WLOG, assume that the circle is the unit circle centered at the origin. Then the points $A$ and $B$ have coordinates $(-a,b)$ and $(a,b)$ respectively and $X$ and $Y$ have coordinates $(r,s)$ and $(-r,-s)$. Note that these coordinates satisfy $a^2 + b^2 = 1$ and $r^2 + s^2 = 1$ since these points are on a unit circle. Now we can find equations for the lines: \begin{align*} AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\\ BY \longrightarrow y &= \frac{(s+b)x+rb-sa}{r+a}. \end{align*} Solving these simultaneous equations gives coordinates for $P$ in terms of $a, b, r,$ and $s$: $P = \left(\frac{as}{b},\frac{1 - ar}{b}\right)$. These coordinates can be parametrized in Cartesian variables as follows: \begin{align*} x &= \frac{as}{b}\\ y &= \frac{1 - ar}{b}. \end{align*} Now solve for $r$ and $s$ to get $r = \frac{1-by}{a}$ and $s = \frac{bx}{a}$ . Then since $r^2 + s^2 = 1, \left(\frac{bx}{a}\right)^2 + \left(\frac{1-by}{a}\right)^2 = 1$ which reduces to $x^2 + (y-1/b)^2 = \frac{a^2}{b^2}.$ This equation defines a circle and is the locus of all intersection points $P$. In order to define this locus more generally, find the slope of this circle function using implicit differentiation: \begin{align*} 2x + 2(y-1/b)y' &= 0\\ (y-1/b)y' &= -x\\ y' &= \frac{-x}{y-1/b}. \end{align*} Now note that at points $A$ and $B$, this slope expression reduces to $y' = \frac{-b}{a}$ and $y' = \frac{b}{a}$ respectively, values which are identical to the slopes of lines $AO$ and $BO$. Thus we conclude that the complete locus of intersection points is the circle tangent to lines $AO$ and $BO$ at points $A$ and $B$ respectively.

DaChickenInc wrote:

Umm, <PXY=180-AXO... I don't understand. Are there any hints for a different synthetic sol?