Τρίτη, 26 Φεβρουαρίου 2013

CONCURRENT CIRCUMCIRCLES

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

B'a, C'a = the reflections of B',C' in AA', resp.

A'a = (perpendicular to BB' from B'a) /\ (perpendicular to CC' from C'a).

C'b, A'b = the reflections of C',A' in BB', resp.

B'b = (perpendicular to CC' from C'b) /\ (perpendicular to AA' from A'b).

A'c, B'c = the reflections of A',B' in CC', resp.

C'c = (perpendicular to AA' from A'c) /\ (perpendicular to BB' from B'c).

The circumcircles of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c are concurrent (?).

Let Ia,Ib,Ic be the incenters of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c, resp.

Are the triangles ABC, IaIbIc perspective?

Antreas P. Hatzipolakis, 27 Feb. 2013

CIRCLES OF QUADRILATERAL

PROBLEM:

Let ABCD be a quadrilateral. To construct two EQUAL and TANGENT circles (K),(L) such that: (K) touches the sides of angleA, and (L) the sides of angleC of ABCD.

Analysis:

K lies on the bisector of angA, and L on the bisector of angC (K,L: the centers of (K),and (L), resp.) Let O be the intersection point of these bisectors. Draw KQ _|_ AB, LI _|_ BC and OZ _|_ AB, OE _|_ BC (Q, Z on AB; I,E on BC).

We have: KQ = LI, and KL = KQ + LI = 2KQ = 2LI (1)

AQK ~ AZO ==> KQ/AK = OZ/OA (2)

Draw Ox // KL, and let M be the intersection point of Ox,AL.

AKL ~ AOM ==> KA/KL = OA/OM (3)

[(1) and (3)] ==> KA/2KQ = OA/OM (4)

[(2) and (4)] ==> OA/2OZ = OA/OM ==> OM = 2OZ.

Draw My // AC, and let H be the intersection point of My and CO.

CLI ~ COE ==> CL/LI = CO/OE (5)

AC // MH ==> CL/CH = AL/AM = KL/OM (6)

(5) ==> CL = LI*OC/OE (7)

(6) and (7) ==> LI* OC/OE / CH = KL/OM (and since KL = 2LI) ==> OC/2OE = CH/OM ==> (since OM = 2OZ) 2OE/OC = 2OZ/CH ==> OE/OC = OZ/CH, that is CH is the 4th proportional of OE,OC,OZ.

Synthesis:

Let O be the intersection point of bisector angA and bisector of angC. Draw OZ _|_ AB, OE _|_ BC. Draw the circle (O, 2OZ). Take on OC the segment CH such that CH = the 4th proportional of OE,OC,OZ. Draw Hy//AC, which intersects the circle (O, 2OZ) at M near to C. Draw MA, intersecting OC at L. Draw Lx//OM, intersecting OA at K. The points K and L are the centers of the circles in question.

Proof: For the reader

Bibliography:

ΠΑΡΑΡΤΗΜΑ του ΔΕΛΤΙΟΥ της ΕΛΛΗΝΙΚΗΣ ΜΑΘΗΜΑΤΙΚΗΣ ΕΤΑΙΡΕΙΑΣ [=Supplement of the Bulletin of the Greek Mathematical Society], March 1967, pp. 160 - 161

Δευτέρα, 25 Φεβρουαρίου 2013

FEUERBACH POINT

Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal triangles of I, resp.

Denote:

Ab, Ac = the reflections of A' in BB', CC'

Bc, Ba = the reflections of B' in CC', AA'

Ca, Cb = the reflections of C' in AA', BB'

Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at I)

1. Antipode of Feuerbach Point.

Denote:

A2,A3 = the orthogonal projections of A on Eb, Ec, resp.

B3,B1 = the orthogonal projections of B on Ec, Ea, resp.

C1,C2 = the orthogonal projections of C on Ea, Eb, resp.

The Euler lines L1,L2,L3 of AA2A3, BB3B1, CC1C2, resp. are concurrent at the antipode of the Feuerbach point.

