#how many trianles can be formed by conceting the dots in the diagram?
96 messages · Page 1 of 1 (latest)
- Do not ping the Moderators, unless someone is breaking the rules.
- Do not ping the Helper Moderators, unless there is a conflict between helpers.
- Do not ping other members randomly for help.
- Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
- Wait patiently for a helper to come along.
- If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:
+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:
What is required to make a triangle?
You need to convey the dots but I’m not 100%sure
Which dots?
You're always wrong to assume.
Because we do not assume in math, we prove.
,rotate
I believe so. It's asking how many unique triangles exist that have the given points as vertices.
So then the question is, what are the necessary and sufficient conditions for a set of vertices to form a triangle?
no clue
Do you know what a vertex is?
yeah
What is a vertex?
the point where they conect?
Yes or the corners
Try looking at the diagram. How many vertices should there be such that a triangle can form? Think about how one vertex can account for more than one triangle
So then what is the requirement for a group of vertices to form a triangle?
Lets classify the dots as
1 Junction dot
4 Horizontal dots (H)
3 vertical dots (V)
Ypu need to pick 3 vertices for the triangle
Case 1: If you dont use junction dot, you need to pick 2 H dots and 1 V dot or 2V dots and 1 H dot.
Case 2: If you use Junction dot, you need to pick 1 H dot and 1 V dot.
But there is no way to make it with 2h and the junction dot right
yea which is why you need to pick one H and one V
do you know combinatorics?
stop trying to manually count and just use that
this is already a huge hint here
Why not?
Because it is now a straight line
Yeah I do
just use combi
I don’t know how to
Right, and that's exactly the condition on the vertices of a triangle I was trying to get you to realize; a set of three noncolinear points.
actually better idea let techie cook
Like what formula would I use
I understand how to do the formula but I’m not sure on how to apply it to this shape
That’s if I want a diagonal
@still valve
... what?
So if I want so let’s say make a straight line from a octogong
I would use this method
I have the wander by counting but I need to figure out the method
It’s 24
It's not 24.
Why?
How because I can’t make another different triangle
That’s what my teacher is saying and I’m not sure why
Are you interested in learning why?
Yeah I am
Okay, so, how many vertices does an octogon have?
8
And how many vertices does a line segment have?
2
Get it?
Yeah but how would I use that with this question
(Of course, not all of those pairs of vertices are diagonals, some of them are sides.)
What are the necessary conditions for a set of vertices to form a triangle?
3
...and?
But they can’t All be in the same line
Right, so how many lines do we have here, with how many points on them?
What do you mean
...I mean... the points... can't all... be in a line. So which points... are all in a line?
You have 1 junction point
3 vertical ones going down
And 4 horizontal going to the right
...can you... count?
Yeah we are excluding the junction point
Why? I didn't say to.
Ok then it’s a 5x4
And if we exclude the shared point from one of the lines, what do we get?
I’m not sure what you mean
We want. To count. Sets of three points. So we do not. Want to count. The same point twice.
Yes I got that
You are counting a point twice.
Am I counting the junction twice
What is "the junction"?
+close
Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.
Thank you for your feedback! !𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪! has been awarded 1
. They now have 865
. They have 3
daily left for today.