How many triangles are there within a nonagon?
In: Geometry [Edit categories] [Edit]
[Edit]
to calculate the number of possible triangles within a nonagon, one must use combinatorics.
Understand that at first you have to pick the first vertex from 9 points, then the second from eight other possible points, and the third from 7 possible points. And finally that by this process any three vertices could be picked in 6 different orders.
triangles possible are: 9*8*7/6 = 84 possible triangles in a nonagon
Wow u guys are so retarded. No wonder there is such a high percentage of wrong ppl. The answer is 10. There is no formula. The little ones are diamonds. There are 10 triangles. There is no other correct answer so stop wasting ur time trying to seem smart when u really have absolutely no idea.
Its 50. No other answer. Ur formulas are stupid. Just count the freakin triangles and ull find that there is 50 triangles. No more, no less. Its 50 and only 50.
I get 79. Here is how I got it. Every triangle uses one of the bottom vertices of the large triangle. Let’s start by considering all triangles that use the bottom right vertex. Look at the triangles that form using a single line emanating from the left bottom vertex of the large triangle.
There are 5 points of intersection, forming 4 segments. You can form a triangle with just the top segment, or the top two segments, the top 3 segments or all 4 segments. There are 3 triangles that can be formed containing the second top most segment but not the top segment.
So the total number of triangles along a segment emanating from one of the segments from the bottom left vertex is 4+3+2+1. Doing this for all four segments gives 10 + 10 +10 + 10 = 40 triangles.
Now reverse roles of the two bottom vertices, counting all triangles that include the left bottom vertex. This gives another 40 triangles. But the whole triangle has been counted twice since it includes both of the bottome vertices of the large triangle, giving 40 + 40 -1 = 79 triangles.
I have to correct my previous answer. There are four triangles along each segment from the other vertex that is double counted. So that would give 40 + 40 – 16 = 64 triangles. I don’t see any additional triangles. What am I missing?
Good Job guys. Really, obviously, it’s 64.
But for those who do not understand WHY it is because of the formula, i’m gonna explain it to you.
First, let’s give names to summit of the triangle:
A for the upper one;
B for the lower left one;
C last but not least upper right.
Second, you oughta admit that in order to have a triangle, there are two possibilities:
- B and C are in this triangle
- B or C is in this triangle
(A is not necessary in this demonstration ^^)
So, let’s count triangles having B and C in their summits.
In fact, it’s the easiest part !
You just have to cout all the dots created by 2 lines crossed (execpt B and C obviously)
And you can notice that you have 4*4 dots
That’s to say (3+1)*(3+1) … that’s (n+1)^2 dots for a first time
Next let’s count triangles with only B (those with only C are as numerous as those one for a question of symetry)
i.e. 2*Triangle(having B) = Triangle(having B or C)
So here, it’s more complicated, but here we go.
Let’s take a line incoming from C and cutting the lines from B …
We can have 3+2+1 = 6 triangles on this line
And … we have (3+1) lines like this
So … (3+1)*(3+2+1) triangles with only B
With n lines, we have so (n+1)*(n+ … + 2 +1) triangles with B
Or (n + … + 2 + 1) = n * (n + 1) / 2
(google that if you do not believe ^^)
So we have (n+1)*(n+1)*n/2 triangles with B
That’s to say we have n*((n+1)^2) triangles with B or C
Counting all the triangles, we have:
(n+1)^2 + n((n+1)^2)
= ((n+1)^2) * (1 + n)
That’s to say (n+1)^3 … (hope you will understand that 64 is the answer :p)
If you’re still sceptic, try with n = 0,1,2 … (you’ll find 1, 8, 27 if you’re smart enough)
EVERYONE IS WRONG THE ANSWER IS 1 DUHH ITS VERY SIMPLE…… WHENI LOOK AT IT I SEE ONE TRIANGLE…RETARDSS IM 12 ND I GOT IT RIGHT (I DONT FEEL LKIKE COUNTING HEHE IM LAZY)
Yea… It’s definately 64… I seriously cannot believe the stubbornness of what I can only hope are a bunch of 7 year olds posting here… It takes no longer than a minute to count at LEAST half of the 64… If you want more proof…go to the site posted above… I’ll post it for anyone who can’t find it…. It literally shows you each triangle one by one… You don’t even have to worry about doing math…
I’m not sure if anyone still cares, but the answer is 22 three-sided polygons….ahem, “triangles”. This is assuming that you are only looking at the big triangle and are not using anything else but your eyes!
I take the question to mean how many *types* of triangles are present in the diagram. In that case, there are three: scalene, isosceles, and equilateral.
Ok, I don’t think that was the real question, but really folks. Trust the formula. The answer is 64.
Hey jake there’s ten smart one. count the triangle you divided into three and count it as one big one. therefore it is ten
I counted 50 since Every 3 sided object that can be created in this design is a triangle, there may even be more, LOL.
30
it is clearly 74, im in a math academy, i can count
what the hell triangles only have 3 sides
its 8
79, so whats the point of not supplying the answer?
How many triangles are there within a nonagon?
In: Geometry [Edit categories] [Edit]
[Edit]
to calculate the number of possible triangles within a nonagon, one must use combinatorics.
Understand that at first you have to pick the first vertex from 9 points, then the second from eight other possible points, and the third from 7 possible points. And finally that by this process any three vertices could be picked in 6 different orders.
triangles possible are: 9*8*7/6 = 84 possible triangles in a nonagon
74
125
there’s def. 16
srry theres 54
now there’s 56
final answer let me know if anybody else gets 56
there are 77 srry 4 changing my answer so much this time i double scratch that triple checked
64. final answer.
C:1+1+1+1=4
L: 6+6+6+6=24
R: 6+6+6+6=24
CL:3+2+1=6
CR:3+2+1=6
Total:4+24+24+6+6=64
C=centre; L=left; R=right
23
theres ONE. always making things harder than they should be.
57
First count I get 70
44
They’d be (10-4)x4x2+(4)x4x1=64 different triangles. Greeting from Sapienza:Università di Roma.
It’s 8 the others are a dimond of some sort
so quit counting them with the triangles
Wow u guys are so retarded. No wonder there is such a high percentage of wrong ppl. The answer is 10. There is no formula. The little ones are diamonds. There are 10 triangles. There is no other correct answer so stop wasting ur time trying to seem smart when u really have absolutely no idea.
Its 50. No other answer. Ur formulas are stupid. Just count the freakin triangles and ull find that there is 50 triangles. No more, no less. Its 50 and only 50.
16, defenitely.. Who knows the correct answer??
Such confidence in assuming that your answer is correct.
You call people names and you still get the answer wrong.
you have to remove the lines and then count how many new ones you got.
So those so called diamonds are now a triangle.
it is more than 64 and less than 100!
I get 79. Here is how I got it. Every triangle uses one of the bottom vertices of the large triangle. Let’s start by considering all triangles that use the bottom right vertex. Look at the triangles that form using a single line emanating from the left bottom vertex of the large triangle.
There are 5 points of intersection, forming 4 segments. You can form a triangle with just the top segment, or the top two segments, the top 3 segments or all 4 segments. There are 3 triangles that can be formed containing the second top most segment but not the top segment.
So the total number of triangles along a segment emanating from one of the segments from the bottom left vertex is 4+3+2+1. Doing this for all four segments gives 10 + 10 +10 + 10 = 40 triangles.
Now reverse roles of the two bottom vertices, counting all triangles that include the left bottom vertex. This gives another 40 triangles. But the whole triangle has been counted twice since it includes both of the bottome vertices of the large triangle, giving 40 + 40 -1 = 79 triangles.
I have to correct my previous answer. There are four triangles along each segment from the other vertex that is double counted. So that would give 40 + 40 – 16 = 64 triangles. I don’t see any additional triangles. What am I missing?
Solution: 64
http://www.iwpcug.org/davidbro/puz0603.htm
I counted 27 triangles
oops counted again and there are definitely 32 triangles
ahh okay i found 2 more my FINAL answer is 34 triangles that is THE answer
no way i found 4 more 38 HAHAHAHHA OKAY NO MORE LOOKING THAT IS MY FINAL FINAL END OF THE WORLD ANSWER AND IT IS RIGHT. 38 I AM SO SURE.
31
64
76
oops! i counted the 9 small quads. sorry, it’s 64
I counted 12
68 triangles.
i counted way off there are 64
Good Job guys. Really, obviously, it’s 64.
But for those who do not understand WHY it is because of the formula, i’m gonna explain it to you.
First, let’s give names to summit of the triangle:
A for the upper one;
B for the lower left one;
C last but not least upper right.
Second, you oughta admit that in order to have a triangle, there are two possibilities:
- B and C are in this triangle
- B or C is in this triangle
(A is not necessary in this demonstration ^^)
So, let’s count triangles having B and C in their summits.
In fact, it’s the easiest part !
You just have to cout all the dots created by 2 lines crossed (execpt B and C obviously)
And you can notice that you have 4*4 dots
That’s to say (3+1)*(3+1) … that’s (n+1)^2 dots for a first time
Next let’s count triangles with only B (those with only C are as numerous as those one for a question of symetry)
i.e. 2*Triangle(having B) = Triangle(having B or C)
So here, it’s more complicated, but here we go.
Let’s take a line incoming from C and cutting the lines from B …
We can have 3+2+1 = 6 triangles on this line
And … we have (3+1) lines like this
So … (3+1)*(3+2+1) triangles with only B
With n lines, we have so (n+1)*(n+ … + 2 +1) triangles with B
Or (n + … + 2 + 1) = n * (n + 1) / 2
(google that if you do not believe ^^)
So we have (n+1)*(n+1)*n/2 triangles with B
That’s to say we have n*((n+1)^2) triangles with B or C
Counting all the triangles, we have:
(n+1)^2 + n((n+1)^2)
= ((n+1)^2) * (1 + n)
That’s to say (n+1)^3 … (hope you will understand that 64 is the answer :p)
If you’re still sceptic, try with n = 0,1,2 … (you’ll find 1, 8, 27 if you’re smart enough)
EVERYONE IS WRONG THE ANSWER IS 1 DUHH ITS VERY SIMPLE…… WHENI LOOK AT IT I SEE ONE TRIANGLE…RETARDSS IM 12 ND I GOT IT RIGHT (I DONT FEEL LKIKE COUNTING HEHE IM LAZY)
Yea… It’s definately 64… I seriously cannot believe the stubbornness of what I can only hope are a bunch of 7 year olds posting here… It takes no longer than a minute to count at LEAST half of the 64… If you want more proof…go to the site posted above… I’ll post it for anyone who can’t find it…. It literally shows you each triangle one by one… You don’t even have to worry about doing math…
http://www.iwpcug.org/davidbro/puz0603.htm
64 I know it is right!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
i say 8
I’m not sure if anyone still cares, but the answer is 22 three-sided polygons….ahem, “triangles”. This is assuming that you are only looking at the big triangle and are not using anything else but your eyes!
I take the question to mean how many *types* of triangles are present in the diagram. In that case, there are three: scalene, isosceles, and equilateral.
Ok, I don’t think that was the real question, but really folks. Trust the formula. The answer is 64.
Jarjar… You’re an i_i_t… Did you (does anyone) read any previous posts?
27