Question

Given ΔABC, where
A = (x1, x1 TanФ1) , B=(x2, x2 TanФ2) and C = (x3, x3 TanФ3)
Circumcenter O = (0, 0).
Orthocenter H = (x₄, y₄) = $(\bar{x}, \bar{y})$.

To show that y4/x4 = [SinФ1 + sinФ2 + SinФ3] / [CosФ1 + CosФ2 + CosФ3]

Answer 1

Given ΔABC, where
     A = (x1, x1 TanФ1)  ,  B=(x2, x2 TanФ2)  and   C = (x3, x3 TanФ3)
     Circumcenter O = (0, 0).
     Orthocenter H = (x₄, y₄) = .

To show that  y4/x4 = [SinФ1 + sinФ2 + SinФ3] / [CosФ1 + CosФ2 + CosФ3]

Solution:
Now we know that OA² = OB² = OC² = R² , where R = circumradius
 
=>  x1² (1 + Tan²Ф1)  =  x2² (1+Tan²Ф2)  =  x3² (1 + Tan²Ф3)  = R²
=>  x1 / Cos Ф1   =   x2 / Cos Ф2   =   x3 / Cos Ф3 = R
=>  x1 = R CosФ1,  x2 = R CosФ2,  x3 = R CosФ3

Now, Points    A = (R CosФ1, R SinФ1),  B = (R CosФ2, R SinФ2)
            and     C = (R CosФ3, R Sin Ф3)

Centroid of the circle = G
   = [ R(CosФ1 +CosФ2 + CosФ3)/3 ,  R (SinФ1 + SInФ2 + SInФ3)/3 ]

Slope of line OG = (SinФ1 + SinФ2 + SinФ3) / (CosФ1 + CosФ2+CosФ3)

We know that in any triangle the three points circumcenter O, Centroid G and the Orthocenter H are always collinear.

=> Slope of OH = 
             = y₄ / x₄ = slope of OG 
             = (SinФ1 + SInФ2 +SinФ3) / (CosФ1 + CosФ2 + CosФ3)