point onto the sidelines AB,BC,CD,DA lie on a line perpendicular to the Newton-Gauss line.

Proof.
Let the points K_i be defined as in our previous property. Then it can be seen that \displaystyle \frac{\tau(K_1,\odot(FCD))}{\tau(K_1, \odot(FAB))}=\frac{\tau(K_2, \odot(FCD))}{\tau(K_2, \odot(FAB))}\Longrightarrow K_1,K_2,F,M are concyclic. So, M is the Miquel point of both ABK_2K_1 and CDK_3K_4.
Using this, we see that projection of M on K_1K_2 is collinear with projections of M on AB,BC,AD. This line is the pedal line of ABCD. The fact that this is orthogonal with the Newton-Gauss line also immediately follows.

Corollary.
Using this property, it can be shown that the Miquel point lies on the Nine Point Circle of the triangle GYZ where Y=AC\cap EF and Z=BD\cap EF. This triangle GYZ is the diagonal triangle of ABCD.


\Box

Property 9.

We define the Hewer Point as the centre of the circle that passes through the orthocentres of O_iO_jO_k. Thus this point is the pseudo-orthocentre of O_1O_2O_3O_4.

Proof.
Since the point H_P is defined as the circumcentre of H_1'H_2'H_3', therefore it is forced that the circle \odot(H_1'H_2'H_3') is congruent to the circle \odot(O_1O_2O_3), and the points H_i' correspond to the vertices O_i, meaning that the segments O_iH_i' are bisected at a common point, so that O_1OH_1'H_P is a parallelogram. Therefore, \vec{OO_1}=\vec{H_1'H_P}. This gives us \vec{OH_P}=\vec{OH_1'}+\vec{OO_1} =\vec{OO_1}+\vec{OO_2}+\vec{OO_3}+\vec{OO_4}, where the last line follows from \vec{OH}=\vec{OA}+\vec{OB}+\vec{OC} in any triangle ABC. So this establishes that H_P is the pseudo-orthocentre of O_1O_2O_3O_4.

import graph; size(18.52cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real...

\Box

Property 10.
If H_P is the Hewer point, O the centre of the Miquel circle, H_1' the orthocentre of O_2O_3O_4; then we have O_1,H_P,O,H_1' and H_1', H_2', O_1, O_2 form a parallelogram.

Proof.
Since H_P is the pseudo-orthocentre of O_1O_2O_3O_4, we already have that
\displaystyle \overrightarrow{OH_P}=\overrightarrow{OO_1}+\overrightarrow{OO_2}+\overrightarrow{OO_3}+\overrightarrow{OO_4}=\overrightarrow{OO_1}+\overrightarrow{OH_1'}\Longrightarrow\overrightarrow{O_1H_P}=\overrightarrow{OH_1'};
And all these similar properties give us our desired results.

import graph; size(18.52cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real...

\Box

Property 11.
If the Euler lines of the four triangles formed, ie FAB, FCD, EBC, EDA meet the Miquel circle again at points S_i, then O_1H_P\perp MS_1 and similars.

Proof.
Due to the spiral similarity that maps O_1 and H_1' to O and H_1, note that we have \angle (MO, MO_1) =\angle (OH_1' , O_1H_1) , and also \vec{OH_1'}=\vec{O_1H_P}, which leads us to the fact that O_1H_P and O_1O are isogonal conjugates wrt \angle MO_1S_1. Because O_1O passes through the circumcentre, so its isogonal passes through the orthocentre of MO_1S_1. In other words, O_1H_P\perp MS_1, leading to the fact that H_P is the point of concurrence of the four altitudes of MO_iS_i through the vertices O_i .

DIAGRAM TO BE ADDED HERE.

Property 12.
For an arbitrary quadrilateral ABCD, the nine-point circles of triangles BCD, CDA, DAB, ABC and the pedal circles of the corresponding remaining point, are concurrent.

Proof.
The proof makes use of a known property of conics, ie the pedal circle of a fixed point chosen on a conic wrt all the triangles inscribed in it pass through a fixed point. For rectangular hyperbola, this fixed point coincides with the centre. Thus every conic through the orthocentre and vertices of a triangle is a rectangular hyperbola, and because the centres of all these conics lie on the nine-point circle, therefore we are done. Refer to Ref. 4 for details. (P123.)

