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$A,B$ 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.

Franklin Pezzuti Dyer
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Rajesh D
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3 Answers3

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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:

enter image description here

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.

enter image description here

Franklin Pezzuti Dyer
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  • any points not on image $O$ will not satisfy the condition? You mean, $O$ is the set of all possible solutions? – Rajesh D Jul 04 '17 at 14:15
  • @RajeshDachiraju Maybe this doesn't generate all possible points $O$, but it is a way to generate infinitely many such points. – Franklin Pezzuti Dyer Jul 04 '17 at 14:16
  • Firstly thanks for the valuable answer. I am interested in all possible solutions. Is it just a curve, or its an area? – Rajesh D Jul 04 '17 at 14:18
  • Well, it looks like you could turn it into an area as well... but the solution I gave just generates a curve of possible points. Glad to help! If I answered your question, don't forget to $\checkmark$! :D – Franklin Pezzuti Dyer Jul 04 '17 at 14:19
  • The complement of the right bisector $r_{A,B}$ of the segment $AB$ is two half-planes $ p_A , p_B$ with $A\in p_A$ and $B\in p_B.$ Any point in $ p_A$ is closer to $A$ than to $B.$..... The complement of the right bisector $r_{P,B}$ of the segment $PB$ is two half-planes $q_P, q_B$ with $P\in q_P$ and $b\in q_B$. Any point in $q_P$ is closer to $P$ than to $A.$ .... If $P$ is not on the line thru $AB$ then $r_{A,B}, r_{P,B}$ are not parallel so the complement of $r_{A,B}\cup r_{P,B}$ is 4 non-empty regions. One of them is $p_A\cap q_P,$ and any point $O$ in it will satisfy $AO<PO<BO.$ – DanielWainfleet Jul 04 '17 at 14:34
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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.

enter image description here

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

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.

enter image description here

robjohn
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  • What does $C$ correspond to? (in the question) – Rajesh D Jul 04 '17 at 14:34
  • @RajeshDachiraju: the question was essentially whether all order relations were possible. In that case, the letters used are irrelevant. However, to match the question, I have changed the letters in the diagrams. – robjohn Jul 04 '17 at 15:34
  • Thanks rob. I knew the theme of our answer, just wanted to clear my own confusion. – Rajesh D Jul 05 '17 at 02:37
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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.

CiaPan
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