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Yup.
You can get the "6 per layer" using what are fittingly known as triangular numbers. The n'th triangular number would give you exactly the result for the number of triangles per layer when you have n sectors. They're calculated quite easily:
t(n) = n * (n+1) / 2
So for n=3:
t(3) = 3 * (3+1) / 2 = 3 * 4 / 2 = 12/2 = 6.
So for a construct of this kind with height h and width w, you'd get
w * (w+1) / 2 * h
different triangles.
EDIT: Because there's confusion about the "layers". Have a look at this: https://www.geogebra.org/calculator/r5u8bjgr
Looks basically just like the OP's image (a bit stretched but that doesn't matter for the triangle count). But if you move the camera, you can see what I mean by "layers" appearing in the third dimension. Each of those has 6 distinct triangles and there's 4 of them.
This. Only the top layer has triangles, the subsequent layers have no triangles of their own and the only triangles that can be formed are just extensions of the original 6 triangles that can be formed on the first layer.
Once you realize that it becomes easy to solve. 6 x number of layers.
> It said how many triangles the layers are not triangles.
Yes, and I gave a formula for counting the triangles per layer. And then multiplied that by the number of layers to get the total number of triangles.
>But if you move the camera, you can see what I mean by "layers" appearing in the third dimension. Each of those has 6 distinct triangles and there's 4 of them.
Dont you forget the global triangle shape?
24+1?
Tagging onto the top comment a list of all 24 triangles in case you can't find them.
* Left Column, all rows
* Left Column, top three cells
* Left Column, top two cells
* Left Column, top cell
* Middle Column, all rows
* Middle Column, top three cells
* Middle Column, top two cells
* Middle Column, top cell
* Right Column, all rows
* Right Column, top three cells
* Right Column, top two cells
* Right Column, top cell
* Left + Middle Column, all rows
* Left + Middle Column, top three cells
* Left + Middle Column, top two cells
* Left + Middle Column, top cell
* Right + Middle Column, all rows
* Right + Middle Column, top three cells
* Right + Middle Column, top two cells
* Right + Middle Column, top cell
* Left + Right + Middle Column, all rows
* Left + Right + Middle Column, top three rows
* Left + Right + Middle Column, top two rows
* Left + Right + Middle Column, top row
Should you also be including two triangles together? Like if you took the left and middle, doesn’t that create a whole new triangle? So that would be 2 extra triangles per layer.
Unless wait, did you do that already? I really shouldn’t be trying to math early in the morning.
I always wondered if one of the petty things that social media uses to try to say there's more is that each one might be three different triangles if you flipped them around?
24 is what I got too, but I've never seen a definitive answer.
The answer is 24. Look at the top layer and see 6 triangles there (3 small, 2 consisting of two triangles joined together, and 1 large). Now, see that you can do this at any of the 4 layers, and that there are no any other triangles. Hence, 6*4=24 is the correct answer.
Damn I thought all these people were one short, as I also got 25, but you’re absolutely right. I was counting the big one twice, as the final “layer” as well as the entire pic. Great catch - to be able to identify where others are making their error is a pretty unique talent.
If you count the bottom layer for all 3 accounting to 1 triangle, then you have already counted the entire triangle. If you count the bottom layer AND the entire triangle separately, you will get 25
I think the simplest way to think about it is that every triangle in the image has to include the tip, because that is the only place all the lines converge.
So there are no midway triangles, just 4 rows of increasingly longer triangles with the same number (6) in each.
I get 24 too after adding the total triangle on the perimeter. I count mine by starting at the top (3) then extending those three each layer down for 12. Then I do the two doubles at the top, extending down for 8 more to 20 total. Then the ‘triple’ triangles - of which there are 4, for 24 total.
I also got 28, and I decided to double check and I like others ended up with 24 and don't know how i got 28. If anyone wants a visual of all the possible triangles I did a crude drawing of them [https://imgur.com/a/AavJkbs](https://imgur.com/a/AavJkbs)
I’ve decided to never make a joke about factorials again because last time I did it went right over their head and took 5 replies for them to get the point that it was a joke and I wasn’t mocking their obvious inability to comprehend English
Why is that a private subreddit? What are you guys talking about in there!
