ADGEOM 5139 * ADGEOM 5140 * ADGEOM 5141 * ADGEOM 5142

#5139

Let ABC be a triangle.

 

OI line of ABC meets BC, CA, AB at A', B', C'.

 

Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

 

Then X(56) of A"B"C" lies on OI line of ABC.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#5140

X(56) of A"B"C" lies on OI line of ABC, is X(23981)

 

 

Let P1 and P2 be the two points on line X(1)X(3) whose trilinear polars are parallel to line X(1)X(3). X(23981) is the barycentric product P1*P2.

 

Angel Montesdeoca

 

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#5141

 

Thank you very much. I see the same problem, I hope you will help me again.

 

Let ABC be a triangle.

 

X(1)X(4) line of ABC meets BC, CA, AB at A', B', C'.

 

Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

 

Then X(84) of A"B"C" lies on OI line of ABC.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#5142

[Tran Quang Hung]:
    
    Let ABC be a triangle.

    X(1)X(4) line of ABC meets BC, CA, AB at A', B', C'.

    Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

    Then X(84) of A"B"C" lies on OI line of ABC.

    Which is this point?

*** W =  X(1)X(3) /\ X(109)X(13138)

 X(1) = incenter
 X(3) = circumcenter
 
 X(109) = Point where the line that passes through incenter and the antipode of anticomplement of Feuerbach point intersects the circumcircle again.
 X(13138) =  Trilinear pole of the line that passes through circumcenter and  Mittenpunkt.
 
W = a (a-b) (a^2-(b-c)^2) (a-c) (2 a^7-a^6 (b+c)+a^4 (b-c)^2 (b+c)-(b-c)^2 (b+c)^5-6 a^3 (b^2-c^2)^2+4 a (b^2-c^2)^2 (b^2+c^2)+a^2 (b-c)^2 (b^3+7 b^2 c+7 b c^2+c^3)) : ... : ...

Angel Montesdeoca

 

 

 

 

ADGEOM 5028 * ADGEOM 5029

#5028

Dear friends, "Circumconics with asymptotes making a given angle" is the title of my recent article published in Forum Geometricorum

 

An update can be found at my personal blog. This update introduces a new center, the focus of a parabola:

 

Briefly, the locus of perspectors of circumconics whose asymptotes make a given angle is a conic. This conic becomes a parabola for some particular angle related to Brocard angle. The focus F of this parabola is the point I introduce. Some properties of this point are:

 

First barycentric coordinate: a^2*(2*a^6 - b^6 - c^6 - 3*a^4*(b^2 + c^2) - 6*a^2*(b^4 - 3*b^2*c^2 + c^4)) : :

 

ETC search numbers: {1.10017518584542021, 4.49336719756577036, 0.0220986439717287574}

 

On lines X(i)X(j) for these {i,j}: {{6,110},{126,3589},{141,6719},{182,14688},{187,1084},{511,14650},{518,11721},{524,5914},{543,597},{1296,5085},{1428,3325},{1503,5512},{1576,21309},{2330,6019},{2492,2780},{3618,14360},{5027,6088},{5050,11258},{5166,9019},{5480,23699},{6094,11166},{6096,21448},{10748,14561},{14654,14853},{14666,20423}}

 

Combos: {X[126]-2*X[3589], X[141]-2*X[6719], 2*X[182]-X[14688], X[1296]-3*X[5085], 5*X[3618]-X[14360], 3*X[5050]+X[11258], X[10748]-3*X[14561], X[14654]+3*X[14853], X[14666]+X[20423]}

 

Midpoint of X(i) and X(j) for these {i,j}: {{6,111},{14666,20423}}

 

Reflection of X(i) in X(j) for these {i,j}: {{126,3589},{141,6719},{14688,182}}

 

Best regards,

Francisco Javier Garcia Capitan

 

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#5029

Hi Francisco Javier,

 

Another property of this point: it is the radical trace of circles {{X(6),X(13),X(16)}} and {{X(6),X(14),X(15)}}.

 

Best regards,

Randy Hurson

ADGEOM 4996 * ADGEOM 4999 * ADGEOM 5001 * ADGEOM 5004

#4996

Let ABC be a triangle.

 

We construct three Malfatti squares of a ABC as in figure.

 

A',B',C' are centers of these square.

 

Sides of squares bound triangle A''B''C''.

 

Then Euler lines of ABC, A'B'C', and A''B''C'' are concurrent. Which is this point wrt ABC, A'B'C' and A''B''C''?

 

Best regards,

Tran Quang Hung.

 

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#4999

Dear Tran Quang Hung

Very complicated coordinates

-61849*a^18 - 37319*a^16*(b^2 + c^2) + 8*a^14*(203696*b^4 + 447393*b^2*c^2 + 203696*c^4) - 8*a^12*(365409*b^6 + 669475*b^4*c^2 + 669475*b^2*c^4 + 365409*c^6) - 2*a^10*(132509*b^8 + 6082112*b^6*c^2 + 11389446*b^4*c^4 + 6082112*b^2*c^6 + 132509*c^8) + 2*a^8*(1610089*b^10 + 7337661*b^8* c^2 + 7479514*b^6*c^4 + 7479514*b^4*c^6 + 7337661*b^2*c^8 + 1610089*c^10) - 8*a^6*(180261*b^12 - 504013*b^10*c^2 - 3200137*b^8*c^4 - 4501662*b^6*c^6 - 3200137*b^4*c^8 - 504013*b^2*c^10 + 180261*c^12) + (b^2 - c^2)^2* (4589*b^14 + 82127*b^12*c^2 - 214779*b^10*c^4 + 50239*b^8*c^6 + 50239*b^6*c^8 - 214779*b^4*c^10 + 82127*b^2*c^12 + 4589*c^14) - 16*a^4*(16511*b^14 + 222377*b^12*c^2 - 94725*b^10*c^4 - 829475*b^8*c^6 - 829475*b^6*c^8 - 94725*b^4*c^10 + 222377*b^2*c^12 + 16511*c^14) + a^2*(139387*b^16 - 474096*b^14*c^2 - 1265660*b^12*c^4 + 1776560*b^10* c^6 + 565122*b^8*c^8 + 1776560*b^6*c^10 - 1265660*b^4* c^12 - 474096*b^2*c^14 + 139387*c^16) - 4*Sqrt[-a^4 - (b^2 - c^2)^2 + 2*a^2*(b^2 + c^2)]*(10312*a^16 - 4277*b^16 + 20294*b^14*c^2 + 10528*b^12*c^4 - 92326*b^10*c^6 + 123370*b^8*c^8 - 92326*b^6*c^10 + 10528*b^4*c^12 + 20294*b^2*c^14 - 4277*c^16 - 127405*a^14* (b^2 + c^2) + a^12*(74761*b^4 - 16612*b^2*c^2 + 74761*c^4) + a^10*(554099*b^6 + 2123601*b^4*c^2 + 2123601*b^2*c^4 + 554099*c^6) - a^8*(523055*b^8 + 490706*b^6*c^2 - 462786*b^4*c^4 + 490706*b^2*c^6 + 523055*c^8) - a^6*(156559*b^10 + 2242475*b^8*c^2 + 4527238*b^6* c^4 + 4527238*b^4*c^6 + 2242475*b^2*c^8 + 156559*c^10) + a^4*(158227*b^12 + 14928*b^10*c^2 - 2020107*b^8*c^4 - 3217072*b^6* c^6 - 2020107*b^4*c^8 + 14928*b^2*c^10 + 158227*c^12) + a^2*(5705*b^14 + 194087*b^12*c^2 - 141483*b^10*c^4 - 421829*b^8*c^6 - 421829*b^6*c^8 - 141483*b^4*c^10 + 194087*b^2*c^12 + 5705*c^14)) : ... : ...

 (6 - 9 - 13) - search numbers   (5.38491118901805, 4.50118573443656, -1.96088465224926)

Best regards
Angel Montesdeoca

 

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#5001

 

Note: There are 4 triads of squares for this configuration. See tri-squares and tri-squares central triangles in the Index of triangles referenced in ETC. The given Euler lines concur only for the 1^st (2^nd) tri-squares/1^st (2^nd) tri-squares-central triangles.

 

·        With respect to ABC, the points of concurrence are:

For the 1^st tri-squares and 1^st tri-squares central triangles:

P1 = EULER LINE INTERCEPT OF X(1285)X(19054)

= (41*a^4-44*(b^2+c^2)*a^2-13*b^4+10*b^2*c^2-13*c^4)*S+18*(c^2+a^2+b^2)*(a^2-b^2-c^2)*a^2 : : (barys)

= 6*S^2+2*SW*S-9*SB*SC : : (barys)

= 4*(SW+3*S)*X(3)+(2*SW-3*S)*X(4)

= As a point on the Euler line, this center has Shinagawa coefficients (E+F+3*S, -9*S/2)

= on lines: {2, 3}, {524, 9541}, {1285, 19054}, {3068, 13662}, {3595, 6451}, {5860, 9741}, {6221, 13639}, {12158, 12256}, {13663, 23249}, {13757, 23273}

= reflection of X(i) in X(j) for these (i,j): (13639, 6221), (23249, 13663)

= [ 5.3849111890180510, 4.5011857344365620, -1.9608846522492640 ]

 

For the 2nd tri-squares and 2nd tri-squares central triangles:

P2 = EULER LINE INTERCEPT OF X(1285)X(19053)

= -(41*a^4-44*(b^2+c^2)*a^2-13*b^4+10*b^2*c^2-13*c^4)*S+18*(c^2+a^2+b^2)*(a^2-b^2-c^2)*a^2 : : (barys)

= 6*S^2-2*SW*S-9*SB*SC : : (barys)

= 4*(SW-3*S)*X(3)+(2*SW+3*S)*X(4)

= As a point on the Euler line, this center has Shinagawa coefficients (E+F-3*S, 9*S/2)

= on lines: {2, 3}, {597, 9541}, {1285, 19053}, {3069, 13782}, {3593, 6452}, {5861, 9741}, {6398, 13759}, {12159, 12257}, {13637, 23267}, {13783, 23259}

= reflection of X(i) in X(j) for these (i,j): (13759, 6398), (23259, 13783)

= [ -5.5596450227196770, -6.4144984616681650, 10.6474611197021900 ]

 

·        With respect to A’B’C’, the points of concurrence are:

For the 1^st tri-squares and 1^st tri-squares central triangles:

P’1 = EULER LINE INTERCEPT OF X(99)X(1270)

= -(a^2+b^2+c^2)*S+5*a^4-4*(b^2+c^2)*a^2-(b^2-c^2)^2: : (barys)

= 2*(SW+4*S)*X(3)+(SW-2*S)*X(4) = 4*X(1151)-X(12222) = 2*X(12313)+X(12510)

= As a point on the Euler line, this center has Shinagawa coefficients (E+F+4*S, -6*S)

= on lines: {2, 3}, {99, 1270}, {193, 9541}, {488, 13712}, {490, 5861}, {591, 12221}, {1151, 12222}, {1271, 14907}, {5860, 8716}, {6409, 12323}, {6462, 13678}, {6567, 13639}, {7585, 9675}, {12313, 12510}

= reflection of X(488) in X(13712)

= [ 5.9375873907285220, 5.0524039623720300, -2.5975785954555750 ]

 

For the 2nd tri-squares and 2nd tri-squares central triangles:

P’2 = EULER LINE INTERCEPT OF X(99)X(1271)

= (a^2+b^2+c^2)*S+5*a^4-4*(b^2+c^2)*a^2-(b^2-c^2)^2: : (barys)

= 2*(SW-4*S)*X(3)+(SW+2*S)*X(4) = 4*X(1152)-X(12221) = 2*X(12314)+X(12509)

= As a point on the Euler line, this center has Shinagawa coefficients (E+F-4*S, 6*S)

= on lines: {2, 3}, {99, 1271}, {487, 13835}, {489, 5860}, {1152, 12221}, {1270, 14907}, {1991, 12222}, {5861, 8716}, {6410, 12322}, {6463, 13798}, {6566, 13759}, {12314, 12509}

= reflection of X(487) in X(13835)

= [ -13.6341382212094600, -14.4676909432220400, 19.9494373908501600 ]

 

·        With respect to A”B”C”, the points of concurrence are:

For the 1^st tri-squares and 1^st tri-squares central triangles:

P”1 = EULER LINE INTERCEPT OF X(488)X(7582)

= 5*a^4-4*(b^2+c^2)*a^2-(b^2-c^2)^2-4*(a^2+b^2+c^2)*S: : (barys)

= 4*(SW+S)*X(3)+(2*SW-S)*X(4)

= As a point on the Euler line, this center has Shinagawa coefficients E+F+S, -3*S/2)

= on lines: {2, 3}, {141, 9541}, {371, 5861}, {372, 13712}, {488, 7582}, {490, 7581}, {492, 23273}, {591, 1588}, {1271, 6221}, {1285, 3068}, {1384, 8974}, {3593, 13785}, {5490, 14226}, {5590, 6561}, {5591, 6200}, {6202, 12305}, {7586, 14482}, {9738, 10517}, {12323, 13886}, {13789, 13794}, {13950, 15048}, {19054, 19103}, {23263, 23311}

= {X(11292), X(11294)}-harmonic conjugate of X(4)

= [ 3.8087187372085070, 2.9291513181132340, -0.1450796939594791 ]

 

For the 2nd tri-squares and 2nd tri-squares central triangles:

P”2 = EULER LINE INTERCEPT OF X(487)X(7581)

= 5*a^4-4*(b^2+c^2)*a^2-(b^2-c^2)^2+4*(a^2+b^2+c^2)*S: : (barys)

= 4*(SW-S)*X(3)+(2*SW+S)*X(4)

= As a point on the Euler line, this center has Shinagawa coefficients E+F-S, 3*S/2)

= on lines: {2, 3}, {371, 13835}, {372, 5860}, {487, 7581}, {489, 7582}, {491, 23267}, {1270, 6398}, {1285, 3069}, {1384, 13950}, {1587, 1991}, {3589, 9541}, {3595, 13665}, {5491, 14241}, {5590, 6396}, {5591, 6560}, {6201, 12306}, {7585, 14482}, {8974, 15048}, {9739, 10518}, {12322, 13939}, {13669, 13674}, {19053, 19104}, {23253, 23312}

= {X(11291), X(11293)}-harmonic conjugate of X(4)

= [ 0.9822875999522916, 0.1101763775635536, 3.1110250205393240 ]

 

César Lozada

 

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#5004

 

Here is my generalization for this problem, I posted to Anopolis, thank Mr Antreas P. Hatzipolakis to rewrite this problem.

 

Let ABC be a triangle.

We construct the external similar rectangles based on the sides of the triangle:

BCCaBa, CAAbCb, ABBcAc with centers A', B', C', resp.

CaBa, AbCb, BcAc bound a triangle A"B"C"
AbAc, BcBa, CaCb bound a triangle A0B0C0

The Euler lines of A'B'C', A"B"C", A0B0C0 are concurrent.

Which is the point of concurrence with term of Kierpert angles of rectangles?

 

Best regards,

Tran Quang Hung.

ADGEOM 4942 * ADGEOM 4943

#4942

Dear friends,

 

Consider a triangle ABC and let P be its incenter. A perpendicular to AP, at P, cut sides AB and AC at E and F, respectively. Similarly, define the segments GH and IJ ciclycally. 

 

Circles (EGJ), (GJF) and (FHI) are concurrent at Q.

 

Is this well-known? What is the nature of this point?

 

Best regards,

Emmanuel José García

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#4943

Dear Emmanuel, I changed the labels of the points of your problem:

Let  ABC be a triangle of incenter I, the perpendicular by I to AI intersects AC and AB in Ab and Ac, respectively.

The points Bc, Ba, Ca and Cb are defined cyclically.

The circles (AbBaBc), (BcCbCa) and (CaAcAb) concur in P.
The circles (AbBaAc), (BcCbBa) and (CaAcCb)  cuncur in U.

Points P and U form a bicentric pair (http://faculty.evansville.edu/ck6/encyclopedia/BicentricPairs.html),

P = f(a,b,c) : f(b,c,a) : f(c,a,b)     and   Q =  f(a,c,b) : f(b,a,c) : f(c,b,a)

 with
f(a,b,c) = a (a-b-c) (b-c) (2 a^2-2 a b-a c+b c-c^2) (a^2+a b-2 b^2-2 a c+b c+c^2).

** The bicentric sum of P and U is:

S = a (b-c)^2 (b+c-a)^2 (5 a^2-4 a (b+c)-(b-c)^2) : .... : ....

S lies on lines X(i)X(j) for these {i, j}: {926,2170}, {2246,4845}, {3119,3900}, {4162,7004}.

 (6 - 9 - 13) - search numbers  of S: (3.01760078443673, 3.25289748021838, -0.00408028952222515).
 
 
** The bicentric difference of P and U is:

D = a (b-c) (4 a^2-5 a (b+c)+b^2+4 b c+c^2) : ...: ....

D is the midpoint of X(i) and X(j), for these {i, j}:  {1,2254}, {145,3716}, {513,4162}, {519,14430}, {663,14077}, {891,3251}, {905,4959}, {1635,3722}, {2814,16200}, {2832,10699}, {3244,3762}, {3295,8648}, {3900,14414}, {8572,20315}.

D is the reflection of X(i) in X(j), for these {i, j}: {2254,14413}, {14413,1}.

D lies on lines X(i)X(j) for these {i, j}: {1,2254}, {145,3716}, {513,4162}, {519,14430}, {663,14077}, {891,3251}, {905,4959}, {1635,3722}, {2814,16200}, {2832,10699}, {3244,3762}, {3295,8648}, {3900,14414}, {8572,20315}.

 (6 - 9 - 13) - search numbers  of D: (5.41971934778287, 0.657342198122180, 0.684172491923085).
 
 Angel Montesdeoca

ADGEOM 4921 * ADGEOM 4925 * ADGEOM 4927 * ADGEOM 4941

#4921

Dear geometers,

 

Let ABC be a triangle then 

 

Centroid G, symmedian point S, circumcenter O and Parry point P are concyclic.

 

Is this result known before?

 

Best regards,

Tran Quang Hung.

