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.

 

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

#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

 

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

#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

 

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

#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.