A place equally distant from three towns (geometry version)

A question was posed about the following problem (modified slightly from the original):

Edison Electric wants to build a wind farm to provide electricity for three towns. Residents are concerned about noise, so Edison Electric wants to build the wind farm equally distant from all three towns.  How can Edison Electric find such a place?

First, we must understand the problem.

What are we asked to find? (Alternately, what is the unknown?)

We are asked to find a place, with reference to three towns.

What are the data?

We are told only that there are three towns.  Notably, we know nothing of the towns’ relative locations.

What is the condition?

Our solution place must be equally distant from each of the three towns.

Is it possible to satisfy the condition?

We should note that the towns cannot lie on a straight line.  If they did, any point equally distant from the two farthest towns would be closer to the town between them.

Three collinear towns, \(A\), \(B\), and \(C\).  \(M\), the midpoint of \(\overline{A C}\), is visibly closer to \(B\) than to \(A\) or \(C\).

As a final step in understanding the problem, we should introduce suitable notation and draw a diagram.

We will call our towns \(A\), \(B\), and \(C\), and we will represent them as noncollinear points in a plane.

Three noncollinear towns, represented as points in a plane.

We will call the equidistant point that we are looking for \(P\).

Having understood the problem, we must next devise a strategy.

Do we know a related problem?

We may not know how to find a point equidistant from three other points, but we do know how to find a point equidistant from two other points: it is the midpoint of the line segment between them. Let's focus on towns \(A\) and \(B\). We will draw a line segment \(\overline{A B}\) and construct its midpoint, which we'll call \(M\).

Line segment \(\overline{A B}\) with midpoint \(M\).

Point \(M\) isn't the only point that's equidistant from \(A\) and \(B\), though.  Any point that lies on the perpendicular bisector of \(\overline{A B}\) (let's call such a point \(N\)) is also equidistant from \(A\) and \(B\).

Point \(N\) is equidistant from \(A\) and \(B\), wherever \(N\) lies on the perpendicular bisector of \(\overline{A B}\), because \(\triangle AMN\) is congruent to \(\triangle BMN\) by the side-angle-side postulate.  \(\overline{BM} = \overline{AM}\), \(\angle BMN \cong \angle AMN\), and \(\overline{MN} = \overline{MN}\).

Now we have an infinite number of points to work with, which are all equidistant from \(A\) and \(B\).  If one of these points is also equidistant from \(C\), then we will have found our solution.  Our strategy will be to use the same process to find such a point.

So let's execute our strategy.

First, we'll redraw our original diagram.

Our original diagram, showing all three towns and what we learned while devising a strategy.

As we did to find a point equidistant from A and B, we'll draw the line segment \(\overline{ B C }\) and construct its midpoint \(Q\).

Segment \(\overline{B C}\), with midpoint \(Q\).

And, again as before, we'll draw the perpendicular bisector \(b\) of \(\overline{B C}\).

Line b, the perpendicular bisector of \(\overline{BC}\).

The point where \(a\) and \(b\) intersect lies on both lines.  Since every point on line \(a\) is equidistant from \(A\) and \(B\), and every point on line \(b\) is equidistant from \(B\) and \(C\), that intersection must be point \(P\), our solution.

Point \(P\), our solution.

But wait, do we need to consider \(\overline{CA}\)?  Nope.  We know \(\overline{PA} = \overline{PB}\), and \(\overline{PB} = \overline{PC}\), so, by the transitive property, \(\overline{PA} = \overline{PC}\).

Circle \(\bigodot{P}\), \(r=\overline{PA}\), circumscribing \(\triangle ABC\).

Can we check our solution?

As always, the check is left for the reader.

Bonus chatter:  I showed a scalene acute triangle to prove this.  Does it matter?

It is left as an exercise why the same proof applies to a right triangle.

Circumcenter of a right triangle.

Or to an obtuse triangle.

Circumcenter of an obtuse triangle.