|
|
|
|
|
|
|
|
|
The Euler Line and the 9-Point CircleThis is a continuation of The Altitudes and the Euler line page, towards the end of which we established existence of the Euler line. In any triangle, three remarkable points - circumcenter, centroid, and orthocenter - are collinear, that is, lie on the same line, Euler's line. Centroid is always located between the circumcenter and the orthocenter twice as close to the former as to the latter. In hindsight, we can see how existence of the Euler line can be shown by purely geometric means. Consider
 The geometric derivation is at least as easy as the application of complex numbers. However, complex numbers gave us a broader outlook on the geometric configuration. For example, recall the formulas for the feet of the altitudes in
where H = x1 + x2 + x3, the orthocenter of the triangle. We obtained those formulas under the assumption that all three vertices x1, x2, x3 lie on the unit circle. Let's rearrange the above identities:
Since all three points lie on the unit circle, we have which says that the feet of the three altitudes in The circle at center H/2 and radius 1/2 is known as the 9-point circle, its center is naturally the 9-point center. Why 9 points? Because, besides the feet of the three altitudes, it also passes through the midside points of The midpoint of the side x1x2 is given by Mx3 = (x1 + x2)/2. Therefore, |H/2 - Mx3| = |x3/2| = 1/2. Similarly, the midpoint of x1H is defined by (x1 + H)/2, so that the distance from that point to H/2 again equals |-x1/2| = 1/2. The same is, of course, true of other midpoints and vertices. Pretty simple. (A more fundamental 8 point circle is associated with quadrilaterals with orthogonal diagonals. In a strange twist, three such quadrilaterals that can be found in any triangle share the same 8 point circle, but the total count of points that lie on it only comes to 9.) But we are not done yet. Select 4 points x1, x2, x3, and x4 on the unit circle. These four points define four triangles: x1x2x3, x2x3x4, x1x2x4, x1x3x4. For each of the four triangles find the center of its 9-point circle. We get, respectively, points (x1 + x2 + x3)/2, (x2 + x3 + x4)/2, (x1 + x2 + x4)/2, and (x1 + x3 + x4)/2. Each of these points is at distance 1/2 from the point (x1 + x2 + x3 + x4)/2. The latter is the center of what is known as the 9-point circle of the four points. Point (x1 + x2 + x3 + x4)/2 is obviously symmetric to the circumcenter in the center of gravity For five points x1, x2, x3, x4, and x5 on the unit circle, there are 5 groups of 4 points. 5 centers of their 9-point circles lie on a circle of radius 1/2 with center at (x1 + x2 + x3 + x4 + x5)/2, and so on. We thus get and infinite sequence of circles and their centers that lie on subsequent circles. This sequence was discovered by J.L.Coolidge (1873-1954). Remark 1The orthocenter x1 + x2 + x3 of triangle x1x2x3 and the vertex x4 are symmetric in the anticenter Remark 2 (Hamilton's theorem)Quite obviously, if References
Copyright © 1996-2008 Alexander Bogomolny |
ABC and its medial triangle MaMbMc. The two triangles are similar; 
| 30747744 |  |
|