If you have a segment $[CA]$ and you want to plot the point $O$ such that $(\vec{OC},\vec{OA})=\frac{\pi}{2} (2\pi)$, and $OA=2OC$. How can you plot $O$?
From the first condition we know that $O$ lies on a semi circle of diameter $[CA]$, but how can we precise exactly where it is from the second condition?
There is a bit complicated way I figured out but I am searching for an easier way.
My way is to draw a triangle similar to $OAC$ that has $C$ and the midpoint $I$ of $[CA]$ as two of its vertices. ($DI=2IC$)
Here is the question by the way (part 2-b), I was told that it could be solved using isobarycenter or something like that which I do not know much about
I just thought of an answer, we can drop a perpendicular from $O$ (assuming we know where it is) to $CA$, let $H$ be the foot of this perpendicular.
The area of $CHO$ + the area of $AHO$ is equal to the area of $COA$, and supposing that $OA=2CO=2$
we get, $\frac{CH.OH}{2} +\frac{AH.OH}{2} =1$
And since $CA=\sqrt{5}$ we get that $OH=\frac{2}{\sqrt{5}}$
Now we are able to calculate $CH=\sqrt{1-(\frac{2}{\sqrt{5}}) ²}=\frac{1}{\sqrt5}=\frac{\sqrt{5}}{5}$
Now we can plot $H$ and draw a perpendicular to $(CA)$ at $H$ and precise the position of $O$ as the intersection between the semi-circle and this perpendicular.
I still believe that there is an easier way though, so if you can help, it will be great.