>>The following problem could be of interest:
>>Let Po be a point in triangle AoBoCo,and A1B1C1 be
>>cevian triangle of Po,P1 be the point in triangle A1B1C1
>>with same position wrt to A1B1C1 as Po wrt to AoBoCo,
>>A2B2C2 be cevian triangle of P1 in triangle A1B1C1,
>>P2 be the point in triangle A2B2C2 with the
>>same position wrt A2B2C2 as Po wrt to AoBoCo and so on.
>>If Po(x:y:z)(in barycentrics),then the limit point P' of
>>sequence of points Po,P1,P2,...,Pn,...
>>is given by(if my calculations are good)
>>From a given point Po it is posible to construct
>>point P' by ruler and compass.
>>Sorry,if this problem and solutions are known.
>I think it would be better to say that point Pj
>have the same Cevian ratios wrt to triangle AjBjCj
>for all j.
>If D1,D2,D3 are the midpoints of B0C0,C0A0,A0B0 and
>if E1,E2,E3 are the midpoints of cevians A0A1,B0B1,C0C1
>then the lines D1E1,D2E2,D3E3 intersect in point P'.
>Special cases are
>1.P=H,point P' is Lemoine point K.
>2.P=Ge (Gergonne point),point P' is incenter I.
>(all wrt to A0B0C0).
If this property of the lines D1E1,D2E2,D3E3
is a method of construction of the point P' then
a second method is by drawing the lines A0M1, B0M2, C0M3
where M1 is the mid point of B1C1 etc
because P=[x : y : z] B1 = [x : 0 : z] = [x(x+y) : 0 :
C1 = [x : y : 0] = [x(x+z) : y(x+z) : 0] and hence the mid
M1 = [x(x+y)+x(x+z) : y(x+z) : z(x+y)] or that the line A0M1
passes through the point P' etc and a third method is by
drawing the complement of the isotomic conjugate of P
since P' = [x(y+z) : y(z+x) : z(x+y)] = [1/y+1/z :1/z+1/x