X₃₂₅₈

CROSSSUM OF X₇₄ AND X₁₁₀

X₃₂₅₈ = midpoint of X₄ and X₄₇₇

5, 6, 7 3, 4, 5 Kepler Kimberling

X₃₂₅₈ (CROSSSUM OF X₇₄ AND X₁₁₀) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {4, 477}

• Related Centers: X₄ (ORTHOCENTER), X₄₇₇ (TIXIER ANTIPODE)

Trilinears: \((\csc(A - B))/(\cos B - 2 \cos C \cos A) + (\csc(C - A))/(\cos C - 2 \cos A \cos B)\)

Search Value (6, 9, 13): \(3.957312674779\)

X₃₂₅₈ lies on the nine-point circle, the cevian circle of X₃₀.

▸ Incident Central Lines11 lines

{2, 476}, {4, 477}, {30, 113}, {114, 858}, {115, 647}, {125, 523}, {131, 2072}, {132, 468}, {133, 403}, {232, 1560}, {868, 1649}

▸ Functional Relations17 relations

• X₃₂₅₈ = midpoint of X₄ and X₄₇₇

• X₃₂₅₈ = reflection of X_i in X_j for these (i,j); (125,3154), (1553,113)

• X₃₂₅₈ = isogonal conjugate of X₁₅₃₉₅

• X₃₂₅₈ = complement of X₄₇₆

• X₃₂₅₈ = complementary conjugate of X₅₂₆

• X₃₂₅₈ = X_i-Ceva conjugate of X_j for these (i,j): (2, 1637), (4, 526), (1138, 523)

• X₃₂₅₈ = crosspoint of X₂ and X₃₂₆₈

• X₃₂₅₈ = crosssum of X₇₄ and X₁₁₀

• X₃₂₅₈ = crossdifference of every pair of points on the line X₂₄₂₀X₂₄₃₃

• X₃₂₅₈ = inverse-in-polar-circle of X₁₃₀₄

• X₃₂₅₈ = inverse-in-{circumcircle, nine-point circle}-inverter of X₈₄₂

• X₃₂₅₈ = inverse-in-Moses-radical-circle of X₁₁₅

• X₃₂₅₈ = X₄₇₇-of-Euler-triangle

• X₃₂₅₈ = reflection of X₁₂₅ in Euler line

• X₃₂₅₈ = perspector of circumconic centered at X₁₆₃₇

• X₃₂₅₈ = center of circumconic that is locus of trilinear poles of lines passing through X₁₆₃₇

• X₃₂₅₈ = intersection, other than X₁₂₆, of the nine-point circle of ABC and the Parry circle of the X₂-Brocard triangle

Notes

Let P = X₄₇₇ and H=X₄; let H_A be the orthocenter of the triangle BCP. Define H_B and H_C cyclically. Then H, H_A, H_B, H_C are the vertices of a quadrilateral that is homothetic to the cyclic quadrilateral having vertices A, B, C, P. The center of homothety is X₃₂₅₈. Moreover, X₃₂₅₈ is the anticenter of the quadrilateral ABCP. (Randy Hutson, 9/23/2011)

Let L_a, L_b, L_c be the lines through A, B, C, resp. parallel to the Euler line. Let M_a, M_b, M_c be the reflections of BC, CA, AB in L_a, L_b, L_c, resp. Let A' = M_b∩M_c, and define B' and C' cyclically. Triangle A'B'C' is inversely similar to, and 3 times the size of, ABC. Let A″B″C″ be the reflection of A'B'C' in the Euler line of A'B'C' (line X₃₀X₇₄). The triangle A″B″C″ is homothetic to ABC, with center of homothety X₃₂₅₈; see Hyacinthos #16741/16782, Sep 2008. (Randy Hutson, March 25, 2016)