Note:

Denote:

12, 13 = the orthogonal projections of A on BB', CC', resp.

23, 21 = the orthogonal projections of B on CC', AA', resp.

31, 32 = the orthogonal projections of C on AA', BB', resp.

The Euler lines of A1213, B2321, C3132 are concurrent at Feuerbach point

(APH, Hyacinthos)

2. Antipode of Feuerbach Point.

Denote:

La = the parallel to Ea through A"

Lb = the parallel to Eb through B"

Lc = the parallel to Ec through C"

The La,Lb,Lc are concurrent at the antipode of the Feuerbach point.

Antreas P. Hatzipolakis, 25 Febr. 2013

Παρασκευή, 22 Φεβρουαρίου 2013

REFLECTING THE CIRCLES (N,NA), (N,NB), (N,NC) ON THE ALTITUDES

Let ABC be a triangle and HaHbHc the orthic triangle.

Denote:

(Na) = the reflection of the circle (N,NA) in AHa

(Nb) = the reflection of the circle (N,NB) in BHb

(Nc) = the reflection of the circle (N,NC) in CHc

1. The circles (Na),(Nb),(Nc) are concurrent at a point P

2. The triangles HaHbHc, NaNbNc are perspective at a point Q.

Antreas P. Hatzipolakis, 22 Febr. 2013

Τετάρτη, 20 Φεβρουαρίου 2013

EQUILATERAL TRIANGLE AND CONCURRENT CENTRAL LINES

Let ABC be an equilateral triangle and P a point.

Which same central lines of the triangles PBC, PCA, PAB

are concurrent for all P's ? The OH (Euler) lines? The OI lines? ......

Antreas P. Hatzipolakis, Hyacinthos #21368

Central lines Ln (OH,OK,OI....) as described in the paper:

Antreas P. Hatzipolakis, Floor van Lamoen, Barry Wolk, and Paul Yiu: Concurrency of Four Euler Lines, Forum Geometricorum Volume 1 (2001) 59–68 [pdf file]

Euler Lines


The Euler line of equilateral is any line passing through the center of the triangle.

So the Euler Lines of ABC, PBC, PCA (and also of ABC, PCA, PAB and of ABC, PAB, PBC) are concurrent on the Euler line of PBC and therefore the Euler lines of PBC, PCA, PAB are concurrent..

Σάββατο, 16 Φεβρουαρίου 2013

A SEQUENCE OF POINTS



hL Line:
Let ABC be a triangle, L a line passing through H, A'B'C' the pedal triangle of H (orthic triangle), A"B"C" the circumcevian triangle of H wrt A'B'C' (aka Euler triangle) and A1,B1,C1 the orthogonal projections of A,B,C, on L, resp. The lines A"A1, B"B1, C"C1 concur at point H11 on the pedal circle of H (NPC of ABC).
Let La,Lb,Lc be the parallels through A,B,C, to A"A1, B"B1, C"C1, resp. They are concurrent at point H12 on the circumcircle of ABC.
The line H11H12 passes through H. Let's call it hL


Concurrent Euler Lines:
Let A2,B2,C2 be the orthogonal projections of A,B,C on hL. The Euler lines of AA1A2, BB1B2, CC1C2 are concurrent at point P1.
Similarly we define hhL, hhhL...... h^nL, and we get a sequence of points Pn: P1 from L and hL, P2 from hL and hhL, etc

The parallels through A,B,C to the Euler lines of AA1A2, BB1B2, CC1C2 concur at point Q1.

Antreas P. Hatzipolakis, 16 Feb. 2013

Τετάρτη, 13 Φεβρουαρίου 2013

REFLECTING PEDAL CIRCLE


Let ABC be a triangle, Q a point, QaQbQc the pedal triangle of Q, (X) the circumcircle of QaQbQc (=pedal circle of Q), P a point on the OQ line and PaPbPc the pedal triangle of P.

Denote:

(X1), (X2), (X3) the reflections of (X) in OPa,OPb,OPc, resp.

A'B'C' = the triangle bounded by the radical axes of ((O),(X1)),((O),(X2)),((O),(X3))

The triangles A'B'C', X1X2X3 are perspective at a point Qp.

Antreas P. Hatzipolakis, 13 Feb. 2013

Δευτέρα, 11 Φεβρουαρίου 2013

REFLECTING NPCs

Let ABC be a triangle.

1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in the sidelines BC,CA,AB, resp. and A'B'C' the triangle bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
The triangles ABC, A'B'C' are perspective.


2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections of the NPC (N) in the altitudes HHa,HHb,HHc, resp. and and A'B'C' the triangle bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
The triangles HaHbHc, A'B'C' are perspective.

3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3) the reflections of the NPC (N) in the perp. bisectors OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
The triangles ABC, A'B'C' are perspective.

Antreas P. Hatzipolakis, 11 Febr. 2013

Κυριακή, 10 Φεβρουαρίου 2013

A PONCELET POINT ON THE EULER LINE


Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of ABC,NBC,NCA,NAB, resp. [concurrent at pN]. The NPCs of the triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur at point ppN on the Euler Line of ABC.

Generalization:

Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs of ABC, PBC, PCA, PAB resp. [concurrent at pP]. If P is on the Euler line of ABC, then the NPCs of N1N2N3, PN1N2, PN2N3, PN3N1 concur at point ppP on the Euler Line of ABC.(??)

Antreas P. Hatzipolakis, 10 Febr. 2013

*********************************************

pP is the center of the rectangular circumhyperbola through P.

ppP does not, in general, lie on the Euler line.

Some results:

P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on lines 1,5 3,962).

P=X(2), ppP=non-ETC 1.690358502447462

P=X(3), ppP=X(140)

P=X(4), ppP=undefined

P=X(5), ppP=X(3628)

P=X(6), ppP=non ETC 0.780037257060191

P=X(7), ppP=non ETC 0.750876768572663

P=X(8), ppP=non ETC 2.966801160450799

P=X(9), ppP=non-ETC 0.972023454564163

P=X(10), ppP=non-ETC 2.238481946743318

P=X(20), ppP=non-ETC 6.363850996796102

P=X(21), ppP=non-ETC -1.717011738240629

P=X(22), ppP=non-ETC -4.036288926987237

Of these, only X(140) and X(3628) lie on the Euler line.Β The ppP for points P on the Euler line do not even lie on the same conic.

Locus?

Randy Hutson, Hyacinthos #21519


Παρασκευή, 8 Φεβρουαρίου 2013

FOUR CONCURRENT LINES, CIRCUMCIRCLES


Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of ABC, IBC, ICA, IAB, resp. and S a point.

Denote:

L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. For which S's the triangles ABC, O1O2O3 are perspective? (Locus)

2. The circumcenter of O1O2O3 is on the line L for all S's (??).

Special Case: L,La,Lb,Lc = Brocard axes

Antreas P. Hatzipolakis, 8 Febr. 2013

************************************************************

Brocard Axes instead of Euler Lines case:

1. The triangles ABC, O1O2O3 are not perspective.

O1O2O3 appears to be perspective to

1/ excentral at

a (3 a^3+2 a^2 b+b^3+2 a^2 c+3 a b c+c^3):: = (r^2-3 s^2) X[1]+4 r (2 r+3 R) X[21] = (3 r^2-s^2) X[171] + 4 r R X[3753],

on lines {{1,21},{36,199},{171,3753},{210,5247},{511,3576},{740,4234},{976,4661},{978,1453},{986,4252},{1104,3742},{1125,1330},{1193,4881},{1247,2363},{1757,4134},{1961,5251},{2308,4511},{2938,4221},{3454,3624},{3880,5255}}

ETC search = 1.0647243333936774920

2/ intouch

3/ Hexyl,

4/ Yff

5/ cicrumcircleMidArc at X(58)

6/ firstCircumPerp

7/ second CircumPerp at X(58)

....

2. The circumcenter of O1O2O3 is on the line L.

Yes, at

a^2 (2 a^5-3 a^3 b^2+a^2 b^3+a b^4-b^5+2 a^3 b c-a^2 b^2 c-2 a b^3 c+b^4 c-3 a^3 c^2-a^2 b c^2-2 b^3 c^2+a^2 c^3-2 a b c^3-2 b^2 c^3+a c^4+b c^4-c^5)::

on lines {{3,6},{140,3454},{540,549},{631,1330},{758,1385},{1046,3576},{1125,2792},{1511,2842},{3794,4189}}.

also P = (4r^2-SW) X[3]-SW X[6]

ETC Search = 3.7671475697791725975

Peter J. C. Moses

ETC X5429

************************************************************

Generalization

1. The locus is the conic:

b^3 c x^2 - b c^3 x^2 - 2 a^3 c x y - a^2 b c x y + a b^2 c x y + 2 b^3 c x y + 2 a c^3 x y - 2 b c^3 x y - a^3 c y^2 + a c^3 y^2 + 2 a^3 b x z - 2 a b^3 x z + a^2 b c x z + 2 b^3 c x z - a b c^2 x z - 2 b c^3 x z + 2 a^3 b y z - 2 a b^3 y z - 2 a^3 c y z - a b^2 c y z + a b c^2 y z + 2 a c^3 y z + a^3 b z^2 - a b^3 z^2=0

2, Yes . The circumcenter of O1O2O3 is on the line L for all S's.

The three centers Oa, Ob, Oc are the mid points of the arcs BC, CA, AB of the circumcircle of ABC. The points O1, O2, O3 are the mid points of SOa, SOb, SOc. If K is the mid point of SO then KO1 = OOa/2 = R/2. Hence K is the center of the circle O1O2O3 with radious R/2.

Nikos Dergiades

Πέμπτη, 7 Φεβρουαρίου 2013

EULER LINES, CIRCUMCIRCLES


Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp. (concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.

Denote:

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. The triangles ABC, O1O2O3 are perspective.

2. The circumcenter of O1O2O3 is on the line L.

Antreas P. Hatzipolakis, 8 Febr. 2013

*******************************************

1. The triangles ABC, O1O2O3 are perspective.

Yes at

Q = a (2 a^3-2 a^2 b-2 a b^2+2 b^3-a^2 c-a b c-b^2 c-2 a c^2-2 b c^2+c^3) (2 a^3-a^2 b-2 a b^2+b^3-2 a^2 c-a b c-2 b^2 c-2 a c^2-b c^2+2 c^3)::

ETC Search = 2.1298183373048648210

Q is on lines {{21,4867},{79,2646},{80,3584},{758,2320},{1389,3746}} and on the Feuerbach hyperbola.

Q = 6 (r+R) X[21] +(2 r-R) X[4867]

Q = R X[79] + 4 (2 r+R) X[2646]

The isogonal of Q, gQ, is on lines {{1,3},{2,4867},{8,3841},{21,4084},{79,3671},{80,226},{81,759},{100,3919},{191,4018},{515,3982},{519,5249},{758,3219},{956,3894},{958,3901},{993,4880},{1001,3899},{1100,5341},{1203,3924},{1411,2003},{2802,3957},{3585,3649},{3636,5330},{3868,5258},{3869,5259},{3874,5288},{3881,4861},{3918,4420},{4067,5260}};

gQ = (4 r + 5 R)X[1] - 2 r X[3]

gQ = (2 r + 3 R) X[21] + 2 R X[4084]

ETC Search = 1.2648627122986571860

O1O2O3 is also perspective to various other triangles too; for example:

