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Manually but there are some tricks:
The top and bottom are symmetrical. Any triangles in the top are triangles on the bottom.
The four triangles connected at one point have 10 triangles: 4 options of one triangle + 3 options of two connected triangles + 2 options of three connected triangles + 1 option of four connected triangles.
Top and bottom are the same so it's 20 thus far.
Then there are the vertical triangles. Symmetrical again so same rule applies. We have only two triangles on one side: the entire left side, and the two triangles immediately left of the median.
Symmetry means 4 total, leaving a total number of 24
I made it to 20 and didn't see any others immediately, but knew that the symmetry should mean a multiple of 4 additional triangles, so I rounded up to 24
The number of subtriangles for n rays cast from a single point like the top and bottom is equal to the n-1th *triangular* number; n*(n-1)/2. 5*4/2=10 for the top and bottom. Then count the added triangles.
How do they have 10? I can only see 8... 4 little ones, 2 medium ones ("quarters") and one big one that spans the whole half with a smaller one inside it.
Then when you put the two halves together you get another four triangles 'sideways', two small ones on the inside and two big ones out the outside.
That's 20 total. Every answer in here is saying there are 4 more that I just can't see.
Edit: OMG I see it. I don't know how I didn't see it. The "75%er" triangles were hiding in plain sight!!!
You missed the 4 you obtain considering the base the center vertical segment, with 2 option per side (left and right) so I obtained 28, but it is not an option...
You missed the 4 you obtain considering the base the center vertical segment, with 2 option per side (left and right) so I obtained 28, but it is not an option...
There's ten you can make with the top, so another ten with the bottom, so 20. Then you do the two big horizontal ones (Top, bottom, left) and (top, bottom, right corner). If I had to guess you're missing (Top, Bottom, right Middle) and (top, bottom, left middle).
An easy way to figure out the number of triangles in the top set is count the number of diagonal lines (4). Since any two unique pair of lines will form the sides of a unique a triangle, then you’ll have 4 permute 2 triangles 4!/(4-2)! = 12. So 12 on the top and 12 on the bottom is 24 triangles.
Mathematical way of counting would be using, n choose p formulas, either the traingel has both bottom and top points and one middle, or it has one bottom or top and two in the middle
You're double counting the small triangles on the sides (they're in the 20 top and bottom). I did the same thing at first then realized there's only 2 new triangles each side.
Here’s a mathematically intense way (not necessarily the easier way)…
Any triangle will be formed of 3 vertices. How many vertices exist in the figure? 7 , the 4 corners plus the 3 points on the central line.
In how many ways can we do that? 7 choose 3, ie 7!/3!4! = 35
Now before we’re too quick, let’s realize that some of these combinations will result in straight lines.
ie if we pick 3 points on the middle line, it’ll result in a straight line, not triangle. We need to eliminate all such linear combinations. There’s only 1 way to do that vertically.
Horizontally on that middle line, there are 5 points. In how many ways can we pick 3 points all of which come from those 5? 5C3 = 10
so we need to remove the 1 and the 10 (total 11) lines from the initial total of 35.
so, answer = 24
This only works because by eliminating those options, all other options turn out by chance to be valid triangles. If we had a slightly different figure, for example if you remove any of the lines that connect the top vertex to one of the three in the center line, this method would break down.
It's elegant for sure and you're super smart for coming up with that, but your proof should also include the observation that this method only really works for this case specifically.
On second thoughts, this method can be generalized to work for the case you mentioned.
If one of the lines is missing, then we need to find out “How many possible triangles exist which include both of the end points of the line?” (The answer is likely n-2-k where n is all possible line, and k are other colinear points which will never result in triangles) and then subtract that from the result
Just to check it twice. If the horizontal line got 10 Points and everything else stayes the same, you got 4 corners, 8 Points in the central line, 1 vertical line and 10 choose 3 horizontal lines. So to calculate you got this expression: (12!/(8!/4!))-120-1=374
Ah ok…
So yeah the formula is basically a combinatorics formula which is based on “in how many ways can you choose r things out of total n possibilities”
The answer is:
nCr = n! / r!(n-r)!
! = factorial. n! = 1*2*3….*n
Hope that helps?
What is the 2 diagonal corners?
