Isoperimetric point



In geometry, the isoperimetric point is a triangle center &mdash; a special point associated with a plane triangle. The term was originally introduced by G.R. Veldkamp in a paper published in the American Mathematical Monthly in 1985 to denote a point $P$ in the plane of a triangle $△ABC$ having the property that the triangles $△PBC, △PCA, △PAB$ have isoperimeters, that is, having the property that

$$\begin{align} & \overline{PB} + \overline{BC} + \overline{CP}, \\ =\ & \overline{PC} + \overline{CA} + \overline{AP}, \\ =\ & \overline{PA} + \overline{AB} + \overline{BP}. \end{align}$$

Isoperimetric points in the sense of Veldkamp exist only for triangles satisfying certain conditions. The isoperimetric point of $△ABC$ in the sense of Veldkamp, if it exists, has the following trilinear coordinates.

$$ \sec\tfrac{A}{2} \cos\tfrac{B}{2} \cos\tfrac{C}{2} - 1 \ : \ \sec\tfrac{B}{2} \cos\tfrac{C}{2} \cos\tfrac{A}{2} - 1 \ : \ \sec\tfrac{C}{2} \cos\tfrac{A}{2} \cos\tfrac{B}{2} - 1$$

Given any triangle $△ABC$ one can associate with it a point $P$ having trilinear coordinates as given above. This point is a triangle center and in Clark Kimberling's Encyclopedia of Triangle Centers (ETC) it is called the isoperimetric point of the triangle $△ABC$. It is designated as the triangle center X(175). The point X(175) need not be an isoperimetric point of triangle $△ABC$ in the sense of Veldkamp. However, if isoperimetric point of triangle $△ABC$ in the sense of Veldkamp exists, then it would be identical to the point X(175).

The point $P$ with the property that the triangles $△PBC, △PCA, △PAB$ have equal perimeters has been studied as early as 1890 in an article by Emile Lemoine.

Existence of isoperimetric point in the sense of Veldkamp
Let $△ABC$ be any triangle. Let the sidelengths of this triangle be $a, b, c$. Let its circumradius be $R$ and inradius be $r$. The necessary and sufficient condition for the existence of an isoperimetric point in the sense of Veldkamp can be stated as follows.


 * The triangle $△ABC$ has an isoperimetric point in the sense of Veldkamp if and only if $$a + b + c > 4R + r.$$

For all acute angled triangles $△ABC$ we have $△ABC$, and so all acute angled triangles have isoperimetric points in the sense of Veldkamp.

Properties
Let $P$ denote the triangle center X(175) of triangle $a + b + c > 4R + r$. $$\frac{2 \triangle}{\bigl| 4R + r - (a+b+c) \bigr|}$$ where $△ABC$ is the area, $P$ is the circumradius, $P$ is the inradius, and $P$ are the sidelengths of $△ABC$.
 * $P$ lies on the line joining the incenter and the Gergonne point of $△ABC$.
 * If $R$ is an isoperimetric point of $△PBC, △PCA, △PAB$ in the sense of Veldkamp, then the excircles of triangles $△ABC$ are pairwise tangent to one another and $r$ is their radical center.
 * If $a, b, c$ is an isoperimetric point of $△PBC, △PCA, △PAB$ in the sense of Veldkamp, then the perimeters of $△$ are equal to

Soddy circles


Given a triangle $△ABC$ one can draw circles in the plane of $△ABC$ with centers at $A, B, C$ such that they are tangent to each other externally. In general, one can draw two new circles such that each of them is tangential to the three circles with $A, B, C$ as centers. (One of the circles may degenerate into a straight line.) These circles are the Soddy circles of $△ABC$. The circle with the smaller radius is the inner Soddy circle and its center is called the inner Soddy point or inner Soddy center of $△ABC$. The circle with the larger radius is the outer Soddy circle and its center is called the outer Soddy point or outer Soddy center of triangle $△ABC$.

The triangle center X(175), the isoperimetric point in the sense of Kimberling, is the outer Soddy point of $△ABC$.