This problem concerns three circles of equal radius $r$ that intersect in a single point $O$. Let $W_1,W_2,W_3$ denote the centers of the three circles and let $\vec w=\overrightarrow {OW_i}$ for $i=1,2,3$. Similarly, let $A,B$, and $C$ denote the remaining intersection points of the circles and set $\vec a= \overrightarrow {OA}$, $\vec b= \overrightarrow {OB}$, and $\vec c= \overrightarrow {OC}$. By numbering the centers of the circles appropriately, write $\vec a$, $\vec b$, $\vec c$ in terms of $\vec w_1$, $\vec w_2$ and $\vec w_3$.
Show that $A,B$ and $C$ lie on a circle of the same radius $r$ as the three given circles. Moreover show that $O$ is the orthocenter of triangle $ABC$.
I need a vector proof; any help would be greatly appreciated.
It might help to look at a GeoGebra picture:
Triangle $COW_1$ has two sides $CW_1$ and $OW_1$ of radius length $r$. The same is true for triangle $OCW_2$. The conclusion is that $\vec c = \vec {OC} = w_1 + w_2$.
Similarly:
$\vec b = \vec {OB} = w_2 + w_3$
$\vec a = \vec {OA} = w_1 + w_3$
Note that $w_1$, $w_2$ and $w_3$ are also located on a circle with radius $r$ as they all have distance $r$ to point $O$.
The center $M$ of the circumcircle of triangle $ABC$ is the intersection of the perpendicular bisectors of $\vec {AB}$ and $\vec {AC}$.
Orthocenter:
You find a triangle’s orthocenter at the intersection of its altitudes. Therefore, you have to show that $\vec a$ is orthogonal to $\vec {CB}$, $\vec b$ is orthongonal to $\vec {AC}$ and $\vec c$ is orthogonal to $\vec {AB}$.