Construct the vertex of an angle given three points on the plane.

Question

Suppose you are given three points A,B,C on the plane. A is on the bisector of angle XOY, B is on the side OX and C is on the side OY. How can you find the locus of the vertices(find all possile points O)?

Answer 1

Taking $P=(x,y)$, the locus of $P$ is given by $$\cos\angle APB=\cos\angle APC\quad\to\quad\frac{\overrightarrow{PA}\cdot\overrightarrow{PB}}{|\overrightarrow{PA}||\overrightarrow{PB}|} = \frac{\overrightarrow{PA}\cdot\overrightarrow{PC}}{|\overrightarrow{PA}||\overrightarrow{PC}|}$$ For the specific case of $A=(a_x,a_y)$, $B=(-d,0)$, $C=(d,0)$ the equation becomes

$$\left(\,x (x - a_x) + y (y - a_y)\,\right) \left(\,a_y (x - a_x) - a_x (y - a_y)\,\right) + d^2 (x - a_x) (y - a_y) = 0\tag{$\star$}$$

representing a cubic curve passing through $A$ (twice), $B$, and $C$:

When $A$ lies on the perpendicular bisector of $\overline{AB}$ (ie, when $a_x=0$ in $(\star)$), the equation factors

$$x \left(\;a_y \left(x^2 + y^2 - d^2\right) - y \left(a_y^2 - d^2\right)\;\right) = 0 \tag{$\star'$}$$

so that the locus is the union of that perpendicular bisector with the circumcircle of $\triangle ABC$: