Παρασκευή 1 Νοεμβρίου 2019

ADGEOM 5172 * ADGEOM 5175 * ADGEOM 5176

#5172

Let ABC be a triangle with orthocenter H.
 
Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.
 
Which is this concurrent point?
 
Best regards,
Tran Quang Hung.


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#5175
 
[Tran Quang Hung]
 
Let ABC be a triangle with orthocenter H.
 
Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.
 
Which is this concurrent point?
 
--------------------------------------------------------------------------------------------
 
 
[Ercole Suppa]
 
Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively concur at point
 
X = midpoint of X(382) and X(13512)
 
= a^16-3 a^14 b^2+a^12 b^4+6 a^10 b^6-10 a^8 b^8+9 a^6 b^10-7 a^4 b^12+4 a^2 b^14-b^16-3 a^14 c^2+4 a^12 b^2 c^2+4 a^10 b^4 c^2-8 a^8 b^6 c^2+11 a^4 b^10 c^2-13 a^2 b^12 c^2+5 b^14 c^2+a^12 c^4+4 a^10 b^2 c^4-3 a^8 b^4 c^4-7 a^4 b^8 c^4+15 a^2 b^10 c^4-10 b^12 c^4+6 a^10 c^6-8 a^8 b^2 c^6+6 a^4 b^6 c^6-6 a^2 b^8 c^6+11 b^10 c^6-10 a^8 c^8-7 a^4 b^4 c^8-6 a^2 b^6 c^8-10 b^8 c^8+9 a^6 c^10+11 a^4 b^2 c^10+15 a^2 b^4 c^10+11 b^6 c^10-7 a^4 c^12-13 a^2 b^2 c^12-10 b^4 c^12+4 a^2 c^14+5 b^2 c^14-c^16 :: (barys)
 
= S^4 + (3 R^4-5 R^2 SB-5 R^2 SC+SB SC+R^2 SW+2 SB SW+2 SC SW-SW^2)S^2 -27 R^4 SB SC+23 R^2 SB SC SW-5 SB SC SW^2 :: (barys)
 
= on circumcircle of JohnsonTriangle
 
= on lines X(i)X(j) for these {i,j}: {3,128},{4,11671},{5,49},{30,930},{137,381},{140,23237},{382,13512},{523,14980},{539,16337},{546,1263},{550,6592},{1209,11016},{1478,3327},{1479,7159},{2070,23181},{2072,23319},{3153,20957},{3627,14073},{3652,10747},{3850,25147},{3851,23516},{3853,23238},{11459,13504},{12111,13505},{12254,14071},{13556,22823},{14050,18439},{14142,21230},{16336,18400}
 
= combo: X[3]-2*X[128], 3*X[4]-X[11671], 2*X[137]-3*X[381], 2*X[140]-3*X[23237], X[382]+X[13512], 2*X[546]-X[1263], X[550]-2*X[6592], X[3627]+X[14073], 4*X[3850]-3*X[25147], 7*X[3851]-6*X[23516], 2*X[3853]+X[23238], 3*X[11459]-X[13504], X[12111]+X[13505], X[12254]-2*X[14071]
 
= midpoint of X(i) and X(j) for these {i,j}: {382,13512},{12111,13505}
 
= reflection of X(i) in X(j) for these {i,j}: {3,128},{550,6592},{930,14072},{1141,5},{1263,546},{14674,14769},{19553,16337}
 
= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: 
 
= (6-9-13) search numbers [1.68062122615268063, -2.17060108179101817, 4.36771697261538524]
 
 
Best regards
Ercole Suppa
 
 
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#5176
 

Let ABC be a triangle with orthocenter H.

 

Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

 

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Q = X(3)X(128) ∩ X(5)X(49)

= S^4-(R^2*(3*R^2+5*SA-4*SW)-2*SA^2+3*SB*SC+SW^2)*S^2+(R^2*(27*R^2-23*SW)+5*SW^2)*SB*SC : : (barys)

= (6*cos(2*A)+4*cos(4*A)+5)*cos(B-C)-2*(cos(A)+cos(3*A))*cos(2*(B-C))+(2*cos(2*A)+1)*cos(3*(B-C))-cos(5*A)-3*cos(A)-cos(3*A) :: (trilinears)

= 3*X(3)-4*X(13372), 3*X(4)-X(11671), 3*X(5)-2*X(12026), 9*X(5)-8*X(25339), 3*X(128)-2*X(13372), 2*X(137)-3*X(381), 2*X(140)-3*X(23237), 4*X(3850)-3*X(25147), 7*X(3851)-6*X(23516), 2*X(3853)+X(23238), 3*X(11459)-X(13504), 3*X(12026)-4*X(25339)

= lies on these lines: {3, 128}, {4, 11671}, {5, 49}, {30, 930}, {137, 381}, {140, 23237}, {382, 13512}, {523, 14980}, {539, 16337}, {546, 1263}, {550, 6592}, {1209, 11016}, {1478, 3327}, {1479, 7159}, {2070, 23181}, {2072, 23319}, {3153, 20957}, {3627, 14073}, {3652, 10747}, {3850, 25147}, {3851, 23516}, {3853, 23238}, {11459, 13504}, {12111, 13505}, {12254, 14071}, {13556, 22823}, {14050, 18439}, {14142, 21230}, {16336, 18400}

= midpoint of X(i) and X(j) for these {i,j}: {382, 13512}, {3627, 14073}, {12111, 13505}

= reflection of X(i) in X(j) for these (i,j): (3, 128), (550, 6592), (930, 14072), (1141, 5), (1263, 546), (12254, 14071), (14674, 14769), (19553, 16337)

= reflection of X(265) in the line X(4)X(14050)

= (Ehrmann-side)-isogonal conjugate of-X(1510)

= (Johnson)-isogonal conjugate of-X(1154)

= [ 1.6806212261526800, -2.1706010817910170, 4.3677169726153860 ]

 

César Lozada

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