In hindsight, we can see how existence of the Euler line can be shown by purely geometric means. Consider ΔABC and its medial triangle MaMbMc. The two triangles are similar; ΔMaMbMc twice as small as ΔABC. Moreover, their corresponding sides are parallel, and, in addition, the two triangles have a common centroid.
ΔMaMbMc can be obtained by rotating ΔABC 180° around its centroid M and then shrinking it towards M to half its size. As a result of this transformation, the segment MH will also rotate 180° around M and, the orthocenter H will be made to coincide with the orthocenter of ΔMaMbMc, which, as can easily be seen, coincides, in turn, with the circumcenter (the common point of the three perpendicular bisectors) of ΔABC.
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 Δx1x2x3:
Hx1 = (H - x2x3/x1)/2
Hx2 = (H - x1x3/x2)/2
Hx3 = (H - x1x2/x3)/2,
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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:
H/2 - Hx1 = x2x3/2x1
H/2 - Hx2 = x1x3/2x2
H/2 - Hx3 = x1x2/2x3.
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Since all three points lie on the unit circle, we have
|H/2 - Hx1| = |H/2 - Hx2| = |H/2 - Hx3| = 1/2,
which says that the feet of the three altitudes in Δx1x2x3 are equidistant from the point H/2. In other words, the circumcenter of ΔHx1Hx2Hx3 is at H/2 - the midpoint of Euler's line. The circumradius of ΔHx1Hx2Hx3 is half that of Δx1x2x3.
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 ΔABC and the midpoints of segments that join the orthocenter with the vertices of the triangle.
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 (x1 + x2 + x3 + x4)/4 and, for this reason, is known as the anticenter of the cyclic quadrilateral x1x2x3x4.
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 1
The orthocenter x1 + x2 + x3 of triangle x1x2x3 and the vertex x4 are symmetric in the anticenter (x1 + x2 + x3 + x4)/2, and similarly for the other three triangles x1x3x4, x1x2x4, and x2x3x4. Therefore, the quadrilateral formed by the orthocenters of the four triangles is the reflection of x1x2x3x4 in the anticenter. (See A Remarkable Line in Cyclic Quadrilateral.)
Remark 2 (Hamilton's theorem)
Quite obviously, if ΔABC is inscribed in a circle of radius R, its 9-point circle has radius R/2. In 1861, W. R. Hamilton (1805-1865) noticed that four triangles ABC, ABH, BCH and CAH share the same 9-point circle. From here it follows that the circumcircles of the four triangles, although different, have the same radius.
References
- Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994
