## A point related to a quadrilateral

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• Dear Hyacinthists with my best greetings for 2011 for you, your family and your friends, I have a little question : Consider a quadrilateral A_1,A_2,A_3,A_4
Message 1 of 12 , Jan 1, 2011
Dear Hyacinthists
with my best greetings for 2011 for you, your family and your friends, I have a little question :
For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
Is there a special name for this point M? Do you know some references?
Friendly. Jean-Pierre
• Dear Jean-Pierre, ... I do not know a special name for point M. However I noticed this. Maybe you know it already. Let T(A_1,A_2,A_3,A_4) = Transform
Message 2 of 12 , Jan 1, 2011
Dear Jean-Pierre,

> For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
> Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
> Is there a special name for this point M? Do you know some references?
> Friendly. Jean-Pierre

I do not know a special name for point M.
However I noticed this. Maybe you know it already.
Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.

Best regards and a creative year to all Hyacinthists!

Chris van Tienhoven
• Dear Chris [JP] ... [Chris] ... Thank you for your nice remark. In fact, if O_1 is the circumcenter of O_2O_3O_4,..., the same homothecy maps A_i to O_i and
Message 3 of 12 , Jan 2, 2011
Dear Chris
[JP]
> > Consider a quadrilateral A_1,A_2,A_3,A_4
> > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
> > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
> > Is there a special name for this point M? Do you know some references?
[Chris]
> I do not know a special name for point M.
> However I noticed this. Maybe you know it already.
> Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
> Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
> T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.

Thank you for your nice remark.
In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
the same homothecy maps A_i to O_i' and B_i to O_i
The point M is characterized by the angular relations
<A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is any permutation of (1,2,3,4)
Friendly. Jean-Pierre
• Dear Jean-Pierre, ... Thank you for your comments. They helped me to search further. 1. The angular relations can be described some more specific as: Aimj =
Message 4 of 12 , Jan 2, 2011
Dear Jean-Pierre,

> [JP]
> > > Consider a quadrilateral A_1,A_2,A_3,A_4
> > > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
> > > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
> > > Is there a special name for this point M? Do you know some references?
> [Chris]
> > I do not know a special name for point M.
> > However I noticed this. Maybe you know it already.
> > Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
> > Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
> > T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.
> [JP]
> In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
> the same homothecy maps A_i to O_i' and B_i to O_i
> The point M is characterized by the angular relations
> <A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is any permutation of (1,2,3,4)

They helped me to search further.
1. The angular relations can be described some more specific as:
Aimj = Aikj + Ailj,
where (i,j,k,l) is any permutation of (1,2,3,4) and
Aimj = Angle Ai.M.Aj
Aikl = Angle Ai.Ak.Aj when A_k on same side of line Ai.Aj as M
Pi - Angle Ai.Ak.Aj when Ak on same side of line Ai.Aj as M
Ailj = Angle Ai.Al.Aj when Al on same side of line Ai.Aj as M
Pi - Angle Ai.Al.Aj when Al on other side of line Ai.Aj as M

2. Conics M.A1.A2.A3.A4 and M.O1*.O2*.O3*.O4* (conics are ellipses when A1.A2.A3.A4 is convex) are mutually tangent in M.
Conics M.B1.B2.B3.B4 and M.O1.O2.O3.O4 (conics are ellipses when A1.A2.A3.A4 is convex) are mutually tangent in M.

3. Let Ma and Mo* be the Centers of Conics M.A1.A2.A3.A4 and M.O1*.O2*.O3*.O4*.
Let Mb and Mo be the Centers of Conics M.B1.B2.B3.B4 and M.O1.O2.O3.O4.
Now M, Ma, Mo* are collinear and M, Mb, Mo are also collinear.
Further M.Mo* / M.Ma = M.Mo / M.Mb.

