HYACINTHOS 27132

[Le Viet An]:

Let ABC be a triangle.

Let AaBaCa be the equilateral triangle such that Aa,Ab,Ac lie on BC, CA, AB, resp. and AbAc // BC
 
Similarly BbBcBa, CcCaCb.

Then

1. AAa, BBb, CCc are concurrent at X.
Point?

2. Let Ia, Ib, Ic be the centers of the triangles AaBaCa, BbBcBa, CcCaCb, resp.
Then AIa, BIb, CIc are concurrent at Y and X, Y, O are collinear.
Point?

3. Let Da = BbBc /\ CcCb, Db = CcCa /\ AaAc, Dc = AaAb /\ BbBa.
Then ADa, BDb, CDc are concurrent at Z.
Point?

4. Bb, Cb, Cc, Ba lie on a circle (Oa). Similarly (Ob), (Oc)
Then Oa, Ob, Oc, Z lie on a circle.
Center of the circle ?

5. (Oa), (Ob), (Oc) are concurrent.
Point of concurrence? 

 

 
[César Lozada]:

 

Several triads of equilateral triangles Ta, Tb, Tc can be built from hypothesis.

 

I. Assuming these triangles are built all inwards ABC.

 

1)      X = X(13) (Aa is the trace of X(13) on BC )

2)      Y= X(17)

3)      Z=X(15)

4)      O*= X(15)X(1337) ∩ X(61)X(1994)

= ((SA+SW)*sqrt(3)+4*S)*(SA+ sqrt(3)*S)*(SB+SC) : : (barys)

= on lines: {15, 1337}, {30, 11555}, {61, 1994}, {154, 3129}

= [ 2.3369930591055660, 2.3362491363113890, 0.9446490525662327 ]

5)      X(5612)

 

II.. Assuming Ta,Tb,Tc are built outwards ABC: (just change S->-S in the above points)

1)      X(14) (Aa is the trace of X(14) on BC)

2)      X(18)

3)      X(16)

4)      O*= X(16)X(1338) ∩ X(62)X(1994)

= ((SA+SW)*sqrt(3)-4*S)*(SA- sqrt(3)*S)*(SB+SC) : : (barys)

= on lines: {16, 1338}, {62, 1994}, {154, 3130}

= [ -5.9307871131524560, 4.8301059889122620, 3.0340313110385630 ]

5)      X(5616)

 

Another property:

In each case, Ab, Ac, Ba, Bc, Ca, Cb are conconic. The center of the conic is:

Case I:

Q = X(323)X(2981) ∩ X(396)X(11063)

= ((45*R^2+8*SA+6*SW)*S^2+4* sqrt(3)*(3*(SA+SW)*R^2+2*SA^2+ 2*S^2)*S+3*(3*R^2+2*SW)*SA^2)* (SB+SC) : : (barys)

= on lines: {323, 2981}, {396, 11063}

= [ 1.9861709716792590, 1.8147731178363090, 1.4675887980149690 ]

 

Case II:

Q = X(323)X(6151) ∩ X(395)X(11063)

= ((45*R^2+8*SA+6*SW)*S^2-4* sqrt(3)*(3*(SA+SW)*R^2+2*SA^2+ 2*S^2)*S+3*(3*R^2+2*SW)*SA^2)* (SB+SC) : : (barys)

= on lines: {323, 6151}, {395, 11063}

= [ -8.8996520507762580, 7.3396841714445220, 2.6668763866502900 ]

 

Other cases, not studied, include one triangle inwards ABC and two triangles outwards ABC  or two triangles inwards ABC and one triangle outwards ABC

 

César Lozada