# Cramer–Castillon problem like

Cramer–Castillon problem being very difficult problem. Related to the Cramer–Castillon problem configuration, I posed a problem as follows:

Let $$ABC$$ be arbitrary triangle and let three collinear points $$E'$$, $$F'$$, $$D'$$ such that $$\frac{E'F'}{E'D'}=k$$ where $$k$$ is positive real number. How can construct points $$D$$ in the circumcircle of $$\triangle ABC$$, $$E$$ in $$AB$$, $$F$$ in $$AC$$ such that $$D, E, F$$ are collinear and $$\frac{EF}{ED}=k$$ ?

This can be dealt with using the $$p,q$$ method, i.e., by assuming that the triangle has vertices $$(0,0)$$, $$(1,0)$$ and $$(p,q)$$. It is then easy to compute the equation of the circumcircle. The points $$E$$ and $$F$$ can be taken as $$(\lambda p,\lambda q)$$ and $$(1-\mu(1-p),\mu q)$$. Now if we substitute $$D=(1-t)E+tF$$ in the equation for the circumcircle, we get an easily computed quadratic equation for $$t$$ whose coefficients depend on $$\lambda$$, $$\mu$$, $$p$$ and $$q$$. (Our $$t$$ is related to you $$k$$ in a very simple manner). This suffices to answer your question. More precisely, given $$k$$ we get a quadric equation in $$\lambda$$ and $$\mu$$ with coefficients depending on $$k$$ and the shape of the triangle. This can easily be computed explicitly but the precise form is too unwieldy for me to type it here. Hence we can choose one of the points $$E$$ and $$F$$ arbitrarily and this will give a quadratic condition to determine the other one. There will be either two solutions (this is because we are allowing $$t$$ to take on negative values) which might coincide or none, depending on the shape of the triangle and the position of the given point. If there is a solution, it will be constructible (and the explicit formula will indicate a construction).