Given a triangle $ABC$, Find a point $P$ such that $PA:PB:PC=1:2:3.$

Question

"Given a triangle $ABC$, Find a point $P$ such that $PA:PB:PC=1:2:3.$

I found this on a Olympiad book, and I was unable to solve it.

Any help will be appreciated.

Answer 1

On the straight line $AB$, let $D$ and $E$ be two distinct points such that $\frac{AD}{DB}=\frac{1}{2}=\frac{AE}{EB}$. On the straight line $BC$, let $F$ and $G$ be two distinct points such that $\frac{BF}{FC}=\frac{2}{3}=\frac{BG}{GC}$. Consider the intersection points $Q$ and $R$ of the circle with diameter $DE$ with the circle with diameter $FG$. Then, a point $P$ satisfies $AP:BP:CP=1:2:3$ iff $P\in\{Q,R\}$.