Special Case.
When the quadrilateral ABCD becomes cyclic, then the point of concurrency is termed as the Euler Point of the Cyclic Quadrilateral or the anticentre, in which case, the pedal circles of A,B,C degenerate into the Simson lines of A wrt BCD and similars. A nice geometric proof is posted here.

NinePointCircles

\Box

Family of Properties 13.
Let X be the Miquel point for the quadruple of lines AB, AC, BD, CD., Y be that for the quadruple of lines AB, AD, BC, CD, and Z be that of BC, AC, BD, AD. Let P_X=AD\cap BC, P_Y =AC\cap BD, P_Z=AB\cap CD. Let K_X and L_X be midpoints of the segments BC and AD respectively. Similarly, define K_Y , L_Y as midpoints of AC, BD, and K_Z, L_Z be that of AB, CD. Let \Gamma _X=K_X L_X, \Gamma _Y =K_Y L_Y, \Gamma _Z=K_Z L_Z be the Newton-Gauss lines for the corresponding quadruples of lines. Then,
Property 13(a)
The lines AX, BY, CZ concur at a point D_0.  Similarly define A_0, B_0, C_0.
Property 13(b)
A_0, B_0, C_0, D_0 are the isogonal conjugates of A, B, C, D with respect to triangle XYZ.
Property 13(c)
X, Y, Z are Miquel points for quadruples of lines joining A_0, B_0, C_0, D_0.
Property 13(d)
The lines AA_0, BB_0, CC_0, DD_0 are parallel.
Property 13(e)
The lines AD, A_0D_0, YZ are concurrent.
Property 13(f)
The points X,Z, P_Y, K_Y lie on a certain circle \omega_Y. Similarly define circles \omega _X, \omega _Z. Then, \omega_X, \omega_Y, \omega_Z have a common point T.
Property 13(g)
T is the point of concurrency of XP_X, YP_Y, ZP_Z.

Proof.
AWAITING FOOL-PROOF COMPLETE PROOFS (I have not been able to solve this yet).

Property 14.
Let the perpendicular from the midpoint of O_1H_1 to CD be denoted as m_1. Then the lines m_1, m_2, m_3, m_4 concur at a point known as the Morley point of the complete quadrilateral, and this point lies on the Miquel-Steiner line. Refer to Ref. 2 for more details.

Proof.
AWAITING PROOFS WITHOUT BARYCENTRICS (Unsolved for me as well)

Property 15.
If the Euler line of one associated triangle is parallel to the corresponding line \ell_n, then this property is true for all the four associated triangles.

Proof.(Yetti)
Let G, H be the centroid and orthocentre of a \triangle ABC and let the Euler line GH meet the sideline BC at a point D. Let a parallel b_1 to GH through $ latexA$ meet BC at C_1. Let H_1 \in AH be the orthocentre of the \triangle HDB. Then DH_1 \parallel CA (both perpendicular to BH). Since (AH_1 \equiv AH) \perp (BC \equiv BC_1) and BH_1 \perp (DH \parallel C_1A), H_1 is also the orthocentre of the \triangle ABC_1. Let a parallel to BC \equiv BC_1 through G meet the line DH_1 at a point G_1. The \triangle AC_1C \sim \triangle DGG_1 are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because G is the centroid of the \triangle ABC and their corresponding sides are oppositely oriented). Therefore, \overline{C_1C} = -3\cdot \overline{GG_1}=3 \cdot \overline{G_1G} and G_1 is the centroid of \triangle ABC_1. As a result, (G_1H_1 \equiv DH_1) \parallel CA is the Euler line of the \triangle ABC_1.

If b_2 is any other line parallel to the Euler line GH of the \triangle ABC (different from GH, b_1 \equiv C_1A), intersecting AB, BC at A_2, C_2, the \triangle A_2BC_2 \sim \triangle ABC_1 are centrally similar with center B and their Euler lines G_2H_2 \parallel G_1H_1 are parallel, hence G_2H_2 \parallel CA as well.
Euler_line GH.GIF
Let the Euler line GH of \triangle ABC meet CA, AB at points E, F. Let a parallel a' to BC through E meet AB at B'. The \triangle AB'E \sim \triangle ABC are centrally similar with center A, hence, their Euler lines G'H' \parallel GH are parallel. In exactly the same way as above, we can show that the Euler lines G'H', G_1'H_1' of the \triangle AB'E, \triangle AFE meet on AB' and that G_1'H_1' \parallel B'E \parallel BC.