4. General rules:
Let T(A1,A2,A3,A4) = Transform A1,A2,A3,A4 --> O1,O2,O3,O4.
Now it comes out that the corresponding circumscribed conics (after T^2-transform) are tangent im M.
Also it appears that there is a fixed similitude-ratio M.Mo*/M.Ma for transform T^2.
Best regards,

Chris van Tienhoven
• Dear Jean-Pierre, I am quite thrilled. As always when I find something beautiful. You described a point M depending on the 4 points A1, A2, A3, A4 of a
Message 5 of 12 , Jan 2, 2011
Dear Jean-Pierre,

I am quite thrilled.
As always when I find something beautiful.
You described a point M depending on the 4 points A1, A2, A3, A4 of a quadrilateral.
Here is the structure behind it.

M = the inverse in the circumcircle of AiAjAk of the isogonal conjugate of Al wrt AiAjAk,
where (i,j,k,l) is any permutation of (1,2,3,4).

Strangely enough this is true for each permutation.
There should be more properties/references relating to this point.
Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
>
>
> Dear Chris
> [JP]
> > > Consider a quadrilateral A_1,A_2,A_3,A_4
> > > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
> > > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
> > > Is there a special name for this point M? Do you know some references?
> [Chris]
> > I do not know a special name for point M.
> > However I noticed this. Maybe you know it already.
> > Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
> > Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
> > T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.
>
> Thank you for your nice remark.
> In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
> the same homothecy maps A_i to O_i' and B_i to O_i
> The point M is characterized by the angular relations
> <A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is any permutation of (1,2,3,4)
> Friendly. Jean-Pierre
>
• Dear Hyacinthists, Happy New Year to all of you. I think we must say something beyond mathematics. I don t care about 2012. I think that this year 2011 comes
Message 6 of 12 , Jan 3, 2011
Dear Hyacinthists,
Happy New Year to all of you.
I think we must say something beyond mathematics.
I think that this year 2011 comes with more semiology.
I think mathematically that 2011 is a subversive year.
2^0 = 1^1.
Begins with one <two> and ends with two <one>.
The known second <two> becomes first and the known first <one> becomes last.
This year puts things upside down.
We can see this in Greece because of wage cuts.
Our 1821 revolution comes to an end.
We'' ll set Greece on fire and
we'' ll spread the light of this fire in all European Union.
But don't be afraid.
Greece never sinks, is like an utricle in ocean.
Greece never dies, is always in intensive teatment unit.
Let us celebrate the beginning of the cycle of this year
by ETC cyclic points exploration.

There are many thousands of cases of concyclic points in ETC.
If we exclude the case of points on the circumcircle or incircle or NPC
known cases are the first Lester circle with points X(3), X(5), X(13), X(14)
or the second Lester circle with points 6, 11, 19, 33.
We'' ll see the case of concyclic points in ETC with X(1) the incenter of ABC.
We consider the special cases of two octagons, five heptagons and six hexagons.
I have checked the case
8a) that is an octagon with circle equation
aayz + bbzx + ccxy - (x + y + z)(Px + Qy + Rz) where
P = -bc(b+c-a)(bb+cc-ab-ac)/ [2(a+b+c)(c-a)(a-b)] and
Q, R the cyclic expressions of P, with center
( a(b-c)(bb + cc - aa - bc) : . . : . . ) in barycentrics not in ETC and case

6a) with center the point X(764).
I have not checked the other cases but I strongly believe that are correct.

