What is the set of points P for which $PA/PA_1+PB/PB_1+PC/PC_1=3$? $A, B$ and $C$ are the points of 6,8,10 triangle. $A_1, B_1, C_1$ are the intersection points of $PA$, $PB$ and $PC$ with the circumcircle of $ABC$. I managed to find the answer but I struggle to come up with an elegant solution. I found it with the help of $3d$ geogebra by writing out the equation for the sum of the ratios minus 3 depending on the coordinates of the point P and seeing for which points the graph is zero. It seemed like the set of points is the circle with a center $I$, such that $I$ is on $OC$ and $IO/OC=1/5$ and with a radius $r=OC/6,$ where $O$ is the center of the circumcircle of $ABC$. That circle goes through $O$ and $G$ - the centroid of $ABC$. The property asked for in the problem statement is obviously true for the point $O$ and I verified it for point $G.$ I checked for some other points on this circle and it was true for them as well. So it seems likely that this is in fact the correct answer but I wonder if there is a more clever and beautiful way to solve the problem without too much algebraic fiddling. A fact that might help, as the first sub task of the problem was to prove it, is that $PA^2+PB^2+PC^2=3*GP^2+GA^2+GB^2+GC^2$.
There was no solution for the problem in the book where I found it and I can’t find anything related in the internet.
By the intersecting chords theorem we know that: $$ PA\cdot PA_1=PB\cdot PB_1=PC\cdot PC_1= |r^2-PO^2|, $$ where $r$ is the radius of the circumcircle. Substituting $PA_1$, $PB_1$, $PC_1$ from this formula into the equation given by the problem yields: $$ PA^2+PB^2+PC^2=\pm3(r^2-PO^2), $$ where the sign on the right hand side depends on point $P$ being inside ($+$) or outside ($-$) the circumcircle.
It is not then difficult to see that in the first case this is the equation of the circle you found, while in the second case we get the equation of a line.