1. ## Help with question about a Triangulated triangle

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

2. Originally Posted by BruceBronson
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
I beleive that the reason that this had no replies when you posted it on the
12th of September is that no one knows what you are talking about.

Either there is some more explanation that goes with this or even a diagram.

RonL

3. 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

4. Originally Posted by BruceBronson

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
Then tell us what course this is from and what topic you were covering at
the time it was set.

RonL

5. 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 $x,\,y,\,z$ arbitarily.
Show that the number of segments on the outside perimeter labled $xy$
has the same parity as the triangulation labled $xyz.$

First of all, I'll guess that a "triangulated triangle" is something like this
. . (not necessarily equilateral):
Code:
              *
/ \
*   *
/ \ / \
* - * - *
/ \ / \ / \
* - * - * - *
/ \ / \ / \ / \
* - * - * - * - *
We can call this a triangulation of "order 4": $n = 4.$

Then each of the smallest triangles is labeled with $x,\,y,\,z$
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:

. . $x-y-z-x-y-z-x-y-z-x-y-z$

That is, $x\!-\!y\!-\!z$ occurs $n=4$ times.

Further sketching convinced me that this always happens.
For a triangulation of order $n$, the $x\!-\!y\!-\!z$ occurs $n$ times.
. . Hence, the segment $xy$ appears $n$ times.

If "the triangulation labeled $xyz$" 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?

6. 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.

7. Originally Posted by BruceBronson
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.
It would make a lot more sense if you could tell us what a "triangulated triangle" is.

-Dan