8a) 1 11 105 109 1054 1155 1362 2254
8b) 1 11 106 108 1319 1361 1769 1785
7a) 1 4 65 1359 1537 1845 3326
7b) 1 100 1155 1282 1357 1477 3021
7c) 1 103 108 1155 1360 1364 1768
7d) 1 105 109 1054 1155 1362 2254
7e) 1 934 1155 1156 2291 3022 3321
6a) 1 2 105 165 901 1022
6b) 1 105 108 243 1283 2078
6c) 1 105 110 147 149 2651
6d) 1 840 1155 2222 3025 3322
6e) 1 934 1319 1358 1362 1477
6f) 1 1319 2718 2720 3025 3319

Finally other examples with ETC cyclic points X(n), n < 1000 are:

1) 1 2 11 392
2) 1 2 23 36
3) 1 3 21 759
4) 1 3 101 840
5) 1 3 106 901
6) 1 3 109 953
7) 1 3 110 501
8) 1 4 36 186
9) 1 4 46 915
10) 1 4 108 953
11) 1 6 36 187
12) 1 6 106 919
13) 1 7 80 885
14) 1 7 840 934
15) 1 8 40 104 901
16) 1 9 104 885 919
17) 1 9 115 442
18) 1 10 40 764
19) 1 10 119 946
20) 1 10 759 901
21) 1 13 14 79
22) 1 15 16 36
23) 1 15 106 202
24) 1 16 106 203
25) 1 22 36 858
26) 1 24 36 403
27) 1 25 36 468
28) 1 33 103 222
29) 1 36 104 108
30) 1 36 105 934
31) 1 36 106 109
32) 1 36 352 353
33) 1 37 759 919
34) 1 40 100 944 953
35) 1 40 104 901
36) 1 40 944 953
37) 1 42 741 901
38) 1 43 727 901
39) 1 55 100 956
40) 1 55 109 840
41) 1 57 105 497
42) 1 57 108 840
43) 1 58 106 501
44) 1 58 741 991
45) 1 74 108 484
46) 1 74 109 399
47) 1 75 759 876
48) 1 78 102 901
49) 1 79 484 501
50) 1 80 109 484 759
51) 1 80 484 759
52) 1 100 165 840
53) 1 100 944 953
54) 1 101 997 999
55) 1 102 219 919
56) 1 103 200 901
57) 1 103 220 919
58) 1 104 125 656
59) 1 104 885 919
60) 1 105 108 243
61) 1 105 390 885
62) 1 106 110 399
63) 1 109 119 915
64) 1 109 243 917
65) 1 109 484 759
66) 1 109 651 934
67) 1 110 147 149
68) 1 110 354 392
69) 1 115 119 399
70) 1 187 663 991
71) 1 213 741 919
72) 1 295 741 932
73) 1 317 435 953
74) 1 731 869 901
75) 1 761 919 984

Best regards
• Dear Nikos [ND] ... TI ROMIOSINI MIN TIN KLAIS ... I think that the following search in ETC would be interesting
Message 7 of 12 , Jan 4, 2011
Dear Nikos

[ND]

> We can see this in Greece because of wage cuts.
> Our 1821 revolution comes to an end.
> We'' ll set Greece on fire and
> we'' ll spread the light of this fire in all European Union.
> But don't be afraid.
> Greece never sinks, is like an utricle in ocean.
> Greece never dies, is always in intensive teatment unit.
>
TI ROMIOSINI MIN TIN KLAIS <http://www.youtube.com/watch?v=vZjtDRVJwdI>

> Let us celebrate the beginning of the cycle of this year
> by ETC cyclic points exploration.
>
>
>
I think that the following search in ETC would be interesting but I guess
it is not easy!

Let (A), (B), (C) be three circles defined with known points listed in ETC
such that they do not have a common KNOWN point.

Not like these circles, for example:

1) 1 2 11 392
2) 1 2 23 36
3) 1 3 21 759

Are there such circles which are concurrent?

If there are, then the point of concurrence is a NEW interesting point !

