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Spherical triangles of area Pi
Recently I stumbled upon this page and found out a very nice result:
If a spherical triangle has area
then four copies of it can tile the sphere.
Here we are talking about triangles on the unit sphere whose edges are geodesics. The above result is a simple consequence of the following facts:
- If a spherical triangle has angles
then its area is
. Therefore if a triangle has area
, then
.
- If
is a triangle of area
and
is obtained by symmetrizing
with respect to the midpoint of
on the sphere, then
and
are congruent triangles.
- Using angles and the fact that on the sphere similar triangles are congruent, we obtain that the triangles
are all congruent.
Here are a few examples of such partitions:
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
Read more…
Application of the isoperimetric inequality
Find the shortest curve which splits an equilateral triangle with edge of length into two regions having equal area.
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