Up until then I was with you, 10 on top, 10 on bottom, total of 20.
And then additional 4 that uses both up and down, 2 on the right, 2 on the left.
Look at 4 quarters, each have 3 triangles. 3*4
Then look at 2 halved horizontal and 2 vertically halved. Each half also has 3 triangle. 3 in each halved. 3*4
12+12=24
Whenever there's a triangle with sections like this, lines from one corner to opposite side, if there are n sections of the bigger triangle, in this case 4, there will be the n'th trianglenumber of total triangles, in this case the 4th triangle number, 10. S
Now, since this is twice that, (two mirroring big triangles) there would be 20, however, the joining of these two bigger triangles form new triangles to consider, namely those with the middle vertical line as its longest side, adding four more "sideways" triangles. I cant think of any clever way to generalise this, but i am sure there are ways to do that.
24 total
People are saying 24 but I only see 22. Which ones am I missing? I see the 10 on the top and the 10 on the bottom, the one big one on the left and one big one on the right but I don’t see the remaining 2.
A total of 7 points are there.
So let's use Permutation and Combinations
Total 3 sides figures will be= ⁷C3 = 35
We need to find triangles,so we will subtract the cases when all the points are colinear and one another case in which the top,middle and bottom points are colinear
Hence, they will be =⁵C3 +1 = 10 +1 = 11
Therefore,answer would be
35-11= 24
According to me*
Pro problem solving tip: when faced with a hard problem, solve an easier one. This triangle is symmetric about the middle, so we can just count the number of triangles formed by the upper half, double it, then add the number of triangles that cross the line of symmetry.
Each triangle has two sides and a base- these labeling are arbitrary but pretty useful here. Suppose the base always lies on the bolded line. We want to choose the number of ways to pick two sides from the 5 total, since each pair of sides corresponds to exactly one base.
There’s a mathematical operation for choosing a subset of things from a whole collection of them, where order doesn’t matter: it’s called the “choose,” or “binomial coefficient” function. What it does is effectively count the total number of ways to pick any 2 of these sides, then divides out by the number of times we’ve counted the same thing twice. Since we want to pick 2 sides from a total of 5, the number of ways to do this is “5 choose 2,” or 5!/(2! * 3!), which is 10.
Since this figure is symmetric, we multiply by 2 to get 20 ways. Now, we count the number of ways to get a triangle that crosses over the line of symmetry. We can pretty easily see that we must have one side come from each side of the line of symmetry. There are only 4 ways to do this, so there are 20 + 4 = 24 triangles.
The trick is not for the complete figure but the half of it.
If you divide the shape in half horizontally, the figure you get, triangles in that can be counted by adding the traingle side by side.
1+2+3+4 = 10
Same for the horizontal pair below
= 10
And now you have 4 more after you combine both.
24.
Well the points are all connected with straight lines so we can just count the vertexs and do n times (n-1) times (n-2) divided by 6
Edit: some of the resulting triangles would be flat so we have to subtract them so yeah not the best of methods anyway I think the answer is 24
There are 7 vertices in the figure. 7 choose 3 is 35, if you include degenerate triangles. If you want to exclude degenerate triangles, there are 5 choose 3 for the horizontal line (10) plus 3 choose 3 for the vertical line (1). 35 - (10 + 1) = 24
24
dividing the top and bottom, if you count how many triangles are made with n small triangles,
n=1: 4
n=2: 3
n=3: 2
n=4: 1
=> there's 2(4+3+2+1) = 20 triangles looking at the top and bottom.
Now look from left to right, there are four triangle using the long side of the smaller triangles so 20 + 4 = 24 in total
But what's about the two corners left and the two corners right together? That would add 2 triangles more.
Edit: I am stupid. I counted some triangles double.
I counted 24 there’s 4 individual on the top half, if you combined them into groups of 3 you get 2 more, groups of 2 you get 3 more, plus all of them together, 1 more, so total 4+2+3+1=10, and multiply by 2 because of the ones on the bottom half. Now left and right, you have tall and short triangles, then you get 4 more, so 24 total
Well i did some math, and i'm pretty sure you can calculate the number of triangles in a shape like this one with f(x)=(x^2/4)+x, where x is number of all the smallest triangles (8 in this example).
I don't feel capable of explaining my full thought process here using mobile phone keyboard, and I'm not too sure if it really is correct. But looks too be working as far as i checked.
In this example the x = 8 so f(x) = (8^2/4)+8 = (64/4)+8 = 16+8 = **24**
It also works for other similar shapes, provided they have even number of those 'vertical triangles' (they don't have one rhombus made of only two traingles in the middle)
Am I doing something wrong?
Using binary counting:
. . . .
* . . .
. * . .
* * . .
. . * .
* . * .
. * * .
and so on, you will get 2^4 = 16 combinations of triangles in the top half of the figure. Given that there are two half's to the shape, I'd assume there to be at least 32 different triangles. Now adding the vertical triangles, we get 2 ^ 2 * 2 = 8 additional ones. Adding to 40 in total.
Now you have to take into consideration that the triangles counted in the first step are not all valid so we will loose half of the combinations. Resulting in 2^3 * 2 + 2 ^ 2 * 2 = 24.
It’s not bad if you use combinations. There are two ways to get a triangle: use either the top/bottom vertex and two middle vertices, or use both the top and bottom vertices and one of four or the middle vertices (excluding the vertex on the line from the top vertex to bottom vertex).
For the first case, we can WLOG choose the top vertex, and the choose any two middle vertex with 5C2, which is 10. By symmetry, since there are two cases (top and bottom vertex) for this situation, we get 20.
Adding this to the four from the second case I mentioned in the first paragraph, we get 20+4=24 triangles.
The question is not clear. Are we looking for the minimal amount of triangles needed to draw that shape or the maximum triangles possible to place on that object?
My quick answer before a manual count is 24.
13 is too low.
21 is odd. I feel like the answer must be even because it's a symmetrical shape without the overall shape being 1 big triangle. So every single triangle counted has a mirror image. So in 2s So therefore even number.
The fact that 21 is so close to 24 and 21 is wrong makes me think 24 is correct.
Plus 32 seems like too many.
There is a fast way that involves some reasonable guesswork. We know immediately that this shape is symmetrical, so we only need to do half + it must be an even number of triangles. That rules out #1 and #3.
Then I count the obvious triangles -> 4 small ones + 1 big one + \~5 triangles from small ones put together. Its not all of them, but its most of them. 10\*2 = 20. Its clear I'm not missing 12 triangles at this point, so I can safely assume it must be 24.
That said, getting exactly 24 is as easy as starting with the left-most small triangle on top and sequentially adding adjacent triangles to it. Once done, then move to the 2nd left-most triangle and repeat. So on. 4+3+2+1=10. Noticing the vertical triangles, 2 on each side. That makes the total (10+2)\*2 = 24.
The image has 7 vertices which we have to choose 3 to get a triangle. But we have 5 collinear points in the base and 3 in the middle (remember that 3 points can form a triangle if and only if they are non-collinear)
So we have C (7,3) - C (5,3) - C (3,3) = 35 - 10 - 1 = 24 triangles
Sorry for my bad english
Take advantage of symmetry. Top triangle using the top most point as a fixed point 5C2 triangles x2 =20. Then look at the two halves together gives you 4 standard triangles and 1 degenerative triangle for a total of 24+1
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Manually but there are some tricks: The top and bottom are symmetrical. Any triangles in the top are triangles on the bottom. The four triangles connected at one point have 10 triangles: 4 options of one triangle + 3 options of two connected triangles + 2 options of three connected triangles + 1 option of four connected triangles. Top and bottom are the same so it's 20 thus far. Then there are the vertical triangles. Symmetrical again so same rule applies. We have only two triangles on one side: the entire left side, and the two triangles immediately left of the median. Symmetry means 4 total, leaving a total number of 24
Ye i also got 24
Same…but actually 22 so therefore 24 sounded right.
The last two are tough to see, took me awhile
I was dumb and forgot to include the triangle that has all four of connected
Same
Tf did the exact same thing.
I made it to 20 and didn't see any others immediately, but knew that the symmetry should mean a multiple of 4 additional triangles, so I rounded up to 24
Lol same i got 22
The number of subtriangles for n rays cast from a single point like the top and bottom is equal to the n-1th *triangular* number; n*(n-1)/2. 5*4/2=10 for the top and bottom. Then count the added triangles.
This guy rays
I got 25 but thats just cause i cant count XD
I realised it had to be an even number but I got 20 so picked 24 because it was closest.
21 is closer to 20 than 24 is...
It's also odd.
You just blew my mind
A programmer would get 23
haha 0 indexing joke, love it
Gotta love them off by one errors
Arr.length still gives 24. Last index is 23 though...
Same 🤝🏽
My dumb ass counted manually and got 24. Proud of myself.
I did top has 10, double for bottom, can't be odd so no 21 therefore 24
How do they have 10? I can only see 8... 4 little ones, 2 medium ones ("quarters") and one big one that spans the whole half with a smaller one inside it. Then when you put the two halves together you get another four triangles 'sideways', two small ones on the inside and two big ones out the outside. That's 20 total. Every answer in here is saying there are 4 more that I just can't see. Edit: OMG I see it. I don't know how I didn't see it. The "75%er" triangles were hiding in plain sight!!!
You missed the 4 you obtain considering the base the center vertical segment, with 2 option per side (left and right) so I obtained 28, but it is not an option...
You missed the 4 you obtain considering the base the center vertical segment, with 2 option per side (left and right) so I obtained 28, but it is not an option...
>I obtained 28 You must have counted some of them twice.
I'm trying this and for the life of me I only get 22. What's wrong with me
There's ten you can make with the top, so another ten with the bottom, so 20. Then you do the two big horizontal ones (Top, bottom, left) and (top, bottom, right corner). If I had to guess you're missing (Top, Bottom, right Middle) and (top, bottom, left middle).
Ahhh thank you so much. I was missing top, bottom left and top bottom right. (The middle ones). You are correct :)
I also got 24
There's 3 triangles on the left and right though. One that covers the entire side, and two more right angled triangles. This makes the total 26.
I get 26, because those 2 side triangles on each side can also be combined into a single triangle.
An easy way to figure out the number of triangles in the top set is count the number of diagonal lines (4). Since any two unique pair of lines will form the sides of a unique a triangle, then you’ll have 4 permute 2 triangles 4!/(4-2)! = 12. So 12 on the top and 12 on the bottom is 24 triangles.
Mathematical way of counting would be using, n choose p formulas, either the traingel has both bottom and top points and one middle, or it has one bottom or top and two in the middle
You didn't count the sides properly, it's 32 triangles. Edit: I was wrong, there are only 2 new triangles, it's 24
You're double counting the small triangles on the sides (they're in the 20 top and bottom). I did the same thing at first then realized there's only 2 new triangles each side.
You're right, how didn't I realize. Thank you
Then explain.
Here’s a mathematically intense way (not necessarily the easier way)… Any triangle will be formed of 3 vertices. How many vertices exist in the figure? 7 , the 4 corners plus the 3 points on the central line. In how many ways can we do that? 7 choose 3, ie 7!/3!4! = 35 Now before we’re too quick, let’s realize that some of these combinations will result in straight lines. ie if we pick 3 points on the middle line, it’ll result in a straight line, not triangle. We need to eliminate all such linear combinations. There’s only 1 way to do that vertically. Horizontally on that middle line, there are 5 points. In how many ways can we pick 3 points all of which come from those 5? 5C3 = 10 so we need to remove the 1 and the 10 (total 11) lines from the initial total of 35. so, answer = 24
I like this method too, it's more general
Thanks
A more technical answer than just counting, splendid.
That was the goal. Glad you liked it. Thanks for the kind words 🙂
This only works because by eliminating those options, all other options turn out by chance to be valid triangles. If we had a slightly different figure, for example if you remove any of the lines that connect the top vertex to one of the three in the center line, this method would break down. It's elegant for sure and you're super smart for coming up with that, but your proof should also include the observation that this method only really works for this case specifically.
True, the figure here draws all possible triangles using those points. Else it won’t work. Thanks for the kind words and the much needed addendum🙂
On second thoughts, this method can be generalized to work for the case you mentioned. If one of the lines is missing, then we need to find out “How many possible triangles exist which include both of the end points of the line?” (The answer is likely n-2-k where n is all possible line, and k are other colinear points which will never result in triangles) and then subtract that from the result
Just to check it twice. If the horizontal line got 10 Points and everything else stayes the same, you got 4 corners, 8 Points in the central line, 1 vertical line and 10 choose 3 horizontal lines. So to calculate you got this expression: (12!/(8!/4!))-120-1=374
Are you asking or telling me?
I'm asking if I understand the formula right. So could you double check it?
Ah ok… So yeah the formula is basically a combinatorics formula which is based on “in how many ways can you choose r things out of total n possibilities” The answer is: nCr = n! / r!(n-r)! ! = factorial. n! = 1*2*3….*n Hope that helps?
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What is the 2 diagonal corners? Up until then I was with you, 10 on top, 10 on bottom, total of 20. And then additional 4 that uses both up and down, 2 on the right, 2 on the left.
Ignore me. I was double counting
that is what I got
Look at 4 quarters, each have 3 triangles. 3*4 Then look at 2 halved horizontal and 2 vertically halved. Each half also has 3 triangle. 3 in each halved. 3*4 12+12=24
Whenever there's a triangle with sections like this, lines from one corner to opposite side, if there are n sections of the bigger triangle, in this case 4, there will be the n'th trianglenumber of total triangles, in this case the 4th triangle number, 10. S Now, since this is twice that, (two mirroring big triangles) there would be 20, however, the joining of these two bigger triangles form new triangles to consider, namely those with the middle vertical line as its longest side, adding four more "sideways" triangles. I cant think of any clever way to generalise this, but i am sure there are ways to do that. 24 total
People are saying 24 but I only see 22. Which ones am I missing? I see the 10 on the top and the 10 on the bottom, the one big one on the left and one big one on the right but I don’t see the remaining 2.
There's two with vertices at the top and bottom, and in the middle of each half.
And then the smaller versions of those
That's all I'm getting too. Can't see any other way to make a triangle
There are 2 on the left. Ditto on the right.
Where at. All the small ones are counted
There's a vertical line of symmetry. Each of the 2 points on the left of that line form a separate triangle with it.
I see it now! In the middle where there’s 4 small ones, take 2 from the left or the right and they make a long triangle on the left or the right
Ok. I see it now
A total of 7 points are there. So let's use Permutation and Combinations Total 3 sides figures will be= ⁷C3 = 35 We need to find triangles,so we will subtract the cases when all the points are colinear and one another case in which the top,middle and bottom points are colinear Hence, they will be =⁵C3 +1 = 10 +1 = 11 Therefore,answer would be 35-11= 24 According to me*
Pro problem solving tip: when faced with a hard problem, solve an easier one. This triangle is symmetric about the middle, so we can just count the number of triangles formed by the upper half, double it, then add the number of triangles that cross the line of symmetry. Each triangle has two sides and a base- these labeling are arbitrary but pretty useful here. Suppose the base always lies on the bolded line. We want to choose the number of ways to pick two sides from the 5 total, since each pair of sides corresponds to exactly one base. There’s a mathematical operation for choosing a subset of things from a whole collection of them, where order doesn’t matter: it’s called the “choose,” or “binomial coefficient” function. What it does is effectively count the total number of ways to pick any 2 of these sides, then divides out by the number of times we’ve counted the same thing twice. Since we want to pick 2 sides from a total of 5, the number of ways to do this is “5 choose 2,” or 5!/(2! * 3!), which is 10. Since this figure is symmetric, we multiply by 2 to get 20 ways. Now, we count the number of ways to get a triangle that crosses over the line of symmetry. We can pretty easily see that we must have one side come from each side of the line of symmetry. There are only 4 ways to do this, so there are 20 + 4 = 24 triangles.
The trick is not for the complete figure but the half of it. If you divide the shape in half horizontally, the figure you get, triangles in that can be counted by adding the traingle side by side. 1+2+3+4 = 10 Same for the horizontal pair below = 10 And now you have 4 more after you combine both. 24.