1. Excentral at

a (3 a^3-a^2 b-3 a b^2+b^3-a^2 c-3 a b c-3 b^2 c-3 a c^2-3 b c^2+c^3) :: = X[1] + 2 X[21]

on lines {{1,21},{30,1699},{35,3753},{36,3742},{100,3968},{210,5251},{214,5284},{442,3586},{484,3919},{1125,2475},{1420,3649},{1698,1837},{2320,3065},{2646,5259},{3158,3679},{3219,4525},{3336,4189},{3337,5267},{3616,4299},{3636,3648},{3683,4867},{3746,3880},{3956,5260},{4316,5249},{4539,5302},{4677,4933}}. ETC search = 2.6815474100289784017

2. Intouch at

a (a+b-c) (a-b+c) (2 a^4-2 a^3 b-2 a^2 b^2+2 a b^3-2 a^3 c-2 a^2 b c+b^3 c-2 a^2 c^2+2 b^2 c^2+2 a c^3+b c^3) :: R (r+4 R) X[7] - r (4 r+7 R) X[21]

on lines {{7,21},{11,30},{12,5251},{100,5172},{191,1420},{392,3647},{758,1319},{1317,2078},{1411,1758},{2771,5126},{3651,5204},{4189,5221}}. ETC search -1.0652570342486130228

2. The circumcenter of O1O2O3 is on the line L.

Yes, at

P = a (2 a^6-2 a^5 b-4 a^4 b^2+4 a^3 b^3+2 a^2 b^4-2 a b^5-2 a^5 c+a^2 b^3 c+2 a b^4 c-b^5 c-4 a^4 c^2+4 a^2 b^2 c^2+4 a^3 c^3+a^2 b c^3+2 b^3 c^3+2 a^2 c^4+2 a b c^4-2 a c^5-b c^5)::

on lines {{2,3},{36,3649},{191,3576},{214,960},{758,1385},{952,5258},{1837,5010},{3579,3754},{3650,5303}}

also

P = 3 R X[2] + (4 r+3 R) X[3]

P = (2 r - R) X[36] + R X[3649]

P = X[191] + 3 X[3576]

P = (2 r +R) X[3579] + 2 R X[3754]

P = R X[3650] - (6 r+R) X[5303]

ETC Search = 4.9797648772556482798.

Peter J. C. Moses, 8 Febr. 2013

ETC X5424, X5426, X5427, X5428

EULER LINE [Generalization]


Let ABC be a triangle, T a fixed point, AtBtCt the circumcevian triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

Antreas P. Hatzipolakis, 7 Feb. 2013

*****************

The locus is the line OT, and the perspector Q lies on OT as well.

If OP : PT = t: 1-t, then

OQ : QT = R^2(1+t) : - (R^2-OT^2) t.

Paul Yiu, Hyacinthos #21597

EULER LINE

Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

Antreas P. Hatzipolakis, 7 Feb. 2013

****************************

A ' = (a^2:-(b^2-c^2):b^2-c^2),

If P = (u:v:w), then

A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) : -c^2(v+w)((b^2-c^2)u-a^2w))

etc.

A''B''C'' and ABC are perspective if and only if P lies on the Euler line. The perspector Q also lies on the Euler line.

If OP P PH = t : 1-t, then OQ : QH = (1+t) : -8t\cos A\cos B\cos C Here are some examples:

P Q

------------

G X(25)

O H

H X(24)

N X(3518)

L O

X(21) X(28)

X(22) G

X(23) X(468)

X(186 X(403)

Paul Yiu, Hyacinthos #12503

Κυριακή, 3 Φεβρουαρίου 2013

REFLECTIONS OF AI,BI,CI IN A LINE THROUGH I


Let ABC be a triangle, L a line through I and La,Lb,Lc the reflections of AI, BI, CI in L.

Denote:

Ab,Ac = the orthogonal projections of A on Lb,Lc, resp.

Bc,Ba = the orthogonal projections of B on Lc,La, resp.

Ca,Cb = the orthogonal projections of C on La,Lb, resp.

The Euler lines of AAbAc, BBcBa, CCaCb are concurrent on a point Q on the pedal circle of I (= incircle of ABC)

Antreas P. Hatzipolakis, 4 Febr. 2013

 

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