1.3 Lines and linear equations 23and in other forms which look quite different. As we shall see in the next prob-
lem, the circumcircle of the mid-triangle of the given triangle P1P2P3 is the famous
nine-point circle of the given triangle. The coordinates in (1) and (2) are there-
fore the coordinates of the center of this nine-point circle. The answers to this
and the two preceding problems are written in such a way that it is very easy to
see that the center of the nine-point circle is the mid-point of the line segment
joining the orthocenter and the circumcenter of the given triangle.
24 To see that there are opportunities to use ideas of the preceding problems
and the rest of this chapter in geometry, we look briefly at a triangle and its
nine-point circle. Figure 1.391 shows a triangle P1P2P3, the points Q1, Q2, Q3 in
which the altitudinal lines intersect the lines containing the sides of the triangle,
and the orthocenter P0. The points M1, M2, M3 are the mid-points of the sides
of the triangle, and the perpendiculars to the sides at these points intersect at a
point C, the circumcenter of the given triangle. The points R1, R2, R3 are the
mid-points of the line segments P1Po, P2Po, P&P0. The famous nine-point-circle
theorem says that the nine points M1, M2, M3, Q1, Q2, Q3, R1, R2, R3 all lie on a
circle. This circle, the nine-point circle, has its center at the mid-point S of the
line segment joining the orthocenter Po and the circumcenter C. The radius
of the nine-point circle is half the radius of the circumcircle. When the triangle
is equilateral, the orthocenter, the circumcenter, the center of the nine-point
circle, and the centroid (intersection of the medians) all coincide. When the
triangle is not equilateral, the four points are distinct but are collinear, and the
line upon which they lie is called the Euler line of the triangle.
25 In this problem we use results of Problems 21 and 23 to obtain a new
formula and a proof of the nine-point-circle theorem. We know from Problem
23 that the mid-points M1, M2, M3 of the sides of the triangle are on the circle;
in fact, these three noncollinear points determine the nine-point circle. The
remaining points R1, R2, R3, Q1, Q2, Qs, which we must prove to be on the nine-
point circle, are not necessarily distinct from each other and from M1, M2, M3i
but our proof will not be a "partial proof" which covers only "general cases."
Our proof will be a proof. Use a result of Problem 21 to show that the x coordi-
nate of the point R1 midway between the vertex P, and the orthocenter Po is
Y1(x3 - x2)(xl + x1) + Y2(x1 - x3)(xl + x2) + y3(x2 - x1)(x1 + x3)X + (Y2 - yl)(Y3 - Y2)(YI - y3)
- 2[y1(xs - x2) + Y2(x1 - xs) + Y3(x2 - x1))
Use this result to show that the x coordinate of the point midway between R1
and the mid-point M1 of the line segment P2P3 is the x coordinate of the center
of the nine-point circle given in Problem 23. Remark: This fact and the associ-
ated fact involving y coordinates imply that the points R1 and M1 are at opposite
ends of a diameter (line segment, not number) of the nine-point circle. Similar
proofs (which are attained by cyclic advances of subscripts) show that R2 and M2
are at opposite ends of a diameter and that R3 and M3 are at opposite ends of a
diameter. This proves that R1, R2, R3, lie on the circle. We recall that Ql is
the point at which the altitudinal line through P1 intersects the line containing
the vertices P2 and P3, that R1 is on the altitudinal line, and that M1 is on the
line containing P2 and P3. In case Ql coincides with M1 or R1, we conclude that
Q1 is on the circle. In the contrary case, the angle R1Q1M1 is a right angle.