cp's OEIS Frontend

Acute integer triangles ABC with longest side BC of length n (A247588) are segregated from obtuse or right integer triangles with the same longest side BC (A236384) by the closed boundary formed by a semicircle and BC as its diameter. The right integer triangles will lie on this boundary and the obtuse integer triangles within this boundary. Define a closed boundary S(q) that is formed by BC and a locus of points whose real distance from B is x, from C is y and x^q + y^q = n^q for integer q > 0. Then S(2) is that closed boundary formed by a semicircle with BC as diameter. Euler proved that there are no integer triangles that lie on S(3) and Wiles for all S(q) where q > 2. This sequence identifies all integer triangles with longest side BC of length n that lie inside S(3).

The boundary formed by BC and a locus of points A' such that triangle A'BC has tan A'/2 + tan B/2 + tan C/2 = 2 approximates a semi-ellipse with major axis BC = n and minor axis 3n/4. See Soddyian triangle link below for exact curve formula.

Integer triangles ABC with semitangents tan A/2 + tan B/2 + tan C/2 = 2 that lie on the boundary have outer Soddy circles that have degenerated into straight lines. Such triangles are Heronian and their sides a <= b <= c are constrained by 1/sqrt(s-c) = 1/sqrt(s-b) + 1/sqrt(s-a) where s is the triangle's semiperimeter. This sequence is the number of integer triangles with base length BC = n such that their semitangents tan A/2 + tan B/2 + tan C/2 >= 2. Integer triangles within the boundary have outer Soddy circles with negative curvature. Those outside the boundary have outer Soddy circles with positive curvature. Those that lie on the boundary have outer Soddy circles with zero curvature, i.e., their outer Soddy circles have degenerated into straight lines.

Furthermore, "An integer triangle ABC has exactly one isoperimetric point J1 and exactly one point of equal detour J2 if and only if tan A/2 + tan B/2 + tan C/2 < 2; it has exactly two equal detour points D1 and D2 and no isoperimetric point if and only if tan A/2 + tan B/2 + tan C/2 > 2; it has exactly one equal detour point D and no isoperimetric point if and only if tan A/2 + tan B/2 + tan C/2 = 2. In the first case, J1 and J2 coincide with the centers of the outer and inner Soddy circles, respectively. In the second case, D1 and D2 are the centers of the two Soddy circles. In the third case, D is the center of the inner Soddy circle, while the outer Soddy circle is a straight line." (Quoted from Hajja, Yff paper - see link below.)