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#4925

Center = X(9175)

Square-radius = 4*R^2* S^2* (3*S^2-SW^2)* (9*R^2-2*SW)*(-SW^3+3*(9*R^2-SW)*S^2)/K^2, where K=3*SW*∏(|SB-SC|)

Through ETC’s: {2, 3, 6, 111, 691, 5653, 9173, 9174, 9178, 11579, 11632, 11637, 11638, 14174, 14180, 14699, 14700, 15546, 15744}

 

Some others:

Antipode of X(2) =

P01 = X(3)X(523) ∩ X(98)X(111)

= (3*a^10-4*(b^2+c^2)*a^8-(2*b^4-7*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(4*b^4-5*b^2*c^2+4*c^4)*a^4-(b^8+c^8-b^2*c^2*(7*b^4-16*b^2*c^2+7*c^4))*a^2-(b^4-c^4)*(b^2-c^2)*b^2*c^2)*(b^2-c^2) : : (barys)

= 3*X(3545)-2*X(18309)

= on lines: {2, 9137}, {3, 523}, {4, 2489}, {6, 1499}, {30, 9178}, {98, 111}, {182, 5652}, {512, 11179}, {542, 5653}, {690, 11579}, {691, 2407}, {804, 11632}, {1352, 11182}, {2492, 19912}, {3545, 18309}, {5996, 9744}, {6088, 14666}, {7709, 8704}, {8371, 15928}, {9171, 20423}, {12106, 21006}

= reflection of X(i) in X(j) for these (i,j): (2, 9175), (1352, 11182)

= [ 3.6274797921220480, 9.0737346583668690, -4.3153732625567060 ]

 

Antipode of X(6) =

P02 = X(3)X(512) ∩ X(74)X(111)

= a^2*(a^8+3*(b^2+c^2)*a^6-(11*b^4+5*b^2*c^2+11*c^4)*a^4+(b^2+c^2)*(9*b^4-2*b^2*c^2+9*c^4)*a^2-2*b^8+3*b^6*c^2-14*b^4*c^4+3*b^2*c^6-2*c^8)*(b^2-c^2) : : (barys)

= 3*X(5054)-2*X(11183), 3*X(10516)-2*X(18309)

= on lines: {2, 1499}, {3, 512}, {6, 9175}, {74, 111}, {381, 11182}, {511, 9178}, {525, 16220}, {526, 11579}, {549, 5652}, {690, 11632}, {691, 2421}, {804, 19905}, {924, 10249}, {1351, 9171}, {5054, 11183}, {5653, 5663}, {6785, 14700}, {9126, 9135}, {9137, 15066}, {10516, 18309}

= reflection of X(i) in X(j) for these (i,j): (6, 9175), (381, 11182), (1351, 9171)

= [ 5.2639400895450970, 9.3372844440941770, -5.2531201745631710 ]

 

César Lozada

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#4927

TCCT  Table 8.1  Central circles that pass through the Parry point

 

Angel Montesdeoca

 

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#4941

 

Dear César 

in the lists reflection of X(i) in X(j) op points P01 and P02 I have find other pairs:

 

Antipode of X(2) = P01 

 = reflection of X(i) in X(j) for these (i,j): {2, 9175}, {1352, 11182}, {5652, 182}, {19912, 2492}, {20423, 9171}

 

 

Antipode of X(6) = P02 

 = reflection of X(i) in X(j) for these (i,j):  {6, 9175}, {381, 11182}, {1351, 9171}, {5652, 549}, {9135, 9126}

 

 

Best regards

Ercole Suppa

ADGEOM 4922 * ADGEOM 4923 * ADGEOM 4924

#4922

Dear geometers,

 

I see five concyclic points are

 

X(3), X(187), X(115), X(131) and X(399)

 

Is this circle known before?

 

Best regards,

Tran Quang Hung.

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#4923

Center = X(6132)

Radius = 2*S*R*|OH|*sqrt(|2*S^2*SW -(12*S^2-SW^2)*R^2|)/K, where K=∏(|SB-SC|)

Through ETS’s: {3, 115, 131, 187, 399, 11799, 14981, 15478}

Another:

 

Antipode of X(3) = ISOGONAL CONJUGATE OF X(18878)

= a^2*((b^2+c^2)*a^4-2*(b^4-b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))*(b^2-c^2) : : (barys)

= 3*X(351)-4*X(6140), 3*X(351)-X(9409), 3*X(351)-2*X(14270), X(3005)-3*X(15451), 4*X(6140)-X(9409), 3*X(15451)-2*X(18117)

= on lines: {3, 6132}, {6, 2510}, {113, 131}, {115, 2971}, {187, 237}, {399, 526}, {523, 11799}, {684, 690}, {804, 6033}, {878, 14601}, {1576, 1625}, {2780, 8552}, {2872, 9142}, {2881, 11641}, {9517, 11615}, {9934, 15478}, {11060, 14398}

= reflection of X(3) in X(6132)

= Gibert circumtangential conjugate of X(10420)

= isogonal conjugate of X(18878)

= X(14687)-of-1st Parry triangle

= X(15919)-of-2nd Parry triangle

= [ 3.7959076033302890, -13.4459239914496000, 11.1974237359893600 ]

 

César Lozada

 

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

#4924

 

 Additional note in:

https://amontes.webs.ull.es/otrashtm/HGT2018.htm#HG250818


Angel Montesdeoca

ADGEOM 4864 * ADGEOM 4898 * ADGEOM 4901

#4864

Dear geometers,

 

I have a conjecture with Equal detour point X(176).

 

Let ABC be a triangle.

 

Let P0=X(176) of ABC.

 

Let Pa1=X(176) of P0BC.

 

Let Pa2=X(176) of Pa1BC.

 

....

 

Let Pan=X(176) of Pa(n-1)BC.

 

Define similarly, the points Pbn and Pcn.

 

Then triangle ABC and PanPbnPcn are perspective?

 

Best regards,

Tran Quang Hung.

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

#4898

 

I will keep your notation P₀ = X(176)

The perspector (ABC, Pa1Pb1Pc1) = P₁ = X(482)

Now I will add another conjecture to your configuration:

If the perspector (ABC, PanPbnPcn) = P_n then

P_n = { X(7), P_n-1}-harmonic conjugate of P_n-2, for n>=2

= (a-b+c)*(a+b-c)*((n+1)*S+a*(-a+b+c)) : :  (barys)

= on the Soddy line X(1)X(7)

A short sequence of Pn:

 

P₂ = {X(7), X(482)}-HARMONIC CONJUGATE OF X(176)

= (a+b-c)*(a-b+c)*(3*S+a*(-a+b+c)) : : (barys)

= on lines: {1, 7}, {226, 3591}, {8965, 17092}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (7, 176, 175), (7, 482, 176), (7, 17802, 1374), (7, 17805, 481), (175, 17804, 176), (176, 482, 17804), (481, 482, 1371), (481, 1371, 17805), (482, 1373, 7), (1371, 17805, 176), (1374, 17802, 17801), (17801, 17802, 175)

= [ 0.9611927715408633, 0.9994235592967669, 2.5051284309139110 ]

 

P₃ = X(1373) = {X(7), P₂}-HARMONIC CONJUGATE OF X(482)

 

P₄ = {X(7), X(1373)}-HARMONIC CONJUGATE OF P₂

= (a+b-c)*(a-b+c)*(5*S+a*(-a+b+c)) : : (barys)

= on the line {1, 7}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 17804, 176), (7, 482, 175), (7, 17804, 17801), (7, 17805, 1374), (175, 482, 176), (176, 17801, 1), (481, 482, 17806), (482, 1374, 17805), (1374, 17805, 175)

= [ 0.8875351968257142, 0.9296281795105022, 2.5874441129421630 ]

 

P₅ = {X(7), P₄}-HARMONIC CONJUGATE OF X(1373)

= (a+b-c)*(a-b+c)*(6*S+a*(-a+b+c)) : : (barys)

= on lines: {1, 7}, {226, 10194}, {553, 5393}, {3982, 13389}, {4114, 13388}, {4654, 5405}, {5589, 7613}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (7, 176, 1374), (7, 482, 481), (7, 1373, 482), (17802, 17804, 17806)

= [ 0.8667352026601208, 0.9099188217431049, 2.6106890502421740 ]

 

César Lozada

PD: Do you have an easy method for finding X(176) in a given triangle?

Note: It seems that a sequence of perspective triangles is found if X(176) is instead of X(1).

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

#4801

Dear Cezar,

 

Your formula

P_n = { X(7), P_n-1}-harmonic conjugate of P_n-2, for n>=2

= (a-b+c)*(a+b-c)*((n+1)*S+a*(-a+b+c)) : :  (barys)

 

is true also if we designate the point X(1) as P_-1

because X(482) = P₁ = { X(7), P₀}-harmonic conjugate of  P_-1=X(1).

Another conjecture is :

If  I_a0 is the incenter of triangle P₀BC then the line I_a0P_a1 passes through A

and

 

If Ia_n-1 is the incenter of triangle Pa_n-1BC then the line Ia_n-1Pa_n passes through Pa_n-2

Best regards

Nikos Dergiades

 

 

ADGEOM 4866 * ADGEOM 4867

#4866

Dear geometers,

 

Let ABC be a triangle with orthocenter H.

 

P=X(110) of ABC.

 

da be the reflection of Euler line of PBC in line HA.

 

Define similarly, the lines db and dc.

 

Then da,db,dc are concurrent.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

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

#4867

[ Tran Quang Hung]:

Let ABC be a triangle with orthocenter H.
P=X(110) of ABC.
da be the reflection of Euler line of PBC in line HA.

Define similarly, the lines db and dc.

Then da,db,dc are concurrent.
Which is this point?

*** da,db,dc are concurrent at W = midpoint of X(265) and X(14989)

W = 4 a^16
-9 a^14 (b^2+c^2)
+a^12 (-6 b^4+44 b^2 c^2-6 c^4)
+14 a^10 (2 b^6-3 b^4 c^2-3 b^2 c^4+2 c^6)
-3 a^8 (5 b^8+15 b^6 c^2-44 b^4 c^4+15 b^2 c^6+5 c^8)
-a^6 (9 b^10-71 b^8 c^2+63 b^6 c^4+63 b^4 c^6-71 b^2 c^8+9 c^10)
+a^4 (b^2-c^2)^2 (4 b^8+14 b^6 c^2-63 b^4 c^4+14 b^2 c^6+4 c^8)
+6 a^2 (b^2-c^2)^4 (b^6-2 b^4 c^2-2 b^2 c^4+c^6)
-(b^2-c^2)^6 (3 b^4+7 b^2 c^2+3 c^4) : ... : ....



W is the midpoint of X(265) and X(14989).
W is the reflection of X(i) in X(j), for these {i, j}: {14677,12079},
{14934,546}, {16340,10113}.
W lies on lines X(i)X(j) for these {i, j}: {30,74}, {523,3627},
{546,14934}, {10113,16340}, {12295,16168}.
(6 - 9 - 13) - search numbers of W: (-12.3731515365849,
-12.2736427038067, 17.8484870629667).

Angel Montesdeoca

 

 

ADGEOM 4856 * ADGEOM 4859

#4856

Dear geometers,

 

I call line X(3)X(49) of ABC by Sine-triple-angle axis of ABC.

 

Let ABC be a triangle.

 

Orthic triangle A'B'C'.

 

Let da be the reflection of Sine-triple-angle axis of AB'C' in line AA'.

 

Define similarly the line db and dc.

 

Then the lines da,db,dc are concurrent.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

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

#4859

[Tran Quang Hung]:

I call line X(3)X(49) of ABC by Sine-triple-angle axis of ABC.

Let ABC be a triangle.
Orthic triangle A'B'C'.
Let da be the reflection of Sine-triple-angle axis of AB'C' in line
AA'.
Define similarly the line db and dc.

Then the lines da,db,dc are concurrent.
Which is this point?

*** The barycentric equation of the reflection of Sine-triple-angle
axis of AB'C' in line AA' is

da: a^2 (b^2-c^2) (a^6-a^2 (b^2-c^2)^2-a^4 (b^2+c^2)+(b^2-c^2)^2
(b^2+c^2)) x+(-2 a^10+2 (b^2-c^2)^5-a^2 (b^2-c^2)^3 (3 b^2+c^2)+a^8 (5
b^2+c^2)+a^6 (-3 b^4-2 b^2 c^2+c^4)+a^4 (b^6+3 b^4 c^2-5 b^2 c^4+c^6))
y+(2 a^10+2 (b^2-c^2)^5-a^2 (b^2-c^2)^3 (b^2+3 c^2)-a^8 (b^2+5 c^2)+a^6
(-b^4+2 b^2 c^2+3 c^4)-a^4 (b^6-5 b^4 c^2+3 b^2 c^4+c^6)) z=0.

The lines da,db,dc are concurrent at D = midpoint of X(4)X(5962)

D = 2 a^16
-7 a^14 (b^2+c^2)
+4 a^12 (2 b^4+5 b^2 c^2+2 c^4)
-a^10 (3 b^6+17 b^4 c^2+17 b^2 c^4+3 c^6)
-2 a^8 b^2 c^2 (b^4-10 b^2 c^2+c^4)
+3 a^6 (b^2-c^2)^2 (b^2+c^2)^3
-4 a^4 (b^2-c^2)^2 (2 b^8-b^6 c^2+2 b^4 c^4-b^2 c^6+2 c^8)
+a^2 (b^2-c^2)^4 (7 b^6+b^4 c^2+b^2 c^4+7 c^6)
-2 (b^2-c^2)^6 (b^4+b^2 c^2+c^4) : .... : ....


D is the midpoint of X(4) and X(5962)
D is the reflection of X(12095) in X(5)
D lies on lines X(i)X(j) for these {i, j}: {4,52}, {5,12095},
{30,131}, {381,13557}, {403,14769}, {924,13851}, {5203,9880},
{7547,14889}}

(6 - 9 - 13) - search numbers of D: (-1.77451732881372,
-1.23193647064380, 5.31255157488257)

Angel Montesdeoca

 

 

 

ADGEOM 4850 * ADGEOM 4851 * ADGEOM 4853 * ADGEOM 4855 * ADGEOM 4857

#4850

Dear geometers,

 

Let ABC be a triangle.

 

Contact triangle A'B'C' and Nagel point Na.

 

Let La,Lb,Lc be the symmedian points triangles NaB'C', NbC'A' and NcA'B'.

 

Then ABC and LaLbLc are perspective.

 

Which is the perpsector?

 

Best regards,

Tran Quang Hung.

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

#4851

[Tran Quang Hung]:

Let ABC be a triangle.

Contact triangle A'B'C' and Nagel point Na.

Let La,Lb,Lc be the symmedian points triangles NaB'C', NbC'A' and
NcA'B'.

Then ABC and LaLbLc are perspective.

Which is the perpsector?

*** ABC and LaLbLc are perspective at
W = (4 r^2+16 r R-s^2) X(1) - 12 r^2 X(2)

W = (a^3-2 a^2 (b+c)-4 (b-c)^2 (b+c)+a (9 b^2-14 b c+9 c^2) : -4
a^3+b^3-2 b^2 c+9 b c^2-4 c^3+a^2 (9 b+4 c)-2 a (b^2+7 b c-2 c^2) : -4
a^3-4 b^3+9 b^2 c-2 b c^2+c^3+a^2 (4 b+9 c)+2 a (2 b^2-7 b c-c^2))

The (6 - 9 - 13) - search numbers of W are
(3.45104654014144, 1.80769104483599, 0.796395355417256),
which are the same coordinates that the center
X(4691) = (-2 a+5 b+5 c : 5 a-2 b+5 c : 5 a+5 b-2 c).

Angel Montesdeoca

 

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

#4853

 

Hi Angel,

 

Interesting.  So now we have a second pair of points (besides X(3635) and X(15519)) that are exactly the same point when the reference triangle is (6,9,13).  It seems that a second set of search tables for another reference triangle, preferably acute (e.g. (7,9,11)), would be helpful in cases like this.

 

Best regards,

Randy Hutson

 

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

#4855

 

Hi Angel, Randy and Mr Hung,

 

W = complement of X(15519)

= a^3-2*(b+c)*a^2+(9*b^2-14*b*c+9*c^2)*a-4*(b^2-c^2)*(b-c) : : (barys)

= (4 r^2+16 r R-s^2) X(1) - 12 r^2 X(2)

= on lines: {1, 2}, {165, 3021}, {1997, 4929}, {3756, 8056}, {4862, 5274}, {7963, 12625}

= complement of X(15519)

= [3.451046540141442691497607, 1.807691044835993790784461, 0.7963953554172560057302172]

 

César Lozada

 

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

#4857

 

The point W lies also on line {1699, 4902}

Ercole Suppa

 

ADGEOM 4796 * ADGEOM 4798 * ADGEOM 4801 * ADGEOM 4808

#4796

Dear geometers,

I see these two points have this property.

 

A is X(74) of triangle X(74)BC.

 

A is also X(1138) of triangle X(1138)BC.

 

Are these property easy seen?

 

Best regards,

Tran Quang Hung.

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

 

#4798

Dear Mr. Tran Quang Hung,

Yes, they are easily seen with the following properties:

X(74) is an intersection of the circumcircle and the Jerabek hyperbola

=> X(74) is an intersection of the circle ABCX(74) and the hyperbola ABCX(74)X(3).

Hence ABCX(74)X(3) is also the Jerabek hyperbola with respect to the triangles X(74)BC, X(74)CA, X(74)AB.

_______________________

The Euler lines of the triangle ABC, X(1138)BC, X(1138)CA, X(1138)AB are parallel.

 

Sincerely

Ngo Quang Duong

 

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

#4801

 Another property of points  X(74) and X(1138):

Let A' be the intersection point of the Euler lines of triangles X(74)BC and X(1138)BC,
and define B', C' cyclically.
Then triangle ABC and A'B'C' are perspective, with perpsector

W = (a^2-b^2-c^2) (2 a^4-a^2 (b^2+c^2)-(b^2-c^2)^2)/(a^8-4 a^6 (b^2+c^2)+a^4 (6 b^4+b^2 c^2+6 c^4)+a^2 (-4 b^6+b^4 c^2+b^2 c^4-4 c^6)+(b^2-c^2)^2 (b^4+4 b^2 c^2+c^4)) : ... : ....

with  (6 - 9 - 13) - search numbers:  (6.26250855866268, 3.50369184105253, -1.67535689666554).

W lies on lines X(i)X(j) for these {i, j}: {30,146}, {74,18317}, {265,14919}, {1294,14677}, {1494,10264}, {1511,3163}, {6699,8552}, {10272,14920}, {16163,19223}.

Angel Montesdeoca

 

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

#4808

This brings up a question I have often wondered about.  Perhaps it has already been discussed.  Are there any other points besides X(4), X(74) and X(1138) that form 'X-centric systems'?  That is, every point in the set {A,B,C,X} is 'X' of the remaining three points.

 

Best regards,

Randy Hutson

 

 

 

ADGEOM 4790 * ADGEOM 4791

#4790

Dear Geometers,

Let Ge be Gergonne point of ABC. DEF cevian triangle of Ge. K1,K2,K3 symmedian points of GeEF, GeFD, GeDE respectively. AK1, BK2, CK3 concur. (Not listed in ETC).

