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The Euler Arch.

by Peter Y. Woo, 9/99

Biola Univ., La Mirada, Calif.

Introduction. Let ABC be any triangle. The Euler line goes through the circumcenter O, the centroid G, the center '9' of the nine-point circle, and the orthocenter H, with '9' being the mid point of OH and OG = OH/3. I have discovered a curve which I shall call the Euler Arch, which goes through the points O, H, the incenter I, the Fermat point F, the first isodynamic point D, as well as two points which I shall call W-point and Y-point, so that the seven points H,Y,F,I,D,W,O are somewhat equally spaced on the arch, in that order. The points Y and W are so named so that "HYFIDWO" is pronounceable. They are isogonal conjugates of each other, and so are H and O, as are F and D.
Every point P on the arch has symmetric trilinear coordinates, i.e., the ratio of the distances of P from the three sides of the triangle is f(A) : f(B) : f(C) where
f(A) = cos (k p/6 + (1-k)A/2) / cos (-k p/6 + (1+k)A/2), for some parameter k. For k = 3, 2, 1, 0, -1, -2, -3, they correspond to the seven points H,Y,F,I,D,W,O. It is true that for any point P inside the triangle we can pick some function g(A) so that P has trilinear coordinates g(A):g(B):g(C), but this function f(A) is significant.

In particular, if one of the angles of DABC, say B, is 60°, then the arch is simply the circumcircle K of DAFC, where Ð AFC = 120°, and then the lines BH, BY, BF, BI, BD, BW, BO will intersect the circle again at H',Y',F',I',D',W',O', which are equally spaced on the circle, with I'I as its diameter. For the general case, when none of the angles are 60°, the arch does not have a simple equation.

    Angular Coordinates. Given the triangle ABC, each point P on the plane can be uniquely specified by the angles a = ÐA'BC, b = ÐB'CA, g = ÐC'AB, where A', B', C' are the centers of the circumcircles of triangles PBC, PCA, PAB respectively.
Then ÐBPC = p/2 +a, etc., and a+b+g=p/2. For any positive or negative values of a,b, g adding up to p/2, there is therefore a unique point P such that
ÐBPC=a+p/2, ÐCPA=b+p/2, and ÐAPB=g+p/2.
    We shall call (a, b, g) the angular coordinates of P. They have nice values for the known points H, F, I, D, O. In fact, it is a simple high school exercise to show that for these points,
a = f(k,A), where f(k,A) = k p/6 + (1-k)A/2, for k = 3,1,0,-1,-3 respectively, i.e.,
p/2-A, p/6, A/2, A-p/6, 2A-p/2.
    By defining b = f(k,B) and g = f(k,C), we see that a+b+g=p/2 for all real k, so that the points P(k) with f(k,A), f(k,B), f(k,C) as angular coordinates will lie on a curve, which will be called the Euler Arch.
    In particular, for k = 2 and -2, let us call the points P(2) and P(-2) Y-point and W-point, so that we have 7 points, H,Y,F,I,D,W,O lying on seven circular arcs each going through B,C, and the angle between any two arcs is a multiple of A/2-p/6. These 7 arcs, together with other 7 arcs from C to A, and another 7 arcs from A to B form a mesh so that each of the points H,Y,F,I,D,W,O is a point of concurrency of 3 arcs out of the 21.
    One particular case is most beautiful: if one angle, say B, is 60°, then f(k,B) is constant, 30°, for all k, so that all the seven arcs through A,C coincide on the circumcircle K_B of DAIC. All the 7 points lie it and ÐAIC = ÐAFC = etc. = 120°. Under a circle inversion with B as center and Ö(a c) as radius, let X" denote the image of any point X on the plane. Then B becomes B" and C becomes C" so that DABC is congruent to DAB"C". Thus the circle K_B remains invariant under the inversion. Also the seven arcs through BC becomes 7 straight lines through C", at A/2-p/6 from one another, which must hit the circle at H", Y", F", I", D", W", O", which are equally spaced on the circle. [See Diagram 2.]
    Our task is to find the other ratios such as AP:BP:CP and the trilinear coordinates of P in terms of the angular coordinates.

Theorem 1.Let b' = b/ 2 cosb, c' = c/ 2 cosg, then AP = 2 b' c' sin (A +b +g)/ B'C', where
B'C'² = b'² + c'² -2 b' c' cos (A+b+g). Also the trilinear coordinates of P are
cos a/cos (A-a) : cos b/cos (B-b) :cos g/cos (C-g) .
Proof. B'C' is the perpendicular bisector of B'C', hence AP = 4 × area(DAB'C') / B'C'. To find the trilinear coordinates of P we need to find the ratio sinÐPAC : sinÐPAB. Let Q,R be projections of B',C' on AC.
Then sinÐPAC = sinÐQB'C' = QR/B'C' = (AQ-AR) / B'C' = [b' cosb - c' cos(A+g)]/B'C'
= [b - c cos(A+g)/cosg]/B'C' = [b cosg - c cos(A+g)]/B'C' cosg .
Now let R = circumradius of DABC, then the above
= 2 R [sin B cosg - sin C cos (A+g)]/B'C' cosg
= R [sin (B+g) + sin(B-g) - sin (C+A+g) - sin(C-A-g)]/B'C' cosg
= R [sin (B+g) - sin(C-A-g)]/B'C' cosg = 2 R cos ((B+C-A)/2) sin((B+A-C+2g)/2)/B'C' cosg
= 2 R sin A cos (C-g)/B'C' cosg
\ if the distances of P from BC, CA, AB are a₁, b₁, c₁,
then b₁:c₁ = cos(C-g)/cosg : cos(B-b)/cosb =cosb/cos(B-b) : cosg/cos(C-g)
Thus the trilinear coords of P are cosa/cos(A-a) : cosb/cos(B-b) : cosg/cos(C-g). QED