Another proof of six points theorem

March 21 2014 Kitonum 3275

5

This post in reply to the Post, Six points on circumference

In this post we present another compact proof of this remarkable theorem without using  geometry package.
The proof uses a procedure called  Cc , which for three points returns a list of the coordinates of the center and the radius of the circumscribed circle.

restart;

Cc:=proc(A,B,C)

local x1, y1, x2, y2, x3, y3, x, y;

x1,y1:=op(A);  x2,y2:=op(B);  x3,y3:=op(C);

solve({(x2-x1)*(x-(x1+x2)/2)+(y2-y1)*(y-(y1+y2)/2)=0, (x2-x3)*(x-(x2+x3)/2)+(y2-y3)*(y-(y2+y3)/2)=0},{x,y});

assign(%);

[simplify([x,y]), simplify(sqrt((x-x1)^2+(y-y1)^2))];

end proc:

Proof for arbitrary triangle:

A, B, C:=[x1,y1], [x2,y2], [x3,y3]:

A1, B1, C1, M:=(B+C)/2, (A+C)/2, (A+B)/2, (A+B+C)/3:

P1:=Cc(A,M,B1)[1]: P2:=Cc(B1,M,C)[1]: P3:=Cc(C,M,A1)[1]:

P4:=Cc(A1,M,B)[1]: P5:=Cc(B,M,C1)[1]: P6:=Cc(C1,M,A)[1]:

Cc1:=Cc(P1,P2,P3):  Cc2:=Cc(P4,P5,P6):

is(Cc1=Cc2);

                                                  true

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