Can anyone help?
Consider a triangulated Triangle. Let the verticies of all the triangles be labled xy and z arbitarily. (Including the verticies on the outside perimiter). Show that the number of segments on the outside perimiter labled XY has the same parity as the triangulation labled XYZ
Sorry i made a mistake
Consider a triangulated Triangle. Let the verticies of all the triangles be labled xy OR z arbitarily. (Including the verticies on the outside perimiter). Show that the number of segments on the outside perimiter labled XY has the same parity as the NUMBER OF TRIANGLES in the triangulation labled XYZ.
That is exactly how the question is worded, no diagrams
Hello, Bruce!
We could use a lot of clarification.
. . But I'll take a wild guess . . .
Consider a triangulated Triangle.
Let the vertices of all the triangles be labled arbitarily.
Show that the number of segments on the outside perimeter labled
has the same parity as the triangulation labled
First of all, I'll guess that a "triangulated triangle" is something like this
. . (not necessarily equilateral):We can call this a triangulation of "order 4":Code:* / \ * * / \ / \ * - * - * / \ / \ / \ * - * - * - * / \ / \ / \ / \ * - * - * - * - *
Then each of the smallest triangles is labeled with
One arrangement looks like this:Code:x / \ y z / \ / \ z x y / \ / \ / \ x - y - z - x / \ / \ / \ / \ y - z - x - y - z
Reading around the outside perimeter, the sequence of vertices is:
. .
That is, occurs times.
Further sketching convinced me that this always happens.
For a triangulation of order , the occurs times.
. . Hence, the segment appears times.
If "the triangulation labeled " means the order of the triangulation,
then the number of xy-segment not only has the same parity as the order,
. . it is actually equal to the order.
Is this anything like what you meant?
Hi Soroban,
I corrected the question,
Consider a triangulated Triangle. Let the verticies of all the triangles be labled xy OR z arbitarily. (Including the verticies on the outside perimiter). Show that the number of segments on the outside perimiter labled XY has the same parity as the number of triangles in the triangulation labled XYZ.
I hope this makes more sense
thanks.