73-X(4240) =12 EULER LINES POINT

X(4240) in Kimberling Center

Click to downloand detail defined and the figure

Barycentrics   f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b⁴ + c⁴ - 2a⁴ + a²b² + a²c² - 2b²c²)/[(b² - c²)(b² + c² - a²)]

X(4240) is the point of intersection of the Euler lines of nine triangles, constructed as in the next three paragraphs.

Let E be the Euler line of a triangle ABC. Let A₁ = E∩BC, and define B₁ and C₁ cyclically. Let A_B be the reflection of A in B₁, and define B_C and C_A cyclically. Let A_C be the reflection of C in B₁, and define B_A and C_B cyclically. The Euler lines of the four triangles ABC, AA_BA_C, BB_CB_A, CC_AC_B concur in X(4240). (Dao Thanh Oai, Problem 1 in attachment to ADGEOM #1709, September 15, 2014). See also Telv Cohl, 'Dao's Theorem on the Concurrency of Three Euler Lines,' International Journal of Geometry 3 (2014) 70-73.

Continuing, let A*B*C* be the paralogic triangle of ABC whose perspectrix is E. Then X(4240) lies on the Euler line of A*B*C*. (Dao Thanh Oai, noted just after Figure 1 in attachment to ADGEOM #1709, September 15, 2014).

Continuing, redefine A_B as the point on line AC and A_C as the point on line AB such that B₁, A₁, A_B, A_C line on a circle and A₁, A_B, A_C are collinear. Define B_C and B_A cyclically, and define C_A and C_B cyclically. Let A₂ = B_AB_C∩C_AC_B and define B₂ and C₂ cyclically. The Euler lines of the five triangles ABC, A₂B₂C₂, AA_BA_C, BB_CB_A, CC_AC_B concur in X(4240). (Dao Thanh Oai, Problem 2 in attachment to ADGEOM #1709, September 15, 2014).

X(4240) lies on these lines:
{2, 3}, {107, 110}, {112, 1302}, {476, 1304}, {925, 1301}, {2407, 3233