ADGEOM 4383 * ADGEOM 4386 * ADGEOM 4389 * ADGEOM 4392 * ADGEOM 4393

 

#4383

Dear Geometers!

 

Let ABC be a triangle, the lines through X13 perpendicular to BC meets AX14 at A'; the line through X14 perspendicular to BC meets AX13 at A''. Define B', C', B'', C'' cyclically.

Then A'B'C' and A''B''C'' are two new equilateral triangle.

Which is centroids of these triangle  in ETC?

 

Best regards

Sincerely
Dao Thanh Oai

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#4386

The centroids of the triangles  A'B'C' and A''B''C'' are X(15) and X(16), respec.

Angel Montesdeoca

 

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#4389

Dear Mr Dao Thanh Oai,

 

Your triangle A'B'C' is homothetic to the first Napoleon triangle and the homothetic center lies on line OX(14).

 

Best regards,

Tran Quang Hung.

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#4392

Dear Mr Hung and Mr Oai,

Homothetic centers are:

 

P1 = (A’B’C’, outer-Napoleon) = X(2)X(14) ∩ X(3)X(13)

= -2*sqrt(3)*S*a^2+5*a^4-7*(b^2+c^2)*a^2+2*(b^2-c^2)^2 : : (barys)

= 7*S^2+sqrt(3)*(SB+SC)*S-3*SB*SC : : (barys)

= on lines: {2, 14}, {3, 13}, {4, 5352}, {5, 5238}, {6, 5054}, {16, 396}, {18, 3526}, {30, 10645}, {61, 140}, {62, 631}, {203, 5432}, {298, 11132}, {299, 618}, {303, 3643}, {381, 11480}, {397, 3530}, {398, 632}, {547, 5321}, {624, 11299}, {629, 633}, {630, 11290}, {1606, 3131}, {3104, 7786}, {3111, 14182}, {3201, 5012}, {3364, 5420}, {3365, 5418}, {3390, 15765}, {3523, 5237}, {3524, 10646}, {3533, 10187}, {3534, 12816}, {3582, 10638}, {3584, 7051}, {3851, 10188}, {4045, 11298}, {5070, 5339}, {5318, 8703}, {5335, 15692}, {5350, 12103}, {5433, 7005}, {5444, 7052}, {5463, 5569}, {5470, 6772}, {5474, 9735}, {5642, 10658}, {6669, 11303}, {6694, 11308}, {6774, 9117}, {6778, 12042}, {7619, 9761}, {7761, 11297}, {9116, 9885}, {10124, 11543}, {10182, 11244}, {11134, 13339}, {11268, 15330}, {11481, 15693}, {11485, 15694}, {11486, 15701}, {11489, 15709}, {11542, 12100}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2, 617, 623), (2, 13083, 5464), (396, 549, 16), (397, 3530, 5351), (619, 5981, 5464), (619, 6671, 2)

= [ 2.9030202383909050, 2.5290443728959210, 0.5499321137221704 ]

 

P2 = (A”B”C”, inner-Napoleon) = X(2)X(13) ∩ X(3)X(14)

= 2*sqrt(3)*S*a^2+5*a^4-7*(b^2+c^2)*a^2+2*(b^2-c^2)^2 : : (barys)

= 7*S^2-sqrt(3)*(SB+SC)*S-3*SB*SC : : (barys)

= on lines: {2, 13}, {3, 14}, {4, 5351}, {5, 5237}, {6, 5054}, {15, 395}, {17, 3526}, {30, 10646}, {61, 631}, {62, 140}, {202, 5432}, {298, 619}, {299, 11133}, {302, 3642}, {381, 11481}, {397, 632}, {398, 3530}, {547, 5318}, {623, 11300}, {629, 11289}, {630, 634}, {1250, 3582}, {1605, 3132}, {2306, 5442}, {3105, 7786}, {3111, 14178}, {3200, 5012}, {3364, 15765}, {3389, 5420}, {3390, 5418}, {3523, 5238}, {3524, 10645}, {3533, 10188}, {3534, 12817}, {3851, 10187}, {4045, 11297}, {5070, 5340}, {5321, 8703}, {5334, 15692}, {5349, 12103}, {5433, 7006}, {5464, 5569}, {5469, 6775}, {5473, 9736}, {5642, 10657}, {6670, 11304}, {6695, 11307}, {6771, 9115}, {7619, 9763}, {7761, 11298}, {10124, 11542}, {10182, 11243}, {11137, 13339}, {11267, 15330}, {11480, 15693}, {11485, 15701}, {11486, 15694}, {11488, 15709}, {11543, 12100}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2, 616, 624), (2, 13084, 5463), (395, 549, 15), (398, 3530, 5352), (618, 5980, 5463), (618, 6672, 2)

= [ 8.2105967007085710, 5.4179799383431200, -3.8997508757340440 ]

 

·         {P1,P2} are harmonic-conjugates w/r to {X(6), X(5054)}

 

·         The center of inverse-similitude of A’B’C’ and A”B”C” is X(98)

 

César Lozada

 

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#4393

 

Thank you very much Mr Lozada,

 

I see a general problem for problem of Mr Dao as following

 

Let ABC be a triangle. P is any point.

 

(H) is rectangular hyperbola passes through A,B,C,P.

 

Q is a point on (H).

 

Perpendicular lines from P to BC,CA,AB meet QB,QB,QC at A',B',C', reps.

 

Then we get equal angles <B'A'C'=180-<BPC.

 

When P=F1 and Q is any point on Kiepert hyperbola then we get A'B'C' is equilateral triangle, is this triangle homothetic to the first Napoleon triangle and the homothetic center is on line X(2)X(14) ?

 

Best regards,

Tran Quang Hung.