Assignment # 4
Constructing the Nine Point Circle
by Michael Ferra
Proposed Investigation
The Nine-Point circle for any triangle passes through the three mid-points of the sides, the three feet of the altitudes, and the three mid-points of the segments from the respective vertices to the orthocenter. Construct the nine points, locate the center (N) and construct the nine point circle for
i. an acute triangle.
ii. an obtuse triangle.
iii. a right triangle.
i. an acute triangle.
Let's first construct an acute triangle ABC.
Since we know the Nine-Point circle for any triangle passes through the three mid-points of the sides, we can construct these points next and call them D, E and F.
Continuing with our given description of the Nine-Point circle, we can then continue to find the three points defined as the three feet of the altitudes. These points can be found by constructing a line that passes through a vertex and is perpendicular to the opposite side. Let's call these points G, H and I. Also notice there is a point of concurrency among these three lines used to create the altitudes. This point is called the orthocenter which will be labeled J.
Now that we've located the orthocenter we can now construct the 3 midpoints of the segments from the respective vertices to the orthocenter. Let's call these points K, L, and M.
Next we need to find a way to locate the center of the Nine-Point circle in order to construct the circle that passes through all nine points.
One approach to this is to construct a segment between two points on the Nine-Point circle. For the purpose of this construction, we will create a segment FM. Now create another segment between two other points. For this construction we will create segment IL. Now that we've created our two segments, let's construct their midpoints and construct a line perpendicular to each respective segment through its midpoint, i.e. their perpendicular bisectors. In doing this we know that all points on the perpendicular bisector through FM are equidistant to points F and M. Likewise, all points on the perpendicular bisector through IL are equidistant to point I and L. Thus, if we construct the point of intersection of these two perpendicular bisectors, then this point of intersection will be equidistant to the four points F, M, I and L. This point is equidistant to four points on the circle, thus it is the center of our nine-point circle, N.
Now that the center has been located, the Nine-Point circle can be created where N to any of the nine points on the circle defines the radius.
ii. an obtuse triangle.
Let's first construct an obtuse triangle ABC.
Since we know the Nine-Point circle for any triangle passes through the three mid-points of the sides, we can construct these points next and call them D, E and F.
Continuing with our given description of the Nine-Point circle, we can then continue to find the three points defined as the three feet of the altitudes. These points can be found by constructing a line that passes through a vertex and is perpendicular to the opposite triangular side (extended if necessary). Let's call these points G, H and I. Also notice there is a point of concurrency among these three lines used to create the altitudes. This point is called the orthocenter which will be labeled J.
Now that we've located the orthocenter we can now construct the 3 midpoints of the segments from the respective vertices to the orthocenter. Let's call these points K, L, and M.
Next we need to find a way to locate the center of the Nine-Point circle in order to construct the circle that passes through all nine points.
One approach to this is to construct a segment between two points on the Nine-Point circle. For the purpose of this construction, we will create a segment FM. Now create another segment between two other points. For this construction we will create segment IL. Now that we've created our two segments, let's construct their midpoints and construct a line perpendicular to each respective segment through its midpoint, i.e. their perpendicular bisectors. In doing this we know that all points on the perpendicular bisector through FM are equidistant to points F and M. Likewise, all points on the perpendicular bisector through IL are equidistant to point I and L. Thus, if we construct the intersection of these two perpendicular bisectors, then this intersection will be equidistant to the four points F, M, I and L. This point is equidistant to four points on the circle, thus it is the center of our nine-point circle, N.
Now that the center has been located, the Nine-Point circle can be created where N to any of the nine points on the circle defines the radius.
iii. a right triangle.
Let's first construct a right triangle ABC.
We will follow the same construction techniques that we used for the previous two constructions. It may be a little repetitive in some instances, but let's include it because we can make some interesting observations. So following our same routine of steps, we know the Nine-Point circle for any triangle passes through the three mid-points of the sides, so we can construct these points and call them D, E and F.
We can now continue to find the three point defined as the three feet of the altitudes. These points can be found by constructing a line that passes through a vertex and is perpendicular to the opposite side. Notice when we construct a perpendicular line through vertex B to its opposite side AC, the lines intersect at vertex A, thus vertex A represents not only the vertex of △ABC but also represents the foot of an altitude, i.e. altitude BA. When we construct a line that passes through vertex C that is perpendicular to side AB, we notice once again that the lines intersect at A. Hopefully this comes as no surprise since ∠A is our 90° angle of △ABC, which is the resulting angle of two lines perpendicular to each other. Constructing the the perpendicular through vertex A to its opposite side BC does yield a unique intersection so we will call this point G. Let's make one more observation from constructing the feet of the altitudes. In the two previous sections we located the orthocenter to help us continue with our constructions. If we recall how to find the orthocenter, we need to find the point of concurrency of altitude lines, so where are these lines concurrent? Notice they are concurrent at our vertex A, thus A is the orthocenter of △ABC.
Now that we've located the orthocenter we can now construct the 3 midpoints of the segments from the respective vertices to the orthocenter. We've already found the midpoint from the vertex B to orthocenter A because this is midpoint of side AB from our first step. By this same idea, We already know the midpoint from vertex C to orthocenter A as well. Since there is no segment from vertex A to orthocenter A, the midpoint defined here remains as A. So notice that no unique points were created in this step.
Notice for a right triangle, the nine points we previously created to make a nine point circle has now turned into four points that imbed other points. Before we find the center of our circle, let's observe a cleaner image of where we are now.
Let's go about finding the center in the same manner as the first two sections. Let's create segment FG and segment DE, construct their midpoints and construct a line perpendicular to each respective segment through its midpoint, i.e. their perpendicular bisectors. The intersection of the perpendicular bisectors will give us the center of our circle. Let's label this center H.
Now that the center has been located, the Nine-Point circle of our right triangle can be created.
