The nine-point circle of a triangle is a circle going through 9 key points:
- the three midpoints of the sides of the triangle (blue in the below picture),
- the three feet of the altitudes of the triangle (yellow in the below picture), and
- the three midpoints from the vertices to the orthocenter of the triangle (green in the below picture).
The nine-point circle satisfies several important and interesting properties, besides the surprising fact that it exists at all.
Let the midpoint of be , the midpoint of be , and the midpoint of be . Similarly, let the foot of the altitude from be , the foot of the altitude from be , and the foot of the altitude from be . Let the intersection of , , and be H. Finally, let the midpoint of be , the midpoint of be , and the midpoint of be . The goal is to show that and all lie on a circle.
Firstly, since and are midpoints of and respectively, the segment is parallel to the segment . In addition, and are midpoints of and respectively, so the segment is parallel to the segment as well. Therefore, and are parallel.
Similarly, since and are midpoints of and respectively, the segment is parallel to the segment . Analogously, is parallel to , so and are parallel as well.
Hence is a parallelogram. But is parallel to , which is the same line as . Furthermore, is perpendicular to , and since is parallel to , is perpendicular to as well. Therefore, is perpendicular to , meaning that is a rectangle.
By similar logic, and are rectangles as well, so , and all lie on the same circle.
Let and intersect at a point , which is the center of the circle going through the above six points since is a rectangle Therefore, is the midpoint of . But , so is the circumcenter of triangle . Thus lies on the same circle as the above six points do, and by identical logic so do and .
Therefore, the points and all lie on a circle, as desired.
The center of the nine-point circle, known as the nine-point center , lies on the Euler line of the triangle, and is the midpoint of the section of the Euler line from the orthocenter to the circumcenter .
Furthermore, due to symmetry, we have the following:
The nine-point circles of triangles and are all the same.
As a consequence, the circumradii of those four triangles are equal, and as a corollary, we have the following:
, where is the circumradius of the triangle.
Similarly, for any point on the nine-point circle,
Consequently, the radius of the nine-point circle is exactly half the circumradius of the triangle:
Remarkably, if is a point for which , and do not form an orthocentric system, then the nine-point circles of , and all concur at a single point.
Similarly, the nine-point circle is externally tangent to each of the three excircles, and internally tangent to the triangle's incircle. The point at which the nine-point circle is tangent to the incircle is called the Feuerbach point.
For a triangle with sides 13, 15 and altitude 12, find the radius of the circle that passes through the following points:
- the midpoint of each side,
- the foot of each altitude, and
- the midpoint of the line segment from each vertex to the orthocenter.
If the radius can be written as , where and are positive integers, and and are coprime, find .
Note: Assume the given altitude to be through the vertex common to both the given sides.