\Box

Property 16.
If one of the Euler lines is parallel to the corresponding line \ell_n, ie if O_1H_1\parallel AB, then the Morley Point and the Hewer point are the same point.

Proof.
We already know that the perpendiculars through the perpendicular bisectors of O_1H_1 and similars concur at the Hewer point. And if O_1H_1\parallel AB, then by PP15, we already have that these lines are the same as the lines perpendicular to the corresponding sides, and passing through the nine-point centres of the corresponding triangles, so that Morley point and Hewer points coincide for these quadrilaterals.

\Box

Property 17.
The polar circles of the four triangles formed by \ell_1, \ell_2, \ell_3,\ell_4  and the circumcircle of the diagonal triangle GYZ are coaxal.

Proof. (taken from Ref. 3)
Obvious from some harmonic divisions. Observe that each of the four polar circles is orthogonal to the three circles with diameter AU, BV, CW. Moreover, as each of the quadruples $latex  (A, U, B’,C’),(B,V,C’,A’)$ and (C, W, A',B') is harmonic, the circle \odot(A'B'C') is orthogonal to the three circles with diameter AU, BV and CW. So this gives a shorter proof of the PP5, and also, helps us in figuring out that the orthocentric line and the pedal line are both perpendicular to the line A'B'C'.

Family of Property 18.
We have some additional results when ABCD is a cyclic quadrilateral. We will just refer to the file “Cyclic Quadrilaterals – The Big Picture” by Yufei Zhao. [Ref. 1]

REFERENCES

1. Cyclic Quadrilaterals – The Big Picture by Yufei Zhao.
2. Orthocentric Properties of the Plane n-Line by F. Morley.
3. Steiner’s Theorems on the Complete Quadrilateral by Jean-Pierre Ehrmann.
4. A Treatise on the Circle and Sphere by Julian Lowell Coolidge.
5. Some Properties of The Newton-Gauss Line by Catalin Barbu and Ion Patrascu.
6. Cut-The-Knot Page on Complete Quadrilaterals.
7. ENCYCLOPEDIA  OF  QUADRI-FIGURES by Chris Van Tienhoven (The ultimate set of properties everyone searches for).

Advertisements

4 thoughts on “On the Complete Quadrilateral Configurations

  1. It is indeed true that when circles and triangles form a special bond, lines and love are born. But if we take the radius of AD and divide it K1 of M0^2, we can see that in fact some derivatives are indeed–moe. Of course we are awaiting a proof that solidifies our claim that circles and triangles can truly coexist without causing any mathematical irregularities.

  2. Potla: circles and triangles can truly coexist without causing any mathematical irregularities.
    MarukoM: But is still moe?
    Potla: no, it won’t be planar
    Potla: Cause if a graph has a K_{3,3} or K_5 in it somwhere, then it loses planarity
    Potla: Some known theorem or something lol
    MarukoM: That indeed fits in with my theory that tsunderes are part of z plane rather than the x or y plane
    Potla: Well, they’re imaginary
    Potla: They don’t fit into the normal category of people that we humans can deal with.
    MarukoM: But since an imaginary tsundere is a character whose square is less than or equal to zero
    Potla: You need to define seperate operations to deal with them
    Potla: tsunderic addition
    Potla: tsunderic multiplication
    MarukoM: So one must go beyond just x y and z
    MarukoM: And dwell into a new region of mathematics yet unexplored, interesting
    Potla: Yeah!
    Potla: There must be some t axis
    Potla: They become time variant
    MarukoM: I see your point. I wonder if the Gauss-Bodenmiller theorem can still apply in such a new region
    Potla: That field of mathematics is kind of topology depedent
    Potla: And I am yet to explore such stuff…

  3. At first I didn’t understand this post, but now that I understand that this is about understanding tsunderes, who, of course, are ununderstandable, I’ve come to the understanding that it isn’t meant to be understood, at all. Well done.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s