HYACINTHOS 26562

 

[Puiu Braica]:

Let ABC be a triangle.
Denote:
(Oa), (Ob), (Oc) = the circles with diameters the sides BC, CA, AB, resp

La, Lb, Lc = the common tangents of the circles ((Ob), (Oc)), ((Oc), (Oa)), ((Oa), (Ob)), resp. on the positive sides of the sidelines BC, CA, AB, resp.

Tbc, Tcb = the touchpoints of La and  (Ob), (Oc), resp.
Tca, Tac = the touchpoints of Lb and (Oc), (Oa), resp.
Tab, Tba = the touchpoints of Lc and (Oa), (Ob), resp.

The perpendicular bisectors of TbcTcb, TcaTac, TabTba are concurrent.

Point?

Reference: Puiu Braica, Olimpiada pe Şcoală (The School Yard Olympiad)

How about if we take the tangents of the circles which are on the negative sides of BC,CA,AB ?

 

[César Lozada]:

 

Tangents in the positive sides of ABC:

 

Q₁ = Q(a,b,c,S) = F(a,b,c)*sin(A/2)+G(a,b,c)* sin(A/2)+G(a,c,b)*sin(C/2)+H( a,b,c) : :  (barycentrics)

where:

F(a,b,c)= -4*S*sqrt(b*c)*((b^2+c^2)*a^3- (b^3+c^3)*a^2-(b^2-c^2)^2*a+( b^3-c^3)*(b^2-c^2))

G(a,b,c)= -4*S*sqrt(a*c)*(2*S*(a-b+c)*b^ 2-c^2*(2*a^3-(b+c)*a^2-(b^2-c^ 2)*(b-c)))

H(a,b,c)= (2*a^3-(b+c)*a^2-(b^2-c^2)*(b- c))*((b^2+c^2)*a^3-(b^3+c^3)* a^2-(b^2-c^2)^2*a+(b^3-c^3)*( b^2-c^2))

= on lines {Q₂,M}

= [ 4.472118485952783, 3.21050394391414, -0.646046780472845 ]

 

Tangents in the negative sides of ABC:

 

Q₂ = Q(a,b,c,-S), where Q(a,b,c,S) is given above,

= on lines {Q₁, M}

= [ 1.867277382796174, 0.74224626938746, 2.264981349502519 ]

 

 

The midpoint M of Q₁Q₂ is:

 

M = complement of X(508)

= (sin(B/2)*sqrt(c)+sin(C/2)* sqrt(b))*sqrt(a) : : ( (barycentrics)

= on line {2,508}

= complement of X(508)

= [ 3.169697936448507, 1.97637510627965, 0.809467283814538 ]

 

César Lozada