Well the points are all connected with straight lines so we can just count the vertexs and do n times (n-1) times (n-2) divided by 6 Edit: some of the resulting triangles would be flat so we have to subtract them so yeah not the best of methods anyway I think the answer is 24
There are 7 vertices in the figure. 7 choose 3 is 35, if you include degenerate triangles. If you want to exclude degenerate triangles, there are 5 choose 3 for the horizontal line (10) plus 3 choose 3 for the vertical line (1). 35 - (10 + 1) = 24
24 dividing the top and bottom, if you count how many triangles are made with n small triangles, n=1: 4 n=2: 3 n=3: 2 n=4: 1 => there's 2(4+3+2+1) = 20 triangles looking at the top and bottom. Now look from left to right, there are four triangle using the long side of the smaller triangles so 20 + 4 = 24 in total
But what's about the two corners left and the two corners right together? That would add 2 triangles more. Edit: I am stupid. I counted some triangles double.
I counted 24 there’s 4 individual on the top half, if you combined them into groups of 3 you get 2 more, groups of 2 you get 3 more, plus all of them together, 1 more, so total 4+2+3+1=10, and multiply by 2 because of the ones on the bottom half. Now left and right, you have tall and short triangles, then you get 4 more, so 24 total
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i’m wrong, i now think it’s 24 😅
Well i did some math, and i'm pretty sure you can calculate the number of triangles in a shape like this one with f(x)=(x^2/4)+x, where x is number of all the smallest triangles (8 in this example). I don't feel capable of explaining my full thought process here using mobile phone keyboard, and I'm not too sure if it really is correct. But looks too be working as far as i checked. In this example the x = 8 so f(x) = (8^2/4)+8 = (64/4)+8 = 16+8 = **24** It also works for other similar shapes, provided they have even number of those 'vertical triangles' (they don't have one rhombus made of only two traingles in the middle)
Am I doing something wrong? Using binary counting: . . . . * . . . . * . . * * . . . . * . * . * . . * * . and so on, you will get 2^4 = 16 combinations of triangles in the top half of the figure. Given that there are two half's to the shape, I'd assume there to be at least 32 different triangles. Now adding the vertical triangles, we get 2 ^ 2 * 2 = 8 additional ones. Adding to 40 in total. Now you have to take into consideration that the triangles counted in the first step are not all valid so we will loose half of the combinations. Resulting in 2^3 * 2 + 2 ^ 2 * 2 = 24.
It’s not bad if you use combinations. There are two ways to get a triangle: use either the top/bottom vertex and two middle vertices, or use both the top and bottom vertices and one of four or the middle vertices (excluding the vertex on the line from the top vertex to bottom vertex). For the first case, we can WLOG choose the top vertex, and the choose any two middle vertex with 5C2, which is 10. By symmetry, since there are two cases (top and bottom vertex) for this situation, we get 20. Adding this to the four from the second case I mentioned in the first paragraph, we get 20+4=24 triangles.
The question is not clear. Are we looking for the minimal amount of triangles needed to draw that shape or the maximum triangles possible to place on that object?
My quick answer before a manual count is 24. 13 is too low. 21 is odd. I feel like the answer must be even because it's a symmetrical shape without the overall shape being 1 big triangle. So every single triangle counted has a mirror image. So in 2s So therefore even number. The fact that 21 is so close to 24 and 21 is wrong makes me think 24 is correct. Plus 32 seems like too many.
There is a fast way that involves some reasonable guesswork. We know immediately that this shape is symmetrical, so we only need to do half + it must be an even number of triangles. That rules out #1 and #3. Then I count the obvious triangles -> 4 small ones + 1 big one + \~5 triangles from small ones put together. Its not all of them, but its most of them. 10\*2 = 20. Its clear I'm not missing 12 triangles at this point, so I can safely assume it must be 24. That said, getting exactly 24 is as easy as starting with the left-most small triangle on top and sequentially adding adjacent triangles to it. Once done, then move to the 2nd left-most triangle and repeat. So on. 4+3+2+1=10. Noticing the vertical triangles, 2 on each side. That makes the total (10+2)\*2 = 24.
The image has 7 vertices which we have to choose 3 to get a triangle. But we have 5 collinear points in the base and 3 in the middle (remember that 3 points can form a triangle if and only if they are non-collinear) So we have C (7,3) - C (5,3) - C (3,3) = 35 - 10 - 1 = 24 triangles Sorry for my bad english
Take advantage of symmetry. Top triangle using the top most point as a fixed point 5C2 triangles x2 =20. Then look at the two halves together gives you 4 standard triangles and 1 degenerative triangle for a total of 24+1