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X(1), X(2), X(10), X(81), X(191), X(366), X(1029), X(1654)

harmonic associates of X(366)

(contributed by Peter Moses)

K328 is the locus of P such that the circumcenters Oa, Ob, Oc of the similar triangles {I[bc],I[cb],P}, {P,I[ca],I[ac]}, {I[ba],P,I[ab]} are perspective to ABC, where I[ab] and I[ac] are the points on the sidelines of the A-isoscelizer triangle through P. See Peter's note below. The locus of the perspector Q is pK(X1<->X1224, X1224).

K328 has the same asymptotic directions as K317 = pK(X81, X86) and also K455 = pK(X2, X319). The tangents at A, B, C concur at X(81).

K328 is the isogonal transform of pK(X31, X37) and the isotomic transform of pK(X75, X321).

Isoscelizers :

A line perpendicular to the angle bisector of vertex A is the A-isoscelizer. When intersected by the sidelines containing A, it is the base of an isosceles triangle containing vertex A. Isoscelizer is a name coined by Peter Yff in the early 1960's.

Suppose the A-isoscelizer passes through P(p,q,r). This line, (c q + b r) x + (b r - c (p + r)) y +(c q - b (p + q)) z = 0, cuts the sidelines AB and AC in the points I[ab] and I[ac] with barycentrics {c (p + r) - b r, c q + b r, 0} and {b (p + q) - c q, 0, c q + b r} respectively. It also cuts the BC sideline at I[aa] = {0, b (p + q) - c q, b r - c (p + r)}, concurrent if P is on the Feuerbach hyperbola.

The length of the A-isoscelizer |I[ab] I[ac]|, passing through P, as the base of an isosceles triangle may be written as 2 (c q + b r) Sin(A / 2) / (p + q + r). If the 3 isoscelizers of a triangle have the same length and are concurrent, they meet at P = X(173), the congruent isoscelizer point.

The area of the A-isoscelizer triangle, A I[ab] I[ac], is (c q + b r)^2 / (b c (p + q + r)^2) that of the area of ABC. If area of the 3 isoscelizer triangles of a triangle are equal and the isoscelizers are concurrent, they meet at P = X(364).

The three triangles I[bc] I[cb] P, P I[ca] I[ac] and I[ba] P I[ab] are similar and become congruent for P = X(174). The area of I[bc] I[cb] P = (a - b - c) p^2 / (a (p + q + r)^2) that of ABC. Its circumcenter is {a p, b p + (a + b + c) q, c p + (a + b + c) r}.

The Euler lines of these 3 similar triangles are concurrent for P on the line X(2) X(7). The point of concurrence is on the line X(59) X(1319).

The Brocard axes of these 3 similar triangles are concurrent for P on the OI line, X(1) X(3). The point of concurrence is on the line X(36) X(59).

The OI lines of these 3 similar triangles are concurrent for P on the line X(173) X(174).

The circumcenters of the other 3 triangles passing through P, A I[ba] I[ca] lie on their appropriate angle bisectors, and are therefore perspective to ABC at X(1).

Paul Yiu has written notes http://www.math.fau.edu/yiu/Isoscelizers.ps

The figure below is drawn with P = X(366).