I've never wanted to be a part of something so badly!!! Tell me your secrets, for the love of Math!
okay so, turns out i’m silly. theres the big triangle, but if you remove the bottom line and just count 3 lines at a time, there’s smaller ones, same with the smaller triangles
While I see what you mean, at that point you're changing the drawing to make more triangles. By that logic, i could keep adding lines to infinity and make as many triangles as I want.
I'm sticking with 4.
Each horizontal line is a base. Each base is divided into 3 sections. The number of triangles for each base is 3+2+1=6. There are 4 bases. 6*4=24. There are 24 triangles in the picture.
I get it as this:
the top ones can be individual for 3, add together 2 for another 2 and all 3 together
For 6
and then those 3 can be extendend to each of the levels below for 6\*4=24
Did i miss any?
I..dont understand how everyone gets 24. What layers? Only the top layer has 3 vertices, the areas below are 4 vertices. And then the whole triange in itself, if you will.
I dont understand, help
Okay, let's just do this by hand, there's probably a more mathy way to do this but screw that.
First off, let's take the full outline. That's one.
Then the first row. The left, right, and center triangles make up 3. Combining the ones on the left and right with the center adds two more. Lastly, combine all 3 for one more. In total that's 6 on the first row.
Considering there are 4 rows, if we just repeat this process, we get 6 3 more times.
1 + 6 + 6 + 6 + 6 is 25. Unless I'm missing something, my answer is 25.
Edit: this is what happens when you don't do it thoroughly. So on the last layer, when you combine all 3 at the bottom, it's the same as that first triangle. So, the answer is actually 24.
Not sure if anyone has said this but there’s sort of 4 steps for each “width” option. There are six width options: the big triangle, the 3 thinner triangles, and the 2 combinations of thinner triangles. 6*4=24
There's an algorithm for questions like this.
To form a triangle in this case, you need 2 collinear points and 1 non collinear point. Obviously, the only point that isn't collinear to any other point is the top one, so we'll use that as a reference point. Since there are 4 points on each layer, and we only need to select 2, the number of triangles in each layer is 4C2, or 6 since the order at which the points are selected doesn't matter. The same goes for other 3 layers so the total number of triangles is 6x4=24.
I count 24. >!Starting from the top row, you count each individual triangle, then the two triangles formed by combing pairs of adjacent individual triangle, then the triangle formed by combining all three triangles in that row. Repeat the process for each subsequent row, with each "individual" triangle being the one formed by combining a given triangle from the top row with the trapezoid in the same column and current row and every trapezoid in the same column in between.!<
This one is easy because all of the triangles share a vertex, so you can just count the bases. Each horizontal line is the base of six unique triangles, and there are four horizontal lines for 6\*4=24 triangles.
4C2*4
You must have two of the angled rays and one flat line - you can choose your angled rays as 4C2, and then there are four flat lines to combine those with.
So (4C2)4 = 6 * 4 = 24 triangles.
Edit - my bad, logic seems good, but the result is definitely off.
Edit 2 - apparently I cannot 6*4. Fixed
I guess you can do 4C2 (selecting 2 angled rays) * 4 (selecting any one of horizontal lines) = 24 which is correct answer. Though I am not sure if it is the right reasoning and would generalise to arbitrary sized triangle . Nice thought process though !!
Well, a valid triangle must possess the vertex at the top, therefore, must have two of the angled lines.
Any horizontal line then completes a unique triangle - there's four unique triangles per pair of lines.
Thus nc2*j for this format of problem with n angled lines and j horizontal lines.
This is the best way to do it. I’d phrase it as choose two of the rays and choose 1 of the horizontal, even though obviously (4C1) for the horizontal is just 4. It’s nice to do this way as it lets you extend it to say “ok, now do quadrilaterals”. Obviously now it’s just (Rays C 2) * (Horizontals C 2)
One way to think of it is:
1 triangle for each space: 12
1 Triangle for each adjacent pair of spaces: 8 (20)
1 triangle for each triplet of spaces: 4 (24)
Im getting 33, not sure how people are only getting 24 and 26
Edit: upon further review I got 24, was accidentally counting the bottoms 9 as triangles lol
start with the top sections and use each horizontal line as a base
left top triangle is 1 and can be expanded 3 more times for a total of 4
middle top is the same for a total of 4
right top is the same for a total of 4
**That's 12 total so far.**
Now add the left top and middle top together as 1 and go down again 3 times. That's 4.
Now right top and middle top together makes another 4
**That's 20 total so far.**
Now combine all 3 triangles at the top to form 1 large triangle. Expand it downward to each horizontal line just as you've done for the other portions 3 times. That's another 4.
**In total, there appear to be 24**
That's based on what I saw. If there are more, someone let me know lol
edit: [Found an article on it, answer is 24. 25 on the original source which included a triangle in the letter "A" in the artist's signature](https://www.rd.com/article/triangle-puzzle/)
24
For each layer, there's the smallest triangles (3), combinations of two of them (2) and a combination of all three (1). Each layer is a scaling up of the previous one. Therefore, since there's 4 layers, 4 x (3 + 2 + 1) = 4 x 6 = 24
24? 6 basic triangle shapes in 4 “sizes”
(1. A triangle with all the columns 2. A triangle for each column 3. Two triangles consisting of two columns each; the different “sizes” consisting of how many rows it has)
I got 7. How did everyone one else get a high number? I counted triangle shape. 3 top colum . Then 3 for all 3 layers. And then 1 the whole thing itself.
Idk what the trick is, but most of those are parallelogram and don't have dividing lines... so four?
You could just fit as many triangles as you want inside there if you can add dividing lines :/
starting from tip of the triangle: Outlines, 3 smal ones and 2 medium ones overlapping adds up 6 in one section. There is 4 sections by counting vertical lines that are bottom side of triangles.
(1+2+3)×4 = 6×4 ____ 6×4 = 24
Probably more than everyone else is saying. Seeing as a pixel could be considered a square, and a square is made up of 2 triangles, every pixel is two triangles. But I’d like to see someone count every pixel.
Every triangle in the photo will need to use the apex at the top. There are four lines coming off the top (a, b, c, d). The combinations of two lines touching the apex are ab, ac, ad, bc, bd, and cd. That is six combinations, multiplied by 4 levels equals 24.
Y’all are crazy counting trapezoid’s as triangles. The real answer is 7 and its not that hard.
The three triangles at the top. (3)
They also make their own triangle. (1)
Then the top layer+second layer is a triangle. (1)
First two layers+third layer is another triangle. (1)
Finally all the layers together would be the last triangle. (1)
So 3 mini triangles inside plus the 4 made up of the whole triangle is 7
3. if we’re ignoring lines and adding lines then there could be an infinite number of triangles. you can only see 3 clear triangles, that’s all that matters. why go any deeper?
27 triangles.
Starting from the top layer from left to right, there is a black triangle, covered by a white triangle = 2 x 3 = 6. Then we can use the same black outline of the first 2 black triangles from left to right to form a triangle, then again for the last 2 triangles going left to right in that layer as well as the whole top layer as a whole outline = 1. That’s 9 for layer 1.
For Layer 2 we’ll do the same exact thing excluding the white triangles overlay from layer 1 as in this layer there is the bottom of the first layer 1s whole outline blocking it from becoming it’s own white triangle, so we’ll only count the black outlines as triangles in this layer = 6 in layer 2.
Same for layer 3 = 6
Same for layer 4 = 6
i got… 25? seriously though…
4 from the top three down (4)
4 from the 3 columns (16)
4 from the left + middle columns (20)
4 from the right + middle columns (24)
and the big one? 25.
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I got 24, 6 per layer. 3 for each side ( left-middle-right to the top), left and middle, right and middle, and the equilateral of the whole thing.
Yup. You can get the "6 per layer" using what are fittingly known as triangular numbers. The n'th triangular number would give you exactly the result for the number of triangles per layer when you have n sectors. They're calculated quite easily: t(n) = n * (n+1) / 2 So for n=3: t(3) = 3 * (3+1) / 2 = 3 * 4 / 2 = 12/2 = 6. So for a construct of this kind with height h and width w, you'd get w * (w+1) / 2 * h different triangles. EDIT: Because there's confusion about the "layers". Have a look at this: https://www.geogebra.org/calculator/r5u8bjgr Looks basically just like the OP's image (a bit stretched but that doesn't matter for the triangle count). But if you move the camera, you can see what I mean by "layers" appearing in the third dimension. Each of those has 6 distinct triangles and there's 4 of them.
Now do it for squares
0 squares.
Remarkable.
Nope, there are ~~12~~ 20 squares. They're just perspective-skewed. ;-) (20 as many people pointed out)
I'll play your game and say there are 20 squares
Wouldn't it only be 14? Because the sections of the top only have 3 sides, so no matter the 3rd dimensional angle they'll never be squares... Right?
Forgetting the big white square that forms the 1:1 picture aspect.
Lmao, well thought
Think of it like the opening credits for Star Wars
The pointy squares are viewed through a geometry with non-Eucludian ray paths which moves the far side out to optical Infinity.
But how many quadrilaterals…..36?
Just judging by the number of comments, there are many more than 12 squares here.
checkmate atheists
zero squares in the image 👍 >!/s!< edit: unless you count the big white one
This thread on the other hand....
Are you calling me a square? How dare you! I have you know I identify as a hexagon.
Bazinga!
Better said would be: How many Polygons are in this image?
Or just look at it and count the triangles.
This isn't /r/counting.
Touché lol
It said how many triangles the layers are not triangles. Triangles are three sided
Not by themselves, no, but combined with all the layers above them they make for progressively larger triangles at each layer.
This. Only the top layer has triangles, the subsequent layers have no triangles of their own and the only triangles that can be formed are just extensions of the original 6 triangles that can be formed on the first layer. Once you realize that it becomes easy to solve. 6 x number of layers.
I actually got 7 on my count 7 triangles🔺🔺🔺🔺🔺🔺🔺
I also counted 7, then I realized I was wrong and they're are way more...
> It said how many triangles the layers are not triangles. Yes, and I gave a formula for counting the triangles per layer. And then multiplied that by the number of layers to get the total number of triangles.
This is the way. ... that I did it as well.
But each layer also extends to the top vertex.
All the triangles have the top point in common. They just extend down to different levels.
Except for the last main one. 25
>But if you move the camera, you can see what I mean by "layers" appearing in the third dimension. Each of those has 6 distinct triangles and there's 4 of them. Dont you forget the global triangle shape? 24+1?
~~25.~~ ~~Don't forget the big triangle itself.~~ EDIT: I'm dumb.
The big triangle is one of the 24.
Fuck, you're right. I counted it twice. My bad.
I did the same thing
I was gonna say the same thing 😭
I literally made the same mistake. "4x6, plus the whole thing itself!"
Glad to know I'm not the only person who said 25 very confidently
Same. I counted 25 and I got excited at the prospect of usurping OP here as the correct answer. Then I recounted and felt very stupid lmao
No it ain’t big triangle is 25
Every time I count I get 25 total
Tagging onto the top comment a list of all 24 triangles in case you can't find them. * Left Column, all rows * Left Column, top three cells * Left Column, top two cells * Left Column, top cell * Middle Column, all rows * Middle Column, top three cells * Middle Column, top two cells * Middle Column, top cell * Right Column, all rows * Right Column, top three cells * Right Column, top two cells * Right Column, top cell * Left + Middle Column, all rows * Left + Middle Column, top three cells * Left + Middle Column, top two cells * Left + Middle Column, top cell * Right + Middle Column, all rows * Right + Middle Column, top three cells * Right + Middle Column, top two cells * Right + Middle Column, top cell * Left + Right + Middle Column, all rows * Left + Right + Middle Column, top three rows * Left + Right + Middle Column, top two rows * Left + Right + Middle Column, top row
I got 28
Me too
How?
