Brian McCall

**Univ. of Wisconsin**

J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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The point of concurrency of the three perpendicular bisectors of a triangle is the **circumcenter**. It is the center of the circle circumscribed about the triangle, making the circumcenter equidistant from the three vertices of the triangle. The circumcenter is not always within the triangle. In a coordinate plane, to find the circumcenter we first find the equation of two perpendicular bisectors of the sides and solve the system of equations.

One of the four main types of points of concurrency that we find in triangles is the circumcenter. Point of concurrency means you have at least three lines intersecting in one spot. So the circumcenter is where the three perpendicular bisectors of each side intersect. Which also makes it the center of the circle that's circumscribed about the triangle.

So if we take a look at a sketch of what a circumcenter might look like, we notice that it could be a really big circle if you have an obtuse triangle. And that it passes through all three vertices. Which means that the center is equidistant from the three vertices. But when will ever apply this in real life?

Well a common problem that you might see on a test is if they give you three different points and they say, "where's the treasure?" The treasure's located at an equal distance from three random points. Maybe a tree stump, a grave stone and the beach.

So what you would do to find the treasure is you would have to find the circumcenter of the triangle by drawing those lines. So what you would do, if we erased this treasure, is you would draw in your three sides of your triangle and then using your compass you would construct the three perpendicular bisectors of each side. So there will be one perpendicular bisector, and then here would be another perpendicular bisector and again this is just an estimate to show you how you would solve this problem and then here you'd find your perpendicular bisector and ideally by definition these points are concurrent. As you can see I'm a little bit off but that's just because I was sketching.

So you would say that the treasure would have to be right here, which is the center of the circle that passes through the three vertices.