## X(175) (ISOPERIMETRIC POINT)

 Interactive Applet

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 Information from Kimberling's Encyclopedia of Triangle Centers

Trilinears           -1 + sec A/2 cos B/2 cos C/2 : -1 + sec B/2 cos C/2 cos A/2 : -1 + sec C/2 cos A/2 cos B/2
Barycentrics    g(A,B,C) : g(B,C,A) : g(C,A,B), where g(A,B,C) = (sin A)(-1 + sec A/2 cos B/2 cos C/2)

If a + b + c > 4R + r, where R and r denote the circumradius and inradius, respectively, then there exists a point X for which the perimeters of triangles XBC, XCA, XAB are equal. Veldkamp proved that X = X(175), and Yff, in unpublished notes, proved that X(175) is the center of the outer Soddy circle. See also the 1st and 2nd Eppstein points, X(481), X(482).

Clark Kimberling and R. W. Wagner, Problem E 3020 and Solution, American Mathematical Monthly 93 (1986) 650-652 [proposed 1983].

G. R. Veldkamp, "The isoperimetric point and the point(s) of equal detour," American Mathematical Monthly 92 (1985) 546-558.

X(175) lies on these lines: 1,7    8,1270    174,483    226,1131    490,664    651,1335

X(175) = X(8)-Ceva conjugate of X(176)
X(175) = X(664)-beth conjugate of X(175)
X(175) = {X(1),X(7)}-harmonic conjugate of X(176)

This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias.