ADGEOM 2197 * ADGEOM 2198 * ADGEOM 2199

#2197

 

Dear Dr. Garcia Capitan, Dr. Randy Hutson and all geometer,

 

Let ABC be a triangle, let a circle through A_c,A_b,C_b,C_a,B_a,B_c,A_c

 such that the center is Incenter (X1) and radius =2r/sqrt{3). Let A_1,B_1,C_1 be 

intersections of three circle (IB_aB_c), (IC_aC_b), (IA_bA_c) then

A_1B_1C_1 be an equilateral triangle. 

 

1-Which is the center of A1B1C1 in Kimberling center?

 

2-ABC and A_1B_1C_1 are perpective which is the perpector in Kimberling center?

 

Thank to You very much,

 

Best regards

Sincerely

Dao Thanh Oai

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#2198

 

Dear Mr. Dao:

 

Maybe I didn´t understand well, but A1B1C1 is not equilateral.

I supposed:

   {Ab,Ac} = {I,2*r/sqrt(3)} /\ (BC)

   A1 =Circle{ I,Ba,Bc}/\ Circle{ I,Ca,Cb}-{I}

 

Perspector ABC, A1B1C1 is I.

 

Trilinears of A1 are:

A1 = [((b+c)*a^2+3*a*b*c-(b^2-c^2)*(b-c))/(a*(c^2+b*c+b^2-a^2)), 1, 1]

and cyclically for B1, C1

 

César Lozada

 

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#2199

 

Dear Dao,

 

If I am correct the perspector is the point with barycentrics

S/(s-a)+32a/Sqrt[3] : . . .

Search number  1.58801253144064532

 

Best regards

Nikos Dergiades