Optimal triangles with vertices on fixed circles

Let and suppose is a triangle such that there exists a point which satisfies , , . What is the relative position of with respect to the triangle such that

a) The area is maximized;

b) The perimeter is maximized.

This was inspired by this answer on Math Overflow.

Identifying edges and boundary points – 2D Mesh – Matlab

A triangulation algorithm often gives as output a list of points, and a list of triangle. Each triangle is represented by the indexes of the points which form it. Sometimes we need extra information on the structure of the triangulation, like the list of edges, or the list of boundary points. I will present below two fast algorithms for doing this.

Finding the list of edges is not hard. The idea is to go through each triangle, and extract all edges. The algorithm proposed below creates the adjacency matrix in the following way: for each triangle we set the elements (and their symmetric correspondents) to be equal to .

In order to find the points on the boundary (in two dimensions), it is enough to look for the edges which are sides to only one triangle. We can do this using the adjacency matrix. Note that if is the adjacency matrix, then stores the number of paths of length (two sides) between two points of the triangulation. Note that any edge which is not on the boundary will contain the starting and ending point of two paths of length . If is a triangle such that points are on the boundary, then (there is one path of length going through . We also have . Conversely, if and then are connected, and there is a unique path of length going from to . Thus, is an edge on the boundary. Therefore, we just need to identify the indexes such that there exists with .

Below are two short Matlab codes doing these two algorithms. I guess they are close to being optimal, since only sparse and vectorized operations are used.

%p is the list of points
%T is the list of triangles, ap is the number of points
%this computes the adjacency matrix
A = min(sparse(T(:,1),T(:,2),1,ap,ap)+sparse(T(:,2),T(:,3),1,ap,ap)+sparse(T(:,3),T(:,1),1,ap,ap),1);
A = min(A+A',1);
% this finds the boundary points, whose indexes are stored in Ibord
B = A^2.*A==1;
Ibord = find(sum(B,2)>0);

Nice characterization side-lengths of a triangle

Find the greatest such that and implies that are the side-lengths of a triangle.

Continuity in Geometry

Here are a few interesting geometry problems which use continuity problems in their solutions.

Pb 1. Consider three parallel lines in the plane . Prove that there exist points such that the triangle is equilateral.

Pb 2. Consider a triangle . Prove that is equilateral if and only if for every point in the plane we can construct a triangle with sides .

The Existence of a Triangle with Prescribed Angle Bisector Lengths

Prove that for any there exists a unique triangle (up to an isometry) such that are the lengths of bisectors of this triangle
Solved by L. Panaitopol & P. Mironescu in 1994, AMM 101, 58-60

Application of the isoperimetric inequality

Find the shortest curve which splits an equilateral triangle with edge of length into two regions having equal area.

Find the angle 2

In triangle we have and . We take such that and such that . Find .