2
$\begingroup$

Say we have $3 \times 3$ vertex grid like so:

*    *    *

*    *    *

*    *    *

How many triangles exist so that at least one of their sides passes through 3 vertices and the corners of the triangles must be on asterisks? I am not sure if this is even possible to find using combinatorics.

$\endgroup$
  • $\begingroup$ What does "at least one of their sides passes from 3 vertices" mean? Do all the corners have to come from the grid? What is the additional restriction? $\endgroup$ – Ross Millikan Feb 3 '14 at 15:09
  • $\begingroup$ Is this more clear? $\endgroup$ – Veritas Feb 3 '14 at 15:17
  • $\begingroup$ Is this a $3 \times 3$ grid or a $9 \times 9$ grid? You have 9 asterisks $\endgroup$ – qwr Feb 3 '14 at 16:55
  • $\begingroup$ Wow so many hours passed and I've just realized I wrote 9x9. $\endgroup$ – Veritas Feb 3 '14 at 16:58
1
$\begingroup$

Suppose that one line is like

*---*---* 

*   *   *  

*   *   *  

In such a case, we have 6 options for the 3rd vertex. And, there a total of 6 such lines(both horizontal and vertical). So we have $6 \times 6 = 36$ such triangles. However we over counted the cases like:

*---*---* 
|
*   *   *  
|
*   *   *  

There are $4$ extra cases, so we have $36 - 4 = 32$ such triangles. Now if the first line we draw is a diagonal, then we have $4$(not 6 because we dont want to choose the corner) choices for each diagonal i.e. $4\times 2 = 8$ possibilities.
$$32 + 8 = \boxed{40}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The problem is that they don't just have to triangles but one of their sides must also pass from 3 vertices. $\endgroup$ – Veritas Feb 3 '14 at 14:16
  • $\begingroup$ Well since we are choosing the triangles as 3 points from the 9 point set isn't that always the case? $\endgroup$ – Veritas Feb 3 '14 at 14:21
  • $\begingroup$ is the new answer correct? $\endgroup$ – Shaurya Gupta Feb 3 '14 at 14:30
  • $\begingroup$ I wouldn't ask if I knew :/ By the way could you explain the last observation in detail ? $\endgroup$ – Veritas Feb 3 '14 at 14:48
0
$\begingroup$

You can pick three asterisks in $9 \choose 3$ ways. They will form a triangle unless they are in a straight line. How many straight lines of three asterisks are there?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The problem is that every triangle must also have a side that passes from 3 asterisks. $\endgroup$ – Veritas Feb 3 '14 at 15:36
0
$\begingroup$

Starting with the edge passing through 3 vertices, we have a couple of different possibilities. Give each vertex/asterix numbers like:

1 2 3
4 5 6
7 8 9 

If we place the longest side of the triangle along the edge of the grid, e.g. passing through the vertices $1, 2$ and $3$, we have $4$ different possibilities when choosing the third corner (I'm ignoring vertices $7$ and $9$ here, since these triangles would be counted several times otherwise). There are $4$ such edges giving us $16$ possible triangles.

If we place the longest side along the horizontal axis, we can choose $6$ different vertices. The same applies for the vertical axis, which gives us an additional $12$ triangles.

Finally, the longest side, can be placed on a diagonal axis (e.g. through $1,5,9$) where we have $6$ possible choices (notice that we get the triangles we ignored in the first case), and there are $2$ such diagonal axes. We therefore have another $12$ possible triangles.

In total we have: $16+12+12=40$ possible configurations.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.