|
|
|
|
|
|
|
|
|
Remarkable Line in Cyclic Quadrilateral: What is it about?
|
| Buy this applet |
Copyright © 1996-2008 Alexander Bogomolny
Explanation
In a triangle, the orthocenter H, the 9-point center N, the centroid G, and the circumcenter O are collinear. The line that houses these four and other remarkable points is known as Euler's line of the triangle. The four points always occur on the Euler line in that order: H, N, G, O. Furthermore, N is always the midway between H and O
| (1) | HN:NO = 1:1 |
| HG:GO = 2:1 |
| Buy this applet |
Now, let there be a quadilateral ABCD. The four points A, B, C, and D form four triangles: ABC, BCD, CDA, and DAB, each with its own orthocenter, 9-point center, centroid, circumcenter, and Euler line. If the quadrilateral is cyclic, all four triangles share the circumcircle and the circumcenter O. In particular, in that case, the four Euler lines concur at the common circumcenter of the four triangles.
The applet illustrates the latter case of a cyclic quadrlateral. Let HA, NA, GA denote, respectively, the orthocenter, the 9-point center, and the centroid of 
BCD (A omitted!) Similarly introduce HB, NB, ..., and GD. In addition to ABCD, we shall consider three more quadrilaterals: HAHBHCHD, NANBNCND, and GAGBGCGD.
Given the standard relative positions of the four points H, N, G, and O on the Euler line, the three quadrilaterals are not only similar, they are homothetic with the center O. HAHBHCHD is obtained from GAGBGCGD with a factor of 3, while NANBNCND is obtained from GAGBGCGD with a factor of 3/2.
In any quadrilateral ABCD, the quadrilateral of centroids GAGBGCGD is homothetic to ABCD with a factor of -1/3. This is an interesting exercise to show that the product (successive application) of two homotheties is either a homothety or a translation. In the former case, the factor of the product is the product of the factors of the constituent homotheties. Furthermore, the three centers of homothety are collinear.
As an application of this result, we find that HAHBHCHD is the homothetic image of ABCD with the factor of
Let H, N, and G be the centers of homothety of ABCD and HAHBHCHD, NANBNCND, and GAGBGCGD, respectively. From the above remark, the four points H, N, G, and O are collinear. In itself, this is a curious property of a cyclic quadrilateral. We may go a little further and determine the relative locations of the points H, N, G, and O on that line.
Perhaps surprisingly the ratios are different from (1):
| (2) | HN:NO = 1:2 |
| HG:GO = 1:1 |
I'll give two proofs of that fact. One is based on a theory of homothetic transformations, the other that was communicated by Paul Yiu makes use of complex numbers.
Let two homotheties -- the first with center S1 and factor k1 and the second with center S2 and factor k2 -- are carried out successively. Barring exceptional cases, the result is a homothety with factor k1k2. The question is to determine the center S of that homothety. As we know, the three points S1, S2, and S are collinear.
Let S' be the image of S1 under the composite homothety. This means that S'S = k1k2·S1S. On the other hand, the first homothety leaves S1 fixed, while the second expands S1S2 by the factor of k2, such that S'S2 = k2S1S2. Subtraction gives
| S'S - S'S2 = k1k2·S1S - k2S1S2, |
or
| S2S = k1k2·(S1S2 + S2S) - k2S1S2, |
and, finally,
| S2S = (k1 - 1)k2/(1 - k1k2)·S1S2. |
In a somewhat simpler form we have
| (3) | S1S = (1 - k2)/(1 - k1k2)·S1S2. |
Let's apply (3) in our case:
|
which is equivalent to (2).
References
- I. M. Yaglom, Geometric Transformations II, MAA, 1968
For the second approach, let's put the circumcenter O at the origin, and represent the vertices A, B, C, D of the quadrilateral by unit complex numbers a, b, c, d.
It is well known that the orthocenter of triangle xyz is x+y+z etc.
We consistently denote by P(t) the point P on the Euler line of a triangle with
| P(t) of triangle bcd is t(b+c+d), | |
| P(t) of triangle cda is t(c+d+a), | |
| P(t) of triangle dab is t(d+a+b), and | |
| P(t) of triangle abc is t(a+b+c). |
The quadrilateral of P(t)'s of the four triangles bcd, cda, dab, abc is homothetic to the quadrilateral abcd with factor -t. The center of homothety is the point Q(t) such that
| Q(t) = t(a+b+c+d)/(1+t). |
Note that (a+b+c+d)/4 is the centroid G of the quadrilateral, so that this center of homothety traverses the line OG as claimed. Furthermore,
| (4) | H = Q(1) = (a+b+c+d)/2, |
| N = Q(1/2) = (a+b+c+d)/3, and, finally, | |
| G = Q(1/3) = (a+b+c+d)/4, |
which implies (2).
Remark
From (4) it follows that Q(1) -- the center of homethety of ABCD and HAHBHCHD -- serves as the circumcenter of NANBNCND. Indeed, NA is represented by the complex number (b+c+d)/2, so that
We see that the 4 nine point circles meet in H. We also know that the four simsons constructed for the four triangles ABC, BCD, CDA, and DAB with the points D, A, B, and C, respectively, concur with the nine point circles, i.e., at H.
Darij Grinberg [see, The Euler point of a cyclic quadrilateral] has also observed that the maltitudes of ABCD concur in the same point H, which makes it the anticenter of the quadrilateral. To see why this is so, let MXY denote generically the midpoint of segment XY. Then
BHA  CD, as an altitude in ΔBCD,H = MAHA, for H is the center of homothety with k = 1, MABH || BHA, as a midline in ΔABHA. |
Thus MABH CD, and the same holds for other maltitudes.
Now observe that the same four points A, B, C, D define quadrilaterals ABCD, ACBD, and ACDB. The foregoing discussion concerning ABCD applies as well to the latter two. In particular, the maltitudes from AC and BD concur with the maltitudes from AD and BC. Turning our attention back to ABCD, this means that the perpendicular from the midpoint of a diagonal to the other diagonal is also incident to H.
A related result has been established for more general quadrilaterals. Some other four points H', N', G', and O' happen to be collinear, lie in the same order, and obey the same proportions.
Copyright © 1996-2008 Alexander Bogomolny
| 28772678 |  |
|