Tricky question!!!

I thought I'd create some images to go along with Carcass' solution.

Let's first consider 2 points (P and Q)


One obvious point that is equidistant from P and Q is this point...



Also recognize that, if we draw two circles of equal radii with their centers at PQ, the 2 points of intersection will also be equidistant from P and Q


In fact, all points on this red line are equidistant from P and Q. The red line (as Carcass mentioned above) is the perpendicular bisector of points P and Q.


At this point, let's add a 3rd point (R) and draw the perpendicular bisector of points R and P in blue. All points on the blue line will be equidistant from R and P.

So, the green point (where the blue line and red line intersect) must be equidistant from R and P, AND it must equidistant from P and Q.
So, that green point must be equidistant from P, Q and R

Since there is only 1 such green point, the correct answer is B.

Cheers,
Brent
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Another useful way to think about this problem is to realize that three points determine a circle. We can draw a circle such that points P, Q, and R all lie on the circumference, with the circle's center being the same distance (radius) from all three points. Since the points are not colinear only one such circle is possible.
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Solution

First consider just two of the three points, say P and Q, and the set of points in the plane that are the same distance from them. Clearly the midpoint of line segment PQ is such a point. Are there others? You may recall from geometry that the points on the line that bisects PQ and is perpendicular to PQ are all the points that are equidistant from P and Q. Similarly, the points in the plane that lie on the perpendicular bisector of line segment PR are all the points that are equidistant from points P and R.

Because R does not lie on line PQ, line segments PQ and PR do not lie on the same line, and so their respective perpendicular bisectors are not parallel.
Therefore, you can conclude that the two perpendicular bisectors intersect at a point. The point of intersection is on both perpendicular bisectors, so it is equidistant from P and Q as well as from P and R. Therefore, the point of intersection is equidistant from all three points. Are there any other points that are equidistant from P, Q, and R ? If there were, they would be on both perpendicular bisectors, but in fact only one point lies on both lines.

The answer is \(B\)

If R is not on line PQ, then the three points form the vertices of a triangle. Let’s call it triangle PQR. Since there is only one point that is equidistant (the same distance) from the three vertices of the triangle (and the point is actually called the circumcenter), choice B is the correct answer.

Answer: B
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Intricate! Can we expect to see such questions on GRE?

yes. For sure.

it is an official question from the Official Guide.

Regards

Why don't we consider opposite side of the line PQ. I mean we can put D Both side of PQ. In that case we have two points. Please explain.

My question about your explanation :
1- I didn't understand how the green point will be equal to the three sides ? which math rule you used to find out the math relation ?

this question testing which concept exactly ?

Sorry, that last diagram had a mistake.
I've now edited it.

Cheers,
Brent
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Timer only contains one option which is coincidentally the right one.

Thank you

Fixed

I understand the solutions given, but why can't I extrapolate a line in 3-D (imagine z-axis) from the equidistant point? Then the answer would become a line, wouldn't it?

Sir,

Read the stem carefully.
P, Q, and R are three points in a plane, and R does not lie on line PQ.

No 3-D


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