Let ABC be a triangle, P a point and A'B'C' the antipedal triangle of P.
Denote:
Ab, Ac = the orthogonal projections of A on A'C', A'B', resp.
A2, A3 = the reflections of B, C in Ab, Ac, resp.
Bc, Ba = the orthogonal projections of B on B'A', B'C', resp.
B3, B1 = the reflections of C, A in Bc, Ba, resp.
Ca, Cb = the orthogonal projections of C on C'B', C'A', resp.
C1, C2 = the reflections of A, B in Ca, Cb, resp.
Oa, Ob, Oc = the circumcenters of A'A2A3, B'B3B1, C'C1C2, resp.
For P = O, the NPC center of OaObOc lies on the Euler line of ABC.
[Angel Montesdeoca]:
**** If P=(u:v:w) the NPC center of OaObOc is:
(-c^4 u v (v+w) (-u^2+w (v+w))
+c^2 u v w (2 b^2 (v+w)^2 +a^2 (-u^2+3 u v+2 v^2+2 u w+5 v w+3 w^2))
-w (a^2 b^2 u v (u^2-2 u v-3 v^2-3 u w-5 v w-2 w^2)+a^4 v (v+w) (2 u^2+v w+2 u (v+w))+b^4 u (v+w) (-u^2+v (v+w)))
**** For P = O, the NPC center of OaObOc is
W = (2r^2+8rR+7R^2-2s^2) X(3) + (2r^2+8rR+9R^2-2s^2) X(4) =
(a^8 (b^2+c^2)
-2 a^6 (b^4-b^2 c^2+c^4)
-a^4 b^2 c^2 (b^2+c^2)
+a^2 (b^2-c^2)^2 (2 b^4-b^2 c^2+2 c^4)
-(b^2-c^2)^4 (b^2+c^2) : ... : ....),
W is the reflection of X(i) on X(j) for these {i,j}: {3,10125}, {10125,12010}, {10224,5}, {10226,140}, {11250,5498}.
W lies on lines: {2,3}, {113,5876}, {156,9927}, {265,1614}, {1154,5448}, {1568,6101}, {5449,5663}, {5476,11255}, {5944,10113}, {6241,10264}, {6564,11266}, {6565,11265}, {7728,11440}, {7951,8144}, {11459,11805}, {11804,12254}.
with (6,9,13)-search numbers (-1.67191299053365, -2.53702236474700, 6.16871673005552)
Angel Montesdeoca
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