X1144

EHRMANN CONGRUENT SQUARES POINT

Suppose that P is a point inside triangle ABC. Let \(S_{A}\) be the square inscribed in triangle PBC, having two vertices on segment BC, one on PB, and one on PC. Define \(S_{B}\) and \(S_{C}\) cyclically. Then X1144 is the unique choice of P for which the three squares are congruent. The function L(a,b,c) is symmetric, homogeneous of degree 1, and satisfies 0 < L(a,b,c) < min{a,b,c}. Also, X1144 lies on the hyperbola {A, B, C,X1,X6}; indeed, X1144 lies on the open arc from X1 to the vertex of ABC opposite the shortest side. L(a,b,c) is the common length of the sides of the three squares. (Jean-Pierre Ehrmann, 12/16/01)
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_1144 = EHRMANN CONGRUENT SQUARES POINT (0.5617, 0.2242) A B C X1144 (a, b, c) = (5, 6, 7) | AB = 1

X1144 (EHRMANN CONGRUENT SQUARES POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Trilinears: \(a/(a - L) : b/(b - L) : c/(c - L)\), where L = L(a,b,c) is the smallest root of \(a^{2}\)/(a - L) + \(b^{2}\)/(b - L) + \(c^{2}\)/(c - L) = 2D/L, and \(D = \operatorname{area}(ABC)\)
Barycentrics: \(a^{2}/(a - L) : b^{2}/(b - L) : c^{2}/(c - L)\)
Search Value (6, 9, 13): \(1.872006645063\)

Notes

See Jean-Pierre Ehrmann, Congruent Inscribed Rectangles.