A question in plane geometry.

Question

$A,B$ are two fixed points in a plane. Given any third point $P$, can we always find a point $O$ such that either $\overline{OA}<\overline{OP}<\overline{OB}$ or $\overline{OB}<\overline{OP}<\overline{OA}$?

PS : I've asked many questions in this forum, but this is my first question in geometry, I'll be thrilled to learn an answer, as my plane geometry intuition is muddled and I am clueless. I hope the answer is surprisingly simple, elegant and enlightening.

Answer 1

If the points are not colinear, then it is possible: just draw the perpendicular bisectors of the segments connecting each pair of points. This will divide space into six wedges; each wedge corresponds to one of the six possible orderings of the distance relationships.

Note that the red dot at the intersection is the center of the circle passing through the three points.

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If the lines are colinear, however, this may not be possible as the perpendicular bisectors of the segments connecting each pair of points are parallel and divide space into only four regions. Thus, some of the six relations will not be possible.

Answer 2

No. If $P$ is at the $AB$ line segment's bisector, then $A$ and $B$ are equidistant from $P$, hence no $O$ will make an $OP$ distance to be strictly between distances $OA$ and $OB$.

In any other case just draw circles through $A$ and through $B$, both centered at $P$ – each $O$ between the circles satisfies the requested condition.

EDIT

I was wrong.

Answer 3

No. Consider the case in which $A$, $B$, and $P$ are all collinear so that $B$ bisects $\overline{AP}$. This is what it would look like:

As you can see, it is clear that no such $O$ can exist in this case.

If this is not the case, then $\triangle ABP$ exists. In this case, the strategy for finding $O$ would be to first find the circumcenter $C$ of the triangle so that $\overline{AC}=\overline{BC}=\overline{PC}$. Then rotate $C$ about $P$ some number of degrees such that the length $\overline{PC}$ stays the same, but $\overline{AC}$ increases and $\overline{BC}$ decreases, or vice versa. Then let $O$ be the image of $C$ and your conditions will be satisfied.