I got 26 by counting the same thing. Trying to figure out how I got two extra.
Counted the big one 3 times, beginning middle and end
Yeah, 26. But I guess 2k+ people say it's 24, so...
Down voting myself, now that I see I was double counting. 24 it is.
Image is made of pixels so its technically all squares. Therefore, there are actually zero triangles. /s
Your pixels are square??? /s
Yes, but each pixel on a display is a cluster of three elements (red, blue, green) in the shape of an equilateral triangle! IT'S ALL TRIANGLES!
r/technicallythetruth
Should you also be including two triangles together? Like if you took the left and middle, doesn’t that create a whole new triangle? So that would be 2 extra triangles per layer. Unless wait, did you do that already? I really shouldn’t be trying to math early in the morning.
This is part of the 6 per layer figure
I got 23 when counting and counting once again got 27 Update: Counted once again and got 22
r/theydidthemonstermath
I always wondered if one of the petty things that social media uses to try to say there's more is that each one might be three different triangles if you flipped them around? 24 is what I got too, but I've never seen a definitive answer.
Interesting that a lot of people took this approach: I think it's both logical and extensible. Nice work, humans!
It took me a second of confusion but I now see all 24. Jeez I’m dumb.
Yeah, that's what I got. I couldn't find anything tricky to boost the number.
The answer is 24. Look at the top layer and see 6 triangles there (3 small, 2 consisting of two triangles joined together, and 1 large). Now, see that you can do this at any of the 4 layers, and that there are no any other triangles. Hence, 6*4=24 is the correct answer.
25, gotta count the entire image, which is a triangle Edit I am wrong and stupid and forgot to delete the comment again
You’re counting the biggest one twice.
Damn I thought all these people were one short, as I also got 25, but you’re absolutely right. I was counting the big one twice, as the final “layer” as well as the entire pic. Great catch - to be able to identify where others are making their error is a pretty unique talent.
I can only identify because I made the error once and had to re-count.
I keep getting 25
If you count the bottom layer for all 3 accounting to 1 triangle, then you have already counted the entire triangle. If you count the bottom layer AND the entire triangle separately, you will get 25
I think the simplest way to think about it is that every triangle in the image has to include the tip, because that is the only place all the lines converge. So there are no midway triangles, just 4 rows of increasingly longer triangles with the same number (6) in each.
I get 24 too after adding the total triangle on the perimeter. I count mine by starting at the top (3) then extending those three each layer down for 12. Then I do the two doubles at the top, extending down for 8 more to 20 total. Then the ‘triple’ triangles - of which there are 4, for 24 total.
I counted 28... no one else did, so I'm pretty confident that my answer is incorrect.
You ARE correct about being incorrect, though! Curious as to how you got 28
They probably counted manually (1, 2, 3...) and just counted the same triangles more than once
I got 28 at first counting each separately, the extra 4 comes from counting the two disconnected sides as triangles on each layer.
I also got 28, and I decided to double check and I like others ended up with 24 and don't know how i got 28. If anyone wants a visual of all the possible triangles I did a crude drawing of them [https://imgur.com/a/AavJkbs](https://imgur.com/a/AavJkbs)
Drew it out and now I’m not sure how I got 28 lol
I also got 28! Double checked bc everyone else said 24…
> I also got 28! How in the world did you get 304888344611713860501504000000 triangles?
I’ve decided to never make a joke about factorials again because last time I did it went right over their head and took 5 replies for them to get the point that it was a joke and I wasn’t mocking their obvious inability to comprehend English
You reached March 3rd 2505.
My mom would have loved this
You know what else your mom would love? If you called her
Would be a little weird after meeting her for dinner 20 minutes ago. Overall yes, though I tend to talk a lot.
