[1] The area of the pedal triangle is

area(P

where R is the circumradius of ABC and O is the circumcenter (R

[2] For P moving on circles concentric to the circumcircle, the pedal triangles have constant area and vice versa. In particular, for P on the circumcircle the corresponding area is zero, the three projection-points being on the Simson line of P.

[3] Denoting by (x,y,z) the trilinear coordinates of P, and considering signed (oriented) areas:

area(P

is a quadratic form in the trilinear coordinates.

[1] The formula follows by extending PC to cut the circumcircle of ABC at D and observing that angle(P

In addition, since AP

Now using (*) and (**) the area of the pedal triangle is:

area(P

= (R

[2] Follows immediately from [1].

[3] Follows by dividing the area of the triangle in the sum: area(P

[1] Formula (1) is general valid, even when point P is outside the triangle, provided we use oriented areas. Next figure illustrates the corresponding proof.

[2] By equating the expressions in (1) and (3) we get:

sin(A)*y*z+sin(B)*z*x+sin(C)*x*y = (R

a*y*z+b*z*x+c*x*y = (R

By taking P on the circle we get its equation in trilinears (see also CircumcircleInTrilinears.html ).

[3] The previous equation shows that the quadratic form f(x,y,z) = a*y*z+b*z*x+c*x*y, where a, b, c denote the lengths of the sides of the triangle, is positive inside the circumcircle, zero on the circumcircle and negative outside. Notice that (x,y,z) are not independent, but satisfy the equation a*x+b*y+c*z = 2*area(ABC). Points at infinity satisfy a*x+b*y+c*z = 0 and fall "far out" of the circumcircle of ABC, there where f is negative. This is in accordance with the formula for the distance of two points in trilinears (see last formula in BarycentricsFormulas.html ).

CircumcircleInTrilinears.html

Pedal.html

Projection_Polygon.html

Simson.html

Produced with EucliDraw© |