# How many points are possible in the plane that are the same distance from all three points.

P, Q and R are the points on a plane

• How many points are possible in the plane that are equidistant from both P and Q?
• If R is equidistant from both P and Q, then how many points are possible in the plane that are the same distance from all three points.
• If R does lie on line PQ but not equidistant from P and Q, then how many points are possible in the plane that are the same distance from all three points.
• If R does not lie on line PQ, then how many points are possible in the plane that are the same distance from all three points.

For $1$, I find infinite points in up and down through the middle point of line PQ.

For $2$, I find one point

For $3$, I find no point

For $4$, I find only one point

Though this all is my guess after sketching lines on paper. Can anyone explain how to solve these?

I assume that the points $P$, $Q$, $R$ are all distinct. It's easy to enumerate the degenerate cases where two or more of them coincide.

Your answer to the first question is correct: the set of points equidistant from $P$ and $Q$ is the perpendicular bisector of the segment $\overline{PQ}$, namely the line perpendicular to $\overline{PQ}$ and crossing it at its midpoint.

Consequently, if a point is supposed to be equidistant from $P$, $Q$ and $R$, it must belong to three lines simultaneously: the perpendicular bisector of $\overline{PQ}$, that of $\overline{PR}$ and that of $\overline{QR}$.

Now, if $PQR$ is an actual triangle, then the perpendicular bisectors of the sides indeed meet at a single point: the circumcenter. And of course, the circumcenter is indeed the unique point equidistant from the three vertices of the triangle.

If $PQR$ is not a triangle, meaning if $P,Q,R$ lie on a single line, then the perpendicular bisectors are parallel and there's no equidistant point.

Therefore, the condition for there being an equidistant point from three distinct point $P,Q,R$ is that the points don't lie on a single line. If they don't, there's a unique such point. If they do, there's no such point.

This, then, has nothing to do with whether or not $R$ is itself equidistant from $P$ and $Q$. It does depend on whether $R$ lies on the line $PQ$ or not.

For the first problem, the set of points equidistant from $P$ and $Q$ is the perpendicular bisector of the line $PQ$. There are infinitely many points on this line, which thus satisfy the desired property.

For the last problem, $\triangle PQR$ is a non-degenerate triangle. Hence, there is exactly one point equidistant from the three given points, the circumcentre of $\triangle PQR$.

The second and third problems are exceptions to the answer for problem 4. If the three points lie on a line – and $R$ doesn't have to be in the middle of the line $PQ$ – the triangle degenerates into a line and no point on the plane will be equidistant from all three points. (Exception to the exception: if $R=P$ or $R=Q$ then the midpoint of $PQ$ is equidistant from all three points.)

• I don't understand 2nd and 3rd problem. Can you please explain a bit. In second problem I see this way- if R was midpoint on the line PQ, then I guess there is no point. But, if R is equidistant from both P and Q but not midpoint, then one point is possible in opposite side of R. – Mahmudul Hasan Sep 5 '16 at 5:21
• Suppose R is equidistant from P and Q, but not at the midpoint. Then the asked-for point exists. But if I move R closer and closer to the midpoint, the asked-for point goes farther and farther away from PQ, until the time when R is on PQ, at which point the circumcentre is infinitely far from PQ and doesn't exist as a real point. – Parcly Taxel Sep 5 '16 at 5:28