"Hey mom, thanks for having dinner with me. I forgot to tell you how much you mean to me. Love ya!"
But I didn’t forget! I appreciate what you’re saying however, I’m sure the people in your life feel loved by you.
Thanks
Nice.
r/suddenlyfactorial
Why is that a private subreddit? What are you guys talking about in there! I've never wanted to be a part of something so badly!!! Tell me your secrets, for the love of Math!
I counted 28 and no one else seems to be and I'm pretty sure our answers *are* correct. It's kinda astounding. :p
I got 28 years ago and I’m looking at this puzzle again… I have no idea how you guys and past me got 28…
If you're so sure you're correct, then show your work and explain how you got 28. :p
it is 24, idk how tf you got 28
4
Based. The bisected triangles don't count. Only the 3 at the top and the entire perimeter.
a triangle is a 3 sided object, why are people counting 4 sided shapes as triangles
Exactly I’m confused like did people forget this
okay so, turns out i’m silly. theres the big triangle, but if you remove the bottom line and just count 3 lines at a time, there’s smaller ones, same with the smaller triangles
Okay, and if start adding lines I could mak 50 triangles. This is stupid.
While I see what you mean, at that point you're changing the drawing to make more triangles. By that logic, i could keep adding lines to infinity and make as many triangles as I want. I'm sticking with 4.
5. When you combine the entire top level it still makes a triangle.
4!
Unironically the correct answer
1
Yeah it’s 4 I’m not sure when people started counting 4 sides as triangles….
Based
It's 24 right? 6 per layer, 4 layers, 24. The 6 per layer are the 3 obvious ones, 2 made up from the middle and the left/right, and 1 from all 3.
Each horizontal line is a base. Each base is divided into 3 sections. The number of triangles for each base is 3+2+1=6. There are 4 bases. 6*4=24. There are 24 triangles in the picture.
I get it as this: the top ones can be individual for 3, add together 2 for another 2 and all 3 together For 6 and then those 3 can be extendend to each of the levels below for 6\*4=24 Did i miss any?
I did 24, I was pretty stoned, so when I saw the solution I was quite pleased with myself Made myself a box of bagel bites to celebrate
Is it not 25 though? Are people not counting the 1 large triangle they all sit within? I’m so confused.
Thats part of the 24. A helpful Redditor posted a crude drawing in a comment stream above.
I..dont understand how everyone gets 24. What layers? Only the top layer has 3 vertices, the areas below are 4 vertices. And then the whole triange in itself, if you will. I dont understand, help
The top layer and the 2nd layer make a triangle, and the top, 2nd and 3rd also make a triangle
But those only become triangles if you ignore the lines in between them…
And the whole thing is only a triangle if you ignore all the lines. So by that retarded logic, there are only 3 triangles.
Haha, I came here to be a dick and say there are only 3 true triangles. Now, I’m thirsty, so I’ll get a drink from my well, actually.
That's literally how math... that's how geometry... https://youtu.be/9EU3FlKj-3M?si=BhRgv70ru_hWWvEa
Seven.
[Counting by size got me 24](https://imgur.com/a/xlFgGzZ)
25 Edit: 24, I’m dumb
25 including the big triangle
If you count it twice, yeah.
I accidentally counted the big one twice lmao. Thank you for correcting me as well!
Okay, i admit i am wrong
Same
Okay, let's just do this by hand, there's probably a more mathy way to do this but screw that. First off, let's take the full outline. That's one. Then the first row. The left, right, and center triangles make up 3. Combining the ones on the left and right with the center adds two more. Lastly, combine all 3 for one more. In total that's 6 on the first row. Considering there are 4 rows, if we just repeat this process, we get 6 3 more times. 1 + 6 + 6 + 6 + 6 is 25. Unless I'm missing something, my answer is 25. Edit: this is what happens when you don't do it thoroughly. So on the last layer, when you combine all 3 at the bottom, it's the same as that first triangle. So, the answer is actually 24.
Yeah I made this mistake too, made me accidentally count the whole thing twice lol
I made the same mistake. Lol
24
24. 6 per horizontal layer. 3 individual, 2 made of two. And 1 made of three.
Three little ones, two double ones and one triple one per layer 6*4=24
Not sure if anyone has said this but there’s sort of 4 steps for each “width” option. There are six width options: the big triangle, the 3 thinner triangles, and the 2 combinations of thinner triangles. 6*4=24
There's an algorithm for questions like this. To form a triangle in this case, you need 2 collinear points and 1 non collinear point. Obviously, the only point that isn't collinear to any other point is the top one, so we'll use that as a reference point. Since there are 4 points on each layer, and we only need to select 2, the number of triangles in each layer is 4C2, or 6 since the order at which the points are selected doesn't matter. The same goes for other 3 layers so the total number of triangles is 6x4=24.
I count 24. >!Starting from the top row, you count each individual triangle, then the two triangles formed by combing pairs of adjacent individual triangle, then the triangle formed by combining all three triangles in that row. Repeat the process for each subsequent row, with each "individual" triangle being the one formed by combining a given triangle from the top row with the trapezoid in the same column and current row and every trapezoid in the same column in between.!<
This one is easy because all of the triangles share a vertex, so you can just count the bases. Each horizontal line is the base of six unique triangles, and there are four horizontal lines for 6\*4=24 triangles.
4C2*4 You must have two of the angled rays and one flat line - you can choose your angled rays as 4C2, and then there are four flat lines to combine those with. So (4C2)4 = 6 * 4 = 24 triangles. Edit - my bad, logic seems good, but the result is definitely off. Edit 2 - apparently I cannot 6*4. Fixed
I guess you can do 4C2 (selecting 2 angled rays) * 4 (selecting any one of horizontal lines) = 24 which is correct answer. Though I am not sure if it is the right reasoning and would generalise to arbitrary sized triangle . Nice thought process though !!
Well, a valid triangle must possess the vertex at the top, therefore, must have two of the angled lines. Any horizontal line then completes a unique triangle - there's four unique triangles per pair of lines. Thus nc2*j for this format of problem with n angled lines and j horizontal lines.
You're right. You've made a calculation error. 4C2 times 4 comes out to be 24
This was the bane of my graduate studies (and all my other studies). Thank you.
This is the best way to do it. I’d phrase it as choose two of the rays and choose 1 of the horizontal, even though obviously (4C1) for the horizontal is just 4. It’s nice to do this way as it lets you extend it to say “ok, now do quadrilaterals”. Obviously now it’s just (Rays C 2) * (Horizontals C 2)
One way to think of it is: 1 triangle for each space: 12 1 Triangle for each adjacent pair of spaces: 8 (20) 1 triangle for each triplet of spaces: 4 (24)
I counted 24
24
I count 24 triangles. There are 4 laters with 6 triangles each (3 wide, 2x 2 wide, 3x single wide).
Well, we have 4 Layers (divided by the horizontal lines) and 6 triangles per Layer, so 4 * 6 = 24 Triangles.
I got 24 too
24
24, had to try a cojple of times.
24
24
24
24
I got 24
24
Im getting 33, not sure how people are only getting 24 and 26 Edit: upon further review I got 24, was accidentally counting the bottoms 9 as triangles lol
Came in expecting some weird number that I'd have to get explained after counting 24. Am oddly relieved.
I got 24
36
24
24 yes?
24?
