Mean Distance from Vertex to Interior of Plane Figures |

To determine the mean distance from a vertex to the interior points of a given triangle, we can begin by noting the expression for the position of the centroid of an arbitrary triangle relative to one vertex. Let A, B, and C denote vectors emanating from one vertex to the other two vertices and to the centroid of the triangle, respectively, as shown below. |

The triangle can be partitioned into four similar triangles, each with half the linear dimensions and one fourth the area, as shown below. |

By similarity, the positions of the centroids of these four triangles (relative to the original vertex) are given by the vectors C/2, A/2 + C/2, B/2 + C/2, and (A+B)/2 - C/2. The average of these is the vector position of the overall centroid, so we have |

Solving this for C gives the familiar result |

This shows, in particular, that the average height of the points of a triangle above any edge taken as the base is 1/3 the height of the apex, because A and B each contribute one whole vertical component of the height, yielding a centroid 2/3 of the way down from the apex. We can use this result to help determine the mean distance from one vertex of a triangle to the points of the triangle. |

The original triangle can be cut into arbitrarily thin slices, each consisting of a thin isosceles triangle and a small remainder region. A single slice is shown in the drawing below. |

Each slice is drawn so that its projected span on the opposite edge (the dashed line in the figure) has an equal incremental width. Since a triangle's area is a function only of the altitude and the base, we see that each slice has the same incremental area. Also, by the previous result, we know that the mean distance from the upper vertex of the points in each slice is 2/3 of the distance to the opposite edge. Consequently, we can replace all the points of the triangle with a single uniformly weighted line segment at a height of 1/3 the height of the apex, as indicated in the figure above. The mean distance from the apex to the points of this line segment is equal to the mean distance from the apex to all the points in the interior of the triangle. |

It's convenient to place the apex at the origin of rectangular coordinates, and let the coordinates of the other two vertices be (x |

Letting t be a uniform interpolating parameter ranging from 0 to 1 along the "2/3 segment" (the dashed line in this figure), the distance from the origin to the segment at the point corresponding to t is |

Expanding the expression under the radical, and making use of the identities |

we have |

Notice that if we replace t with 1-T (representing interpolation in the opposite direction), the edge lengths a and b are swapped, so this expression is correctly symmetrical in these two edges. The mean distance from the vertex to the points of the triangle is given by the average of D(t) for t ranging from 0 to 1. Carrying out the integration, we arrive at the result |

where u = (a+b)/c and v = (a-b)/c. For the "apex" of an isosceles triangle we have a = b and v = 0. In the case of an equilateral triangle with a = b = c = 1, we have u = 2 and v = 0, and this formula gives |

For any given triangle with edge lengths a,b,c we can compute the mean distances from the points of the triangle to each of the three vertices. The figure below shows a triangle with a circle centered on each vertex with radius equal to the mean distances from the vertex to the points of the triangle. |

Based on examination of a wide range of triangles, it appears that there is always some region covered by all three circles (the area shown in red in the figure above), i.e., there is always at least one point in the interior of a triangle that is less than the mean distance to each of the three vertices. |

Incidentally, the three mean distances can then be taken as the edge lengths of a new triangle, and the process of finding the mean distances of the new triangle can be repeated. Iterating this process, we see that at each stage the linear size of the triangle tends to decrease by a factor of D |

Naturally the mean distance (from the appropriate vertex) of the points of a right triangle is the same as the mean distance for the isosceles triangle constructed by placing two of the right triangles together, "back-to-back". In terms of the function this fact is expressed by the identity |

Equivalently, in terms of the parameterization with s = a/c, this identity is |

which can be verified by substituting into the expression for . |

The above formulae can also be used to find the mean distance from the vertices to the interior of |

We can also apply these results to distances between the points of a circular region. Clearly the mean distance from the center to the points of a circle of radius R is 2R/3, because we can slice up the circle into infintesimal triangular wedges, each of which has a mean distance of (2/3)R. On the other hand, to find the mean distance of the point of a unit circle from given point on the perimeter, we need to perform an integration. One approach would be to integrate doubly in a rectangular pattern as shown below. |

This leads to the formula for the mean distance from the point P on the perimeter |

Unfortunately this double integral is not very well conditioned for numerical evaluation, and not easy to evaluate in closed form. It does give the numerical value 1.131768..., but we can use a different approach to determine the exact value. |

The mean distance of the points in each slice is (2/3)r where r is the distance to the perimeter along the slice. We need to integrate these distances, weighted according to the areas of the respective slices. The incremental area of a slice is dA = (1/2)r(rdq), so we arrive at the integral for the mean distance |