Best regards

Abdilkadir Altıntaş, Emirdağ, Türkiye

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

#4791

[Abdilkadir Altıntas]:

Let Ge be Gergonne point of ABC. DEF cevian triangle of Ge. K1,K2,K3 symmedian points of GeEF, GeFD, GeDE respectively. AK1, BK2, CK3 concur. (Not listed in ETC).

*** AK1, BK2, CK3 concur  at W = 3s^2 X(1) - 4(r+4R)^2 X(7),

 with first barycentric coordinate:  (a+b-c) (a-b+c) (a^2-5 a (b+c)+4 (b-c)^2).

W lies on lines X(i)X(j) for these {i, j}:  {1,7}, {948,4031}, {1358,18421}, {4495,7195}, {4859,7960}.

 (6 - 9 - 13) - search numbers  of W: (0.577822777754951, 0.636155362937831, 2.93356179475590).

Angel Montesdeoca

 

ADGEOM 4765 * ADGEOM 4766 * ADGEOM 4767

#4765

Dear Geometers,

Let Ge Gergonne point of ABC.

DEF cevian triangle of Ge.

Na, NPC of GeFE. Define Nb,Nc cyclically.

Perpendiculars from Na to BC, Nb to CA and Nc to AB concur. (Not listed in ETC).

If DEF is cevian triangle of Nagels point and Na,Nb,Nc are NPC s defined above Perpendiculars from Na to BC, Nb to CA and Nc to AB concur. (Not listed in ETC).

Abdilkadir Altıntaş, Emirdağ, Türkiye

Best regards...

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

#4766

[Abdilkadir Altıntas]:

Let Ge Gergonne point of ABC.
DEF cevian triangle of Ge.
Na, NPC of GeFE. Define Nb,Nc cyclically.

Perpendiculars from Na to BC, Nb to CA and Nc to AB concur. (Not listed
in ETC).

**** Perpendiculars from Na to BC, Nb to CA and Nc to AB concur at
U = (2r+7R) X(7) - (2r+3 R) X(79).

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

U is the midpoint of X(i) and X(j), for these {i, j}: {9,3874},
{10,15185}, {942,5572}, {5542,5728}.

U is the reflection of X(3678) in X(6666)

W lies on lines X(i)X(j) for these {i, j}: {1,1170}, {7,79}, {9,3874},
{10,15185}, {11,118}, {142,3841}, {390,5902}, {516,942},
{518,1125}, {758,1001}, {971,12005}, {997,3243}, {1699,12669},
{2550,5883}, {2802,14563}, {3059,5439}, {3086,11038}, {3305,3873},
{3336,7676}, {3338,7675}, {3670,4343}, {3754,5853}, {3826,3833},
{3889,8583}, {4312,7671}, {4349,14523}, {5425,12758}, {5708,11495},
{5732,10980}, {5762,6583}, {5903,8236}, {5904,18230}, {6684,16216},
{8232,10399}, {12755,16173}, {13751,13995}, {15008,15726},
{15299,18389}.

(6 - 9 - 13) - search numbers of U: (1.34655429409374,
1.43528118005218, 2.02552168305884)


[Abdilkadir Altıntaş]:
If DEF is cevian triangle of Nagels point and Na,Nb,Nc are NPC s defined
above Perpendiculars from Na to BC, Nb to CA and Nc to AB concur. (Not
listed in ETC).

**** Perpendiculars from Na to BC, Nb to CA and Nc to AB concur at
V = (2r+3R) X(8) - (2r-5R) X(13729).

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

V is the midpoint of X(i) and X(j), for these {i, j}: {5,5694},
{10,5887}, {72,946},}, {355,3878}, {960,5777},}, {5693,5884} ,
{5882,14872}, {11362,12672}, {11715,12665}.

V is the reflection of X(i) in X(j), for these {i, j}: {3754,9956},
{3881,5901}, {5690,4015}, {5885,3628}, {6684,5044}, {12005,1125}.

V lies on lines X(i)X(j) for these {i, j}: {1,6920}, {2,5693},
{3,3647}, {4,5692}, {5,758}, {8,13729}, {9,1630}, {10,119},
{21,6326}, {40,3876}, {52,15049}, {65,3614}, {72,946}, {79,6901},
{140,2771}, {191,6905}, {210,11362}, {355,3878}, {392,5882},
{515,960}, {517,546}, {518,13464}, {631,15071}, {912,1125},
{936,1158}, {952,3884}, {956,12059}, {997,5450}, {1071,10165},
{1216,2392}, {1339,10247}, {1376,5780}, {1385,2801}, {1656,5883},
{1768,6940}, {1844,7551}, {1858,13411}, {1898,4304}, {1935,11700},
{2778,5893}, {2779,5907}, {2842,10170}, {3057,18908}, {3090,5902},
{3219,11012}, {3336,6946}, {3428,15650}, {3452,12616}, {3485,18397},
{3576,12528}, {3628,3833}, {3681,7982}, {3743,5396}, {3754,3838},
{3812,10172}, {3817,4067}, {3868,8227}, {3869,5587}, {3873,9624},
{3874,5886}, {3877,5881}, {3881,5901}, {3901,7988}, {3940,11496},
{3951,12704}, {4005,13865}, {4015,5690}, {4127,9955}, {4134,4301},
{4187,10265}, {4661,5734}, {5044,6001}, {5086,6246}, {5225,5697},
{5250,17857}, {5251,12691}, {5535,6915}, {5603,5904}, {5719,12564},
{5720,6796}, {5770,10200}, {5779,12114}, {5811,6256}, {5818,5903},
{5927,14110}, {6839,16125}, {6881,11263}, {6909,7701}, {6913,12635},
{6916,16127}, {6986,16132}, {7686,10157}, {10914,14740},
{11112,16120}, {11375,18389}, {11715,12665}, {12047,15556},
{12736,17606}.

(6 - 9 - 13) - search numbers of V: (1.83745805188466,
0.208007872308717, 2.64860146943927).

Angel Montesdeoca

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

#4767

 

In Addition:

Let Ge Gergonne point of ABC.
DEF cevian triangle of Ge.
Na, NPC of GeFE. Define Nb,Nc cyclically.
*** The orthology center of ABC with respect to NaNbNc is X(1156) = isogonal conjugate of  X(1155).       (X(1155) = SCHRÖDER POINT).

If DEF is cevian triangle of Nagels point and Na,Nb,Nc are NPC s defined above, then:
 *** The orthology center of ABC with respect to NaNbNc is X(1320) = isogonal conjugate of X(1319).    (X(1319) = BEVAN-SCHRÖDER POINT).
 
 Angel Montesdeoca

 

ADGEOM 4653 * ADGEOM 4654 * ADGEOM 4660

#4653

 

Dear geometers,

 

Let ABC be a triangle with centroid G and orthocenter H.

 

P=X(381) is midpoint of GH which is the center of orthocentroidal circle.

 

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

 

Then APa,BPb,CPc are concurrent at.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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

#4654

 

[Tran Quang Hung]:

Let ABC be a triangle with centroid G and orthocenter H.

P=X(381) is midpoint of GH which is the center of orthocentroidal
circle.

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

Then APa,BPb,CPc are concurrent.

Which is this point?

*** APa,BPb,CPc are concurrent at

W = 1/((4 a^4-5 a^2 b^2+b^4-5 a^2 c^2-2 b^2 c^2+c^4)(2 a^4-4 a^2 b^2+2
b^4-4 a^2 c^2-b^2 c^2+2 c^4)) : .... : ....

on the line X(381)X(14483), with (6 - 9 - 13) - search numbers
(0.125506022546977, 4.78594924329040, 0.269389149223193)

Angel Montesdeoca

 

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

#4660

 

 

Dear geometers,

I have seen general problem for this,

G and H be the centroid and orthocenter of ABC respectively.

Let ABC be a triangle and P is a point on Euler line of ABC which has Shinagawa coefficients P=kG+H.

Let Ga,Ha be the centroid and orthocenter of PBC respectively.

Pa is a point on Euler line of PBC which has Shinagawa coefficients Pa=kGa+Ha.
 
Define similarly the points Pb and Pc.

Then APa,BPb and CPc are concurrent. Which is this concurrent point in term of k?

Best regards,
Tran Quang Hung.

 

ADGEOM 4649 * ADGEOM 4650

#4649

Dear Geometers,

1. Let DEF be cevian triangle of Nagels point Na. Ha,Hb,Hc be orthocenters of NaEF, NaFD, NaDE respectively. Perpendiculars from Ha to BC, Hb to CA and Hc to AB concur. (ıts a construction for X(3057).

2. Let DEF be cevian triangle of Nagels point Na. Ga,Gb,Gc be centroids of NaEF, NaFD, NaDE respectively. Perpendiculars from Ga to BC, Gb to CA and Gc to AB concur. (ıts a not listed in ETC).

Best regards

Abdilkadir Altıntaş

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

#4650

[Abdilkadir Altıntas]:

2. Let DEF be cevian triangle of Nagels point Na. Ga,Gb,Gc be centroids
of NaEF, NaFD, NaDE respectively. Perpendiculars from Ga to BC, Gb to CA
and Gc to AB concur. (ıts a not listed in ETC).

**** The perpendiculars from Ga to BC, Gb to CA and Gc to AB concur at

W = (r-2 R) O - (r+4 R)Na

with first barycentric coordinate:

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

W is the reflection of X(i) in X(j), for these {i, j}: {354,10175},
{3576,3740}, {3753,5790}, {3892,10171}, {5603,10157}, {10164,3956}.

W lies on lines X(i)X(j) for these {i, j}: {3,3697}, {4,8}, {5,3555},
{10,1071}, {12,354}, {40,4662}, {84,165}, {200,1012}, {210,515},
{392,952}, {405,5534}, {495,5728}, {518,5587}, {519,15064}, {912,3753},
{936,958}, {942,5261}, {944,5044}, {956,5720}, {960,5881}, {971,5657},
{997,18236}, {1385,5260}, {1532,4847}, {1698,12675}, {1737,17625},
{2836,13214}, {3036,12665}, {3086,17624}, {3090,5045}, {3295,9844},
{3617,12528}, {3621,13600}, {3678,14110}, {3679,6001}, {3698,5884},
{3870,6913}, {3889,5056}, {3892,10171}, {3916,11499}, {3921,10167},
{3935,6912}, {3956,10164}, {3983,6684}, {4015,4297}, {4533,18525},
{4882,12705}, {5049,11374}, {5173,10590}, {5251,5531}, {5290,5902},
{5302,10902}, {5439,9956}, {5603,10157}, {5687,7330}, {5693,5836},
{5726,18412}, {5770,17612}, {5780,10246}, {5791,10786}, {5886,11240},
{5904,7686}, {5919,10950}, {6927,12125}, {6965,18527}, {7580,18528},
{7989,13374}, {8164,11018}, {8168,12703}, {8728,10202}, {9004,12587},
{9708,18446}, {9780,9940}, {10039,12711}, {10176,18250}, {10573,12709},
{10588,16193}, {11362,12688}, {12599,12692}, {12629,12635}

(6 - 9 - 13) - search numbers of W: (1.49964716737654,
-0.605533359430132, 3.36773503810838)

Angel Montesdeoca

 

ADGEOM 4589 * ADGEOM 4613

#4589

 

Dear geometers,

 

Let ABC be a triangle and P,Q are isogonal conjugate points.

 

Ab, Ac are isogonal conjugate of B, C wrt triangle APQ.

 

Define similarly, the points Bc, Ba, Ca, Cb.

 

Then six points Ab, Ac, Bc, Ba, Ca, Cb lie on a conic.

 

When is it a circle?

 

Best regards,

Tran Quang Hung.

 

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

#4613

 

Dear Tran Quang Hung,

 

I think the locus of points P such that the conic is a circle is complicated.  I found several such points on the Euler line, none are in ETC.

 

I did find several interesting observations from this configuration:

 

When P,Q are the Brocard points, PU(1):

The center is the trilinear pole of the line through X(351) parallel to the trilinear polar of X(351).

On lines {3,9217}, {6,694}, {32,249}, {39,512}, {543,598}, {574,805}, {733,12074}, {2086,3229} et al.

Barycentrics: a^2 (2 a^2 - b^2 - c^2) / (a^4 - b^2 c^2) : :

(6,9,13) values: (-0.765454011614075, 7.206375474085383, -0.995078225560544).

 

The perspector of the conic lies on lines {262,6036}, {523,3629}, {576,2065} et al

Barycentrics: a^2 (a^2 b^2 + a^2 c^2 - b^4 - c^4) / (a^4 + 2 b^4 + 2 c^4 - b^2 c^2 - 2 a^2 b^2 - 2 a^2 c^2) : :

(6,9,13) values:(-0.177721420347812, -0.199750198006888, 3.860978351457369).

 

The centroid of AbAcBcBaCaCb is X(10567).

 

The inverse of X(39) in the conic is X(882).

 

 

When P,Q are X(2), X(6):

The center of the conic is X(6).  The conic passes through X(2) and its antipode, X(1992) which are vertices of the conic.

 

The perspector of the conic has barycentrics: (5 a^2 - b^2 - c^2) / (14 a^4 + 2 b^4 + 2 c^4 - 20 a^2 b^2 - 20 a^2 c^2 + 13 b^2 c^2) : :

(6,9,13) values:(-0.654815572698841, -1.061665363990563, 4.677886152069619).

 

 

When P,Q are X(3), X(4), the conic is the Yff hyperbola, with center X(381) and perspector X(13481).

 

The centroid of AbAcBcBaCaCb is X(13448).

 

 

When P,Q are X(13), X(15):

The conic is degenerate, 2 lines intersecting at X(13).

 

The centroid of AbAcBcBaCaCb lies on lines {13,15}, {1989,9140}.

 

 

When P,Q are X(14), X(16):

The conic is degenerate, 2 lines intersecting at X(14).

 

The centroid of AbAcBcBaCaCb lies on lines {14,16}, {1989,9140}.

 

 

Best regards,

Randy Hutson

ADGEOM 4604 * ADGEOM 4608

#4604

 

Dear geometers,

 

Let ABC be a triangle with circumcenter O.

 

P is on its Euler line.

 

Oa, Ob, Oc are isogonal conjugate of O wrt triangles PBC, PCA, PAB.

 

Then centroid of OaObOc lies on Euler line of ABC.

 

Which is this point in term of P?

 

Best regards,

Tran Quang Hung.

 

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

#4608

 

Dear Tran Quang Hung.

If P is on its Euler line such that OP:PH=t , then centroid Q of
OaObOc ( lies on Euler line of ABC) is:

Q = (a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) t (a^16 t^2-(b^2-c^2)^6
(b^2+c^2)^2 t^2-a^14 (b^2+c^2) t (1+3 t)+a^2 (b^2-c^2)^4 (b^2+c^2) t (-2
b^2 c^2 (-2+t)+b^4 (1+3 t)+c^4 (1+3 t))-a^6 (b^2-c^2)^2 (b^2+c^2) (b^4
(-5+t) t+c^4 (-5+t) t+b^2 c^2 (-5+7 t-4 t^2))-a^8 (b^4 c^4 (-7+14 t-12
t^2)+b^6 c^2 (3-9 t+4 t^2)+b^2 c^6 (3-9 t+4 t^2))-a^4 (b^2-c^2)^2 (2 b^8
t (2+t)+2 c^8 t (2+t)+b^6 c^2 (2+5 t-3 t^2)+b^2 c^6 (2+5 t-3 t^2)+b^4
c^4 (5-6 t+6 t^2))+a^10 (b^2+c^2) (b^4 (-5+t) t+c^4 (-5+t) t-b^2 c^2
(1-5 t+8 t^2))+a^12 (2 b^4 t (2+t)+2 c^4 t (2+t)+b^2 c^2 (1+9
t^2)))-(-a^2+b^2+c^2) (a^8 t-2 a^6 (b^2+c^2) t+a^4 b^2 c^2 (-2+3
t)-(b^2-c^2)^2 (b^2 c^2 (-1+t)+b^4 t+c^4 t)+a^2 (b^2+c^2) (b^2 c^2 (1-4
t)+2 b^4 t+2 c^4 t)) ((b^2-c^2) (-a^2 c^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2
b^4+b^2 c^2 (1-2 t))-a^8 t-(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 c^2 t+b^2
(1+t))+a^2 (b^2-c^2) (-b^2 c^2 (-2+t)+2 c^4 t+b^4 (1+t)))+(-b^2+c^2)
(-a^2 b^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2 c^4+b^2 c^2 (1-2 t))-a^8
t+(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 b^2 t+c^2 (1+t))-a^2 (b^2-c^2) (-b^2
c^2 (-2+t)+2 b^4 t+c^4 (1+t)))) : ... : ...

Pairs {P=X(i),Q=X(j)} (safe error or omission) , for {i,j}: {3,1650},
{23,4}, {186,2}, {1113,10720}, {1114,10719}, {1583,11321}, {2060,14020},
{7486,16950}, {7512,14006}, {8359,17686}, {10128,8364}, {11343,16896},
{14142,17531}, {16355,14001}, {16383,11335}, {16399,14461}

*** If P=X(2)

Q = a^16-2 a^14 (b^2+c^2)-3 a^12 (b^4-4 b^2 c^2+c^4)-(b^2-c^2)^4
(b^2+c^2)^2 (2 b^4-3 b^2 c^2+2 c^4)+2 a^10 (4 b^6-5 b^4 c^2-5 b^2 c^4+4
c^6)-2 a^6 (b^2-c^2)^2 (5 b^6-2 b^4 c^2-2 b^2 c^4+5 c^6)+a^8 (b^8-15 b^6
c^2+29 b^4 c^4-15 b^2 c^6+c^8)+a^4 (b^2-c^2)^2 (3 b^8+2 b^6 c^2-15 b^4
c^4+2 b^2 c^6+3 c^8)+4 a^2 (b^2-c^2)^2 (b^10-b^8 c^2+2 b^6 c^4+2 b^4
c^6-b^2 c^8+c^10) : ... : ...

with (6 - 9 - 13) - search numbers (2.25482286270048,
1.37935466015752, 1.64503916516741).


*** If P=X(4) then Q is the reflection of X(2) in X(13448) :

Q = a^12-2 a^10 (b^2+c^2)+2 a^8 (b^4+b^2 c^2+c^4)-(b^2-c^2)^4 (2
b^4+b^2 c^2+2 c^4)-2 a^6 (b^6+c^6)+2 a^2 (b^2-c^2)^2 (2 b^6-b^4 c^2-b^2
c^4+2 c^6)-a^4 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8) : ... : ...,

on lines X(i)X(j) for these {i, j}: {2,3}, {94,9140}, {2088,14846}.
(6 - 9 - 13) - search numbers ( -54.1570303678729, -54.8836826772129,
66.6326126589962).


*** If P=X(5)

Q = (b^2-c^2)^2 (-a^2+b^2+c^2) (a^8+a^4 b^2 c^2-2 a^6
(b^2+c^2)-(b^2-c^2)^2 (b^4+c^4)+a^2 (b^2+c^2) (2 b^4-3 b^2 c^2+2 c^4))
(a^10-4 a^8 (b^2+c^2)+2 (b^2-c^2)^4 (b^2+c^2)+a^6 (4 b^4+b^2 c^2+4
c^4)-a^2 (b^2-c^2)^2 (5 b^4+4 b^2 c^2+5 c^4)+a^4 (2 b^6+3 b^4 c^2+3 b^2
c^4+2 c^6))+(a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) (a^16-4 a^14
(b^2+c^2)-(b^2-c^2)^6 (b^2+c^2)^2-4 a^10 (b^2+c^2) (b^4+b^2 c^2+c^4)+2
a^6 (b^2-c^2)^2 (b^2+c^2) (2 b^4+b^2 c^2+2 c^4)+2 a^2 (b^2-c^2)^4
(b^2+c^2) (2 b^4+b^2 c^2+2 c^4)+2 a^12 (3 b^4+5 b^2 c^2+3 c^4)+a^8 (2
b^6 c^2+5 b^4 c^4+2 b^2 c^6)-a^4 (b^2-c^2)^2 (6 b^8+4 b^6 c^2+5 b^4
c^4+4 b^2 c^6+6 c^8)) : ... : ....

with (6 - 9 - 13) - search numbers (-9.94914860821929,
-10.7924224258757, 15.7042562882303).

*** If P=X(20)

Q = (b^2-c^2)^2 (-a^2+b^2+c^2) (-a^^2 c^2+c^4)-2 a^4 (5 b^6-9 b^4
c^2-9 b^2 c^4+5 c^6))-(a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) (a^16-34 a^8
b^2 c^2 (b^2-c^2)^2-a^14 (b^2+c^2)+11 a^10 (b^2-c^2)^2
(b^2+c^2)-(b^2-c^2)^6 (b^2+c^2)^2+a^12 (-6 b^4+13 b^2 c^2-6 c^4)+a^2
(b^2-c^2)^4 (b^6-9 b^4 c^2-9 b^2 c^4+c^6)-a^6 (b^2-c^2)^2 (11 b^6-8-7 a^4 b^2 c^2+2 a^6
(b^2+c^2)+(b^2-c^2)^2 (b^4+3 b^2 c^2+c^4)-2 a^2 (b^6-2 b^4 c^2-2 b^2
c^4+c^6)) (-5 a^10+8 a^8 (b^2+c^2)+2 (b^2-c^2)^4 (b^2+c^2)+4 a^6 (b^4-8
b^2 c^2+c^4)+a^2 (b^2-c^2)^2 (b^4+14 b27 b^4
c^2-27 b^2 c^4+11 c^6)+a^4 (b^2-c^2)^2 (6 b^8+5 b^6 c^2-38 b^4 c^4+5 b^2
c^6+6 c^8)) : ... : ...

with (6 - 9 - 13) - search numbers (5.41707665942407,
4.53326635151949, -1.99793991195560).

*** If P=X(30)

Q = a^12-a^10 (b^2+c^2)-5 a^6 (b^2-c^2)^2 (b^2+c^2)+a^8 (b^4-b^2
c^2+c^4)-(b^2-c^2)^4 (2 b^4+3 b^2 c^2+2 c^4)+2 a^4 (b^2-c^2)^2 (2 b^4+5
b^2 c^2+2 c^4)+2 a^2 (b^2-c^2)^2 (b^6-3 b^4 c^2-3 b^2 c^4+c^6) : ... :
....

on lines X(i)X(j) for these {i, j}: {2,3}, {125,9530}, {523,1853},
{1899,2452}, {2972,10714}, {3258,11550}}.
(6 - 9 - 13) - search numbers (7.18058761893295, 6.29212512458911,
-4.02953950539252)

Best regards,
Angel Montesdeoca

 

 

ADGEOM 4584 * ADGEOM 4585 * ADGEOM 4587

#4584

 

Dear geometers,

 

Let ABC be a triangle.

 

O and H are circumcenter and orthocenter.

 

I is incenter of ABC.

 

Ja, Jb, Jc are incenters of triangles AHO, BHO, CHO.

 

Then isogonal conjugate of I wrt triangle JaJbJc lies on line OH.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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

#4585

 

I find huge coordinates ( T stands for S·OH where S is twice the area of ABC.

 

Best regards, 

Francisco Javier.

 

{-a^13 + a^12 b + 3 a^11 b^2 - 3 a^10 b^3 - 2 a^9 b^4 + 2 a^8 b^5 - 

  2 a^7 b^6 + 2 a^6 b^7 + 3 a^5 b^8 - 3 a^4 b^9 - a^3 b^10 + 

  a^2 b^11 + a^12 c - 2 a^11 b c - 3 a^10 b^2 c + 2 a^9 b^3 c + 

  3 a^8 b^4 c + 4 a^7 b^5 c - a^6 b^6 c - 4 a^5 b^7 c - 2 a^3 b^9 c + 

  2 a b^11 c + 3 a^11 c^2 - 3 a^10 b c^2 - 5 a^9 b^2 c^2 + 

  4 a^8 b^3 c^2 + 4 a^7 b^4 c^2 - 2 a^6 b^5 c^2 - 4 a^5 b^6 c^2 + 

  4 a^4 b^7 c^2 + a^3 b^8 c^2 - 3 a^2 b^9 c^2 + a b^10 c^2 - 

  3 a^10 c^3 + 2 a^9 b c^3 + 4 a^8 b^2 c^3 - 10 a^7 b^3 c^3 - 

  3 a^6 b^4 c^3 + 4 a^5 b^5 c^3 + 2 a^4 b^6 c^3 + 10 a^3 b^7 c^3 - 

  6 a b^9 c^3 - 2 a^9 c^4 + 3 a^8 b c^4 + 4 a^7 b^2 c^4 - 

  3 a^6 b^3 c^4 + 2 a^5 b^4 c^4 - 3 a^4 b^5 c^4 + 3 a^2 b^7 c^4 - 

  4 a b^8 c^4 + 2 a^8 c^5 + 4 a^7 b c^5 - 2 a^6 b^2 c^5 + 

  4 a^5 b^3 c^5 - 3 a^4 b^4 c^5 - 16 a^3 b^5 c^5 - a^2 b^6 c^5 + 

  4 a b^7 c^5 - 2 a^7 c^6 - a^6 b c^6 - 4 a^5 b^2 c^6 + 

  2 a^4 b^3 c^6 - a^2 b^5 c^6 + 6 a b^6 c^6 + 2 a^6 c^7 - 

  4 a^5 b c^7 + 4 a^4 b^2 c^7 + 10 a^3 b^3 c^7 + 3 a^2 b^4 c^7 + 

  4 a b^5 c^7 + 3 a^5 c^8 + a^3 b^2 c^8 - 4 a b^4 c^8 - 3 a^4 c^9 - 

  2 a^3 b c^9 - 3 a^2 b^2 c^9 - 6 a b^3 c^9 - a^3 c^10 + a b^2 c^10 + 

  a^2 c^11 + 2 a b c^11 + 2 a^10 T - 4 a^9 b T - 4 a^8 b^2 T + 

  10 a^7 b^3 T - 6 a^5 b^5 T + 4 a^4 b^6 T - 2 a^3 b^7 T - 

  2 a^2 b^8 T + 2 a b^9 T - 4 a^9 c T + 6 a^8 b c T - 4 a^6 b^3 c T + 

  8 a^5 b^4 c T - 8 a^4 b^5 c T + 4 a^2 b^7 c T - 4 a b^8 c T + 

  2 b^9 c T - 4 a^8 c^2 T + 4 a^6 b^2 c^2 T - 10 a^5 b^3 c^2 T - 

  4 a^4 b^4 c^2 T + 12 a^3 b^5 c^2 T + 4 a^2 b^6 c^2 T - 

  2 a b^7 c^2 T + 10 a^7 c^3 T - 4 a^6 b c^3 T - 10 a^5 b^2 c^3 T + 

  16 a^4 b^3 c^3 T - 10 a^3 b^4 c^3 T - 4 a^2 b^5 c^3 T + 

  10 a b^6 c^3 T - 8 b^7 c^3 T + 8 a^5 b c^4 T - 4 a^4 b^2 c^4 T - 

  10 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T - 6 a b^5 c^4 T - 6 a^5 c^5 T - 