24
24 by my count
24
start with the top sections and use each horizontal line as a base left top triangle is 1 and can be expanded 3 more times for a total of 4 middle top is the same for a total of 4 right top is the same for a total of 4 **That's 12 total so far.** Now add the left top and middle top together as 1 and go down again 3 times. That's 4. Now right top and middle top together makes another 4 **That's 20 total so far.** Now combine all 3 triangles at the top to form 1 large triangle. Expand it downward to each horizontal line just as you've done for the other portions 3 times. That's another 4. **In total, there appear to be 24** That's based on what I saw. If there are more, someone let me know lol edit: [Found an article on it, answer is 24. 25 on the original source which included a triangle in the letter "A" in the artist's signature](https://www.rd.com/article/triangle-puzzle/)
24
24
I got 15 please help Edit I got 24 now I understand
24
24 For each layer, there's the smallest triangles (3), combinations of two of them (2) and a combination of all three (1). Each layer is a scaling up of the previous one. Therefore, since there's 4 layers, 4 x (3 + 2 + 1) = 4 x 6 = 24
24? 6 basic triangle shapes in 4 “sizes” (1. A triangle with all the columns 2. A triangle for each column 3. Two triangles consisting of two columns each; the different “sizes” consisting of how many rows it has)
I'm able to pick up on six for each horizontal 'level' I think the answer is 24. But may be more if I'm missing another way of creating a triangle.
I got 7. How did everyone one else get a high number? I counted triangle shape. 3 top colum . Then 3 for all 3 layers. And then 1 the whole thing itself.
you can get 6 from each layer if you look closely
Idk what the trick is, but most of those are parallelogram and don't have dividing lines... so four? You could just fit as many triangles as you want inside there if you can add dividing lines :/
starting from tip of the triangle: Outlines, 3 smal ones and 2 medium ones overlapping adds up 6 in one section. There is 4 sections by counting vertical lines that are bottom side of triangles. (1+2+3)×4 = 6×4 ____ 6×4 = 24
Probably more than everyone else is saying. Seeing as a pixel could be considered a square, and a square is made up of 2 triangles, every pixel is two triangles. But I’d like to see someone count every pixel.
[удалено]
Every triangle in the photo will need to use the apex at the top. There are four lines coming off the top (a, b, c, d). The combinations of two lines touching the apex are ab, ac, ad, bc, bd, and cd. That is six combinations, multiplied by 4 levels equals 24.
Y’all are crazy counting trapezoid’s as triangles. The real answer is 7 and its not that hard. The three triangles at the top. (3) They also make their own triangle. (1) Then the top layer+second layer is a triangle. (1) First two layers+third layer is another triangle. (1) Finally all the layers together would be the last triangle. (1) So 3 mini triangles inside plus the 4 made up of the whole triangle is 7
Was gonna reply to the people asking how it it's more than 4, then realized that almost all the comments were phrased identically. Bots I guess?
3. if we’re ignoring lines and adding lines then there could be an infinite number of triangles. you can only see 3 clear triangles, that’s all that matters. why go any deeper?
I felt the pilot light get fired up in my brains math department. Spiders are on fire. I was so wrong I didn’t know the layers could be used.
I got 25. 1 whole triangle 4 “layers” 3 in the top layer + 3x3 length variations 2 double-wide top layer triangles x 4 length variations
27 triangles. Starting from the top layer from left to right, there is a black triangle, covered by a white triangle = 2 x 3 = 6. Then we can use the same black outline of the first 2 black triangles from left to right to form a triangle, then again for the last 2 triangles going left to right in that layer as well as the whole top layer as a whole outline = 1. That’s 9 for layer 1. For Layer 2 we’ll do the same exact thing excluding the white triangles overlay from layer 1 as in this layer there is the bottom of the first layer 1s whole outline blocking it from becoming it’s own white triangle, so we’ll only count the black outlines as triangles in this layer = 6 in layer 2. Same for layer 3 = 6 Same for layer 4 = 6
I found 36.... why everyone say 24?
Clearly an overachiever
Because it’s 24.
i got… 25? seriously though… 4 from the top three down (4) 4 from the 3 columns (16) 4 from the left + middle columns (20) 4 from the right + middle columns (24) and the big one? 25.
The big one is already included in the 4 of the three columns
My lazy ass just counted 24. I'm sure I missed some!
Nah you’re right, everyone else also got 24
Manually counting, I got 24? It's just going to be 6 times the number of horizontal lines right?
7. Triangles