With my Best Wishes for the New Year

Antreas

[Non-text portions of this message have been removed]
• Dear Nikos and friends, I don t know if this set of 10 cyclic points is known: X(11) X(36) X(65) X(80) X(108) X(759) X(1354) X(1845) X(2588) X(2589) Best
Message 8 of 12 , Jan 4, 2011
Dear Nikos and friends,

I don't know if this set of 10 cyclic points is known:
X(11)
X(36)
X(65)
X(80)
X(108)
X(759)
X(1354)
X(1845)
X(2588)
X(2589)

Best Regards,

Chris

> Let us celebrate the beginning of the cycle of this year
> by ETC cyclic points exploration.
• Dear Chris, very good case. I didn t know it. I had worked only with cases with X(1). Now with my computer seems that your points are indeed concyclic. The
Message 9 of 12 , Jan 5, 2011
Dear Chris,
very good case.
I didn't know it.
I had worked only with cases with X(1).
Now with my computer seems that your points
are indeed concyclic.
The center of this circle I think
is the point with barycentrics
( (b-c)(abc(b+c-a)-(cc+aa-bb)(aa+bb-cc) : .. :..)
not in ETC.
Best regards

> Dear Nikos and friends,
>
> I don't know if this set of 10 cyclic points is known:
> X(11)
> X(36)
> X(65)
> X(80)
> X(108)
> X(759)
> X(1354)
> X(1845)
> X(2588)
> X(2589)
>
> Best Regards,
>
> Chris
>
> > Let us celebrate the beginning of the cycle of this
> year
> > by ETC cyclic points exploration.
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear Antreas, ... Yes. It is not easy. ... Yes. There are. In order to reduce the time of calculation by the computer one of the three circles can be the
Message 10 of 12 , Jan 5, 2011
Dear Antreas,
You wrote:
> I think that the following search in ETC  would be
> interesting but I guess
> it is not easy!

Yes. It is not easy.

>
> Let (A), (B), (C) be three circles defined with known
> points listed in ETC
> such that they do not  have a common KNOWN point.
>
> Not like these circles, for example:
>
> 1) 1 2 11 392
> 2) 1 2 23 36
> 3) 1 3 21 759
>
> Are there such circles which are concurrent?

Yes. There are.
In order to reduce the time of calculation by the
computer one of the three circles can be the
circumcircle (O).
Hence one example is the circles
(O) (11, 13, 14), (36, 65, 79)
with common point P not in ETC and a second is
(O), (11, 13, 14), (46, 57, 79)
with the same common point P.
Hence we have found 4 circles concurrent at P.
(11, 13, 14) means the circle passing through
the two Fermat points X(13), X(14)
and the Feuerbach point X(11).

>
> If there are, then  the point of concurrence is a NEW
> interesting point !

I think that this point has not simple coordinates.

Best regards
Nikos
• ... I did this using complex numbers as coordinates, where the formula for M should be symmetric in the 4 complex variables (A1,A2,A3,A4), and have some
Message 11 of 12 , Jan 9, 2011
> Dear Jean-Pierre,
>
> I am quite thrilled.
> As always when I find something beautiful.
> You described a point M depending on the 4 points A1, A2,
> A3, A4 of a quadrilateral. Here is the structure behind it.
>
> M = the inverse in the circumcircle of AiAjAk of the
> isogonal conjugate of Al wrt AiAjAk,
> where (i,j,k,l) is any permutation of (1,2,3,4).
>
> Strangely enough this is true for each permutation.
> There should be more properties/references relating to this point.
> Best regards,
>
> Chris van Tienhoven

I did this using complex numbers as coordinates, where the formula for M should be symmetric in the 4 complex variables (A1,A2,A3,A4), and have some invariance properties. The answer can be expressed as follows:

M is a quotient of two determinants. Each determinant is 4-by-4, and its rows correspond to the variables. In the numerator, the row for x is [1, x, x^2, x x#], where x# means the complex conjugate of x. In the denominator, the row for x is [1, x, x^2, x#].

Can other ETC points be expressed in this way?
--
Barry Wolk
• Dear Jean-Pierre I add some circular properties of your drawing. There is a direct circular map f: A_1 -- B_1, A_2 -- B_2, A_3 -- B_3, A_4 -- B_4 and this
Message 12 of 12 , Feb 25, 2011
Dear Jean-Pierre