A surprising fact about triangles which wasn't known until the 19th century:

Take any triangle, and draw a line from each vertex which meets the opposite side at a right angle. (These lines are called "altitudes", and all pass through a single point in the triangle's interior which is called the "orthocentre".) Then the feet of these altitudes--i.e., the points where they meet the sides of the triangle--and the midpoints of the triangle's sides and the midpoints of the lines joining the vertices to the orthocentre (nine points in total) all lie on a single circle.

A historical research1 revealed that in the early 19th century, the nine-point circle theorem was independently discovered by English, French, German, and Swiss mathematicians! Sometimes the circle is called the "Euler circle," giving credit to Euler, or the "Feuerbach circle," giving credit to Karl Feuerbach, who have also noted that the circle is tangent to the incircle (at the Feuerbach point) and excircles.

Triangle ABC, orthocenter H, and the nine points.
```                                    C
/|\
/  | \
/    |  \
/      |   \
R/        |C"  \
/  \       |     \
/      \     |      \Q
B'  /          \   |    _, \
*              \ |_,-'    * A'
/               _,-H         \
/             _,-'   | \        \
/           _,-'       |   \       \
/       A"_,-'           |     \      \
/       _,-'               |       \B"   \
/     _,-'                   |         \    \
/   _,-'                       |           \   \
/ _,-'                           |             \  \
/,-'                               |               \ \
A-------------------------*---------+-----------------B
C'        S
```
The altitudes are AQ, BR, and CS. The sides BC, AC, AB have midpoints A', B', C'. Points A", B", C" are midpoints of AH, BH, CH. The centroid G and circumcenter D are not shown. The nine-point circle theorem claims that the nine points lie on a circle centered at N, the midpoint of D and H, on the Euler line.

Proof by the eight point circle theorem2

Note that quadrilaterals ABCH, ABHC, and AHBC all have perpendicular diagonals. Thus the eight point circle theorem applies. The sides of the quadrilaterals are perpendicular in a way that two pairs of points overlap, making them six point circles instead. The eight point circle theorem states that the circle center is located at the centroid of a quadrilateral. Since ABCH, ABHC, and AHBC all share the same centroid N, all six point circles share the same center. Since the circles share points on the circumference as well as the center, the circles are equivalent. One circle covers all nine points.

Furthermore, it can be shown that N lies on the Euler line. By the theorem of Snapper we know
-2(D - G) = H - G
-2D = H - 3G
D = -(1/2)H + (3/2)G
and Since N is the centroid of A, B, C, and H we know
N = (1/4)(A + B + C + H)
N = (3/4)(1/3)(A + B + C) + (1/4)H
N = (3/4)G + (1/4)H
N = (1/2)[(3/2)G - (1/2)H] + (1/2)H
N = (1/2)D + (1/2)H
So N is the midpoint of D and H.
QED

Proof by central dilatations3

This proof doesn't rely on the eight point circle theorem, which relies on the theorem of Thales. A central dilatation is a type of mapping that maps an images into a new image as if it was mapped by a lens: it can enlargen, shrink, and/or invert an image. The general equation of a central dilatation with center of dilatation C and a non-zero dilatation coefficient r is
δC,r(X) = r(X - C) + C
Let f = δG,-1/2. This function maps ABC to A'B'C', due to where the centroid G is located in a circle (as explained in the node centroid). It also maps the circumcircle O of ABC into the circumcircle O' of A'B'C'. If the nine-point circle exists, it must be O' because a triangle can only have one circumcircle (as explained in the node circumcenter), and the nine-point circle certainly circumscribes triangle A'B'C'. Hence N = f(D). Along with D = -(1/2)H + (3/2)G derived from the theorem of Snapper above, it can be shown that N is the midpoint of D and H:
N = f(D)
N = (-1/2)(D - G) + G
N = (-1/2)D + (1.5)G
N = (1/2)D - D + (1.5)G
N = (1/2)D - [-(1/2)H + (3/2)G] + (1.5)G
N = (1/2)D + (1/2)H - (3/2)G + (1.5)G
N = (1/2)D + (1/2)H
Since N is equidistant from D and H, N is also equidistant from lines lDC' and lHS by congruent triangles. Hence ΔC'NS is an isosceles triangle, and the distance to C' and S from N is equal. Similarly, Q, R, and S also lie on circle O'.

Consider dilitation g = δH,1/2. This maps points A, B, C to A", B", C". Hence A", B", C" lies on the circle centered at g(D). Then
g(D) = (1/2)(D - H) + H = (1/2)(D + H) = N
This circle has the same center as O' and the same radius (half the radius of the circumcircle of ΔABC). Hence they are the same circle. Points A", B", C" also lie on O'.
QED

Corollary: Hamlton's theorem4

In 1861, W. R. Hamilton (1805-1865) noticed that triangles ABC, ABH, AHC, HBC all have the same nine-point circle. Since the nine-point circle is half the width of the circumcircle, it follows that the four triangles have a circumcircle of same radius (but different centers).
Sources
1. Mackay, J. S. "History of the Nine-Point Circle." Proc. Edinburgh Math. Soc. 11, 19-61, 1892.
2. http://www.cut-the-knot.com/Curriculum/Geometry/SixPointCircle.html
3. "Vectors and Transformations in Plane Geometry" by Philippe Tondeur, Publish or Perish, Inc. 1993.
4. http://www.cut-the-knot.com/triangle/EulerLine.html

The theorem that led to the proof of the Nine Point Circle was first printed in Thomas Leybourn's Mathematical Repository, in 1804, on page 18. It is printed as follows:

VII. QUESTION 67, by Mr. BENJAMIN BEVAN.

In a plane triangle, let w be the centre of a circle passing through x, y, and z; then will Cw = Cc, and be in the same right line; and wx = wy = wz = 2R, or the diameter of the circumscribing circle; where x, y, z, &c. represent the same points and lines as they denote in the Synopsis of Data for the Construction of Triangles.

The proof of the theorem was given by John Butterworth, on page 143 of the same volume. The proof provided the basis of the proof of the Nine Point Circle. It is as follows:

VII. QUESTION 67, by Mr. BENJAMIN BEVAN.

In a plane triangle, let w be the centre of a circle passing through x, y, and z; then will Cw = Cc, and be in the same right line; and wx = wy = wz = 2R, or the diameter of the circumscribing circle; where x, y, z, &c. represent the same points and lines as they denote in the Synopsis of Data, for the Construction of Triangles?

SOLUTION, by Mr. JOHN BUTTERWORTH, Haggate.

Let S, H, and G (fig. 114. pl. 6.) be the points where the lines xc, yc, and zc meet the circumscribing circle, and draw the radii CS, CH, and CG; also draw xw parallel to CS meeting cC produced in w, and join yw, sw; then w is the centre of a circle passing through x, y, and z. For it is now well known that Sc = Sx, therefore Cc = Cw and CH = Hy, consequently yw is parallel to CH. But CH is = CS, therefore yw is = wx = 2CS. In like manner it is proved that zq = xw = 2CS; therefore w is the centre of a circle passing through the points x, y, and z, and consequently the points c, C, w, are in a straight line, and Cc = Cw; and yw = xw = 2CS = 2R. Q.E.D.

Thus nearly was the proposition demonstrated by Messrs. Boole, Dawes, and Johnson.

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