  8 a^4 b c^5 T + 12 a^3 b^2 c^5 T - 4 a^2 b^3 c^5 T - 

  6 a b^4 c^5 T + 12 b^5 c^5 T + 4 a^4 c^6 T + 4 a^2 b^2 c^6 T + 

  10 a b^3 c^6 T - 2 a^3 c^7 T + 4 a^2 b c^7 T - 2 a b^2 c^7 T - 

  8 b^3 c^7 T - 2 a^2 c^8 T - 4 a b c^8 T + 2 a c^9 T + 2 b c^9 T, 

 a^11 b^2 - a^10 b^3 - 3 a^9 b^4 + 3 a^8 b^5 + 2 a^7 b^6 - 

  2 a^6 b^7 + 2 a^5 b^8 - 2 a^4 b^9 - 3 a^3 b^10 + 3 a^2 b^11 + 

  a b^12 - b^13 + 2 a^11 b c - 2 a^9 b^3 c - 4 a^7 b^5 c - 

  a^6 b^6 c + 4 a^5 b^7 c + 3 a^4 b^8 c + 2 a^3 b^9 c - 

  3 a^2 b^10 c - 2 a b^11 c + b^12 c + a^10 b c^2 - 3 a^9 b^2 c^2 + 

  a^8 b^3 c^2 + 4 a^7 b^4 c^2 - 4 a^6 b^5 c^2 - 2 a^5 b^6 c^2 + 

  4 a^4 b^7 c^2 + 4 a^3 b^8 c^2 - 5 a^2 b^9 c^2 - 3 a b^10 c^2 + 

  3 b^11 c^2 - 6 a^9 b c^3 + 10 a^7 b^3 c^3 + 2 a^6 b^4 c^3 + 

  4 a^5 b^5 c^3 - 3 a^4 b^6 c^3 - 10 a^3 b^7 c^3 + 4 a^2 b^8 c^3 + 

  2 a b^9 c^3 - 3 b^10 c^3 - 4 a^8 b c^4 + 3 a^7 b^2 c^4 - 

  3 a^5 b^4 c^4 + 2 a^4 b^5 c^4 - 3 a^3 b^6 c^4 + 4 a^2 b^7 c^4 + 

  3 a b^8 c^4 - 2 b^9 c^4 + 4 a^7 b c^5 - a^6 b^2 c^5 - 

  16 a^5 b^3 c^5 - 3 a^4 b^4 c^5 + 4 a^3 b^5 c^5 - 2 a^2 b^6 c^5 + 

  4 a b^7 c^5 + 2 b^8 c^5 + 6 a^6 b c^6 - a^5 b^2 c^6 + 

  2 a^3 b^4 c^6 - 4 a^2 b^5 c^6 - a b^6 c^6 - 2 b^7 c^6 + 

  4 a^5 b c^7 + 3 a^4 b^2 c^7 + 10 a^3 b^3 c^7 + 4 a^2 b^4 c^7 - 

  4 a b^5 c^7 + 2 b^6 c^7 - 4 a^4 b c^8 + a^2 b^3 c^8 + 3 b^5 c^8 - 

  6 a^3 b c^9 - 3 a^2 b^2 c^9 - 2 a b^3 c^9 - 3 b^4 c^9 + a^2 b c^10 -

   b^3 c^10 + 2 a b c^11 + b^2 c^11 + 2 a^9 b T - 2 a^8 b^2 T - 

  2 a^7 b^3 T + 4 a^6 b^4 T - 6 a^5 b^5 T + 10 a^3 b^7 T - 

  4 a^2 b^8 T - 4 a b^9 T + 2 b^10 T + 2 a^9 c T - 4 a^8 b c T + 

  4 a^7 b^2 c T - 8 a^5 b^4 c T + 8 a^4 b^5 c T - 4 a^3 b^6 c T + 

  6 a b^8 c T - 4 b^9 c T - 2 a^7 b c^2 T + 4 a^6 b^2 c^2 T + 

  12 a^5 b^3 c^2 T - 4 a^4 b^4 c^2 T - 10 a^3 b^5 c^2 T + 

  4 a^2 b^6 c^2 T - 4 b^8 c^2 T - 8 a^7 c^3 T + 10 a^6 b c^3 T - 

  4 a^5 b^2 c^3 T - 10 a^4 b^3 c^3 T + 16 a^3 b^4 c^3 T - 

  10 a^2 b^5 c^3 T - 4 a b^6 c^3 T + 10 b^7 c^3 T - 6 a^5 b c^4 T - 

  4 a^4 b^2 c^4 T - 10 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T + 

  8 a b^5 c^4 T + 12 a^5 c^5 T - 6 a^4 b c^5 T - 4 a^3 b^2 c^5 T + 

  12 a^2 b^3 c^5 T - 8 a b^4 c^5 T - 6 b^5 c^5 T + 10 a^3 b c^6 T + 

  4 a^2 b^2 c^6 T + 4 b^4 c^6 T - 8 a^3 c^7 T - 2 a^2 b c^7 T + 

  4 a b^2 c^7 T - 2 b^3 c^7 T - 4 a b c^8 T - 2 b^2 c^8 T + 

  2 a c^9 T + 2 b c^9 T, 

 2 a^11 b c + a^10 b^2 c - 6 a^9 b^3 c - 4 a^8 b^4 c + 4 a^7 b^5 c + 

  6 a^6 b^6 c + 4 a^5 b^7 c - 4 a^4 b^8 c - 6 a^3 b^9 c + 

  a^2 b^10 c + 2 a b^11 c + a^11 c^2 - 3 a^9 b^2 c^2 + 3 a^7 b^4 c^2 -

   a^6 b^5 c^2 - a^5 b^6 c^2 + 3 a^4 b^7 c^2 - 3 a^2 b^9 c^2 + 

  b^11 c^2 - a^10 c^3 - 2 a^9 b c^3 + a^8 b^2 c^3 + 10 a^7 b^3 c^3 - 

  16 a^5 b^5 c^3 + 10 a^3 b^7 c^3 + a^2 b^8 c^3 - 2 a b^9 c^3 - 

  b^10 c^3 - 3 a^9 c^4 + 4 a^7 b^2 c^4 + 2 a^6 b^3 c^4 - 

  3 a^5 b^4 c^4 - 3 a^4 b^5 c^4 + 2 a^3 b^6 c^4 + 4 a^2 b^7 c^4 - 

  3 b^9 c^4 + 3 a^8 c^5 - 4 a^7 b c^5 - 4 a^6 b^2 c^5 + 

  4 a^5 b^3 c^5 + 2 a^4 b^4 c^5 + 4 a^3 b^5 c^5 - 4 a^2 b^6 c^5 - 

  4 a b^7 c^5 + 3 b^8 c^5 + 2 a^7 c^6 - a^6 b c^6 - 2 a^5 b^2 c^6 - 

  3 a^4 b^3 c^6 - 3 a^3 b^4 c^6 - 2 a^2 b^5 c^6 - a b^6 c^6 + 

  2 b^7 c^6 - 2 a^6 c^7 + 4 a^5 b c^7 + 4 a^4 b^2 c^7 - 

  10 a^3 b^3 c^7 + 4 a^2 b^4 c^7 + 4 a b^5 c^7 - 2 b^6 c^7 + 

  2 a^5 c^8 + 3 a^4 b c^8 + 4 a^3 b^2 c^8 + 4 a^2 b^3 c^8 + 

  3 a b^4 c^8 + 2 b^5 c^8 - 2 a^4 c^9 + 2 a^3 b c^9 - 5 a^2 b^2 c^9 + 

  2 a b^3 c^9 - 2 b^4 c^9 - 3 a^3 c^10 - 3 a^2 b c^10 - 

  3 a b^2 c^10 - 3 b^3 c^10 + 3 a^2 c^11 - 2 a b c^11 + 3 b^2 c^11 + 

  a c^12 + b c^12 - c^13 + 2 a^9 b T - 8 a^7 b^3 T + 12 a^5 b^5 T - 

  8 a^3 b^7 T + 2 a b^9 T + 2 a^9 c T - 4 a^8 b c T - 2 a^7 b^2 c T + 

  10 a^6 b^3 c T - 6 a^5 b^4 c T - 6 a^4 b^5 c T + 10 a^3 b^6 c T - 

  2 a^2 b^7 c T - 4 a b^8 c T + 2 b^9 c T - 2 a^8 c^2 T + 

  4 a^7 b c^2 T + 4 a^6 b^2 c^2 T - 4 a^5 b^3 c^2 T - 

  4 a^4 b^4 c^2 T - 4 a^3 b^5 c^2 T + 4 a^2 b^6 c^2 T + 

  4 a b^7 c^2 T - 2 b^8 c^2 T - 2 a^7 c^3 T + 12 a^5 b^2 c^3 T - 

  10 a^4 b^3 c^3 T - 10 a^3 b^4 c^3 T + 12 a^2 b^5 c^3 T - 

  2 b^7 c^3 T + 4 a^6 c^4 T - 8 a^5 b c^4 T - 4 a^4 b^2 c^4 T + 

  16 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T - 8 a b^5 c^4 T + 4 b^6 c^4 T - 

  6 a^5 c^5 T + 8 a^4 b c^5 T - 10 a^3 b^2 c^5 T - 10 a^2 b^3 c^5 T + 

  8 a b^4 c^5 T - 6 b^5 c^5 T - 4 a^3 b c^6 T + 4 a^2 b^2 c^6 T - 

  4 a b^3 c^6 T + 10 a^3 c^7 T + 10 b^3 c^7 T - 4 a^2 c^8 T + 

  6 a b c^8 T - 4 b^2 c^8 T - 4 a c^9 T - 4 b c^9 T + 2 c^10 T}

 

Frabcisco Javier Garcia Capitan

 

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#4587

Dear Mr Francisco and friends,

 

I see similar problem with excenters.

 

H,O and I are orthocenter, circumcenter and incenter of ABC.

 

If Ja, Jb, Jc are incenters of AOH, BOH and COH.

 

If Ja’, Jb’, Jc’ are A,B,C excenters of AOH, BOH, COH.

 

I1 is isogonal conjugate of I wrt triangle JaJbJc.

 

I2 is isogonal conjugate of I wrt triangle Ja’Jb’Jc’.

 

Then I1, I2 are on line OH and they are reflection in NPC center of ABC.

 

Best regards,

Tran Quang Hung.