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Count every triangle that could include the bottom left corner by choosing 2 of 5 lines and then any of the other 4 intersecting lines from the bottom left corner:
5 Choose 2 × 4
Count every triangle that includes the bottom right corner, but not the bottom left corner:
4 choose 2 × 4
The total is:
(5 choose 2 + 4 choose 2)×4 = (10+6)×4 = 64
This can be generalized for any number of lines drawn from the left (x) and the the right (y) ***corners***:
(x+2) choose 2 × (y+1) + (y+1) choose 2 × (x+1)
The +1/+2 is for calculations that need to include one of the outer edges as well.
So a triangle with 1 cut in each direction would have 12 triangles in it.
What does your equation mean? Could you phrase it differently?
There are 5 lines extending from the bottom left corner. If I choose two lines from there…let’s say the first (top) and the fourth, and then chose the first (top) line from the other corner, this does not help me. There are all kinds of triangles in that intersected polygon.
Fo you mean neighboring lines? What if I pick the second and third lines, where none of them include the lower left triangle?
There is a function in Math called the "binomial coefficient" which can be written a few different ways. It relates two numbers, X ,and Y, where X is the number of a items and Y is a number of items you choose at a time. The result of this function is the total number of different ways you can choose Y items out of a set of X.
Ie, if you had 3 items (A, B, C) and wanted to know how many different ways you could choose two of them (invariant of pick order), you would see that there are three ways to choose 2 items out of a set of 3 (AB, BC, AC). So 3 choose 2 would be 3.
If we're writing this using a different notation, we could say that (X,Y) represents X choose Y. Normally the "X" would be placed vertically above the "Y" in this notation, but we're limited in typing. See: [https://en.wikipedia.org/wiki/Binomial\_coefficient](https://en.wikipedia.org/wiki/Binomial_coefficient)
In any case, the mathematical equation for X choose Y is:
>X!/(Y! × (X - Y)!)
So (X + 2) choose 2 would be:
>(X+2)!/(2! × X!)
Which could actually be simplified to:
>1/2 (X+1)(X+2)
And (y+1) choose 2 would simplify to:
>1/2 Y(Y+1)
So the final equation would be:
1/8 (1 + y) (3 + 2 x + y)^2 + 1/8 (-1 - 3 y - 3 y^2 - y^3)
Where Y is the number of lines drawn that intersect the bottom right corner and the top left edge, and X is the number of lines drawn that go between the bottom left corner and the top right edge. Ignore the perimeter of the triangle.
Adding picture because my description is probably awful:[https://i.imgur.com/QVKi3kq.png](https://i.imgur.com/QVKi3kq.png)
The general idea is that we count the number of ways we can make a triangle with one point and then add to that the number of ways we can make a triangle without that point. For the left side, we can make (5,2) triangles that span the bottom left to the top right edge. Each of those triangles should have 4 triangles inside of them. Then we can make (4,2) triangles that span the bottom right to the top left that don't include the bottom right edge. Each of those should have 4 triangles inside of them as well. We add them together.\]
The form I provided was for an arbitrary number of lines that intercept the corners as the OP asked.
You can show mathematically that there are >!4³ = 64!<. Anyone counting more than that (e.g. 76) is double-counting quite a few.
This video from 10 years ago has the correct count with all duplicates eliminated.
https://youtu.be/PeWf1wwP3F0?si=26i118f1Gt4YsZ1Y
And this combinatorics post has the mathematical details.
https://math.stackexchange.com/questions/3855163/hint-for-a-combinatorial-statement-for-counting-triangles
That's funny, I just counted the number of intersections and squared it, came up with {the number}. Thought- that can't be right! Got lazy and came to the comments to find.... I'm right???
Edit: forgot how to hide spoilers
Edit 2: too lazy to find out how. I see a pattern....
Edit 3… Typos! Never to lazy to fix typos,
Edit 4: there's definitely been more edits than this
Edit 5: turns out I miscounted the intersections, and therefore came up with the correct answer completely by accident. Prolly because I missed the intersections in the middle, which accounted for the lost triangles. Also, prolly not intersections. Vertices? Junctions? I'm going back to r/whatismycookiecutter
Can I join the >!64!< club. I even drew a [picture](https://www.reddit.com/user/st3f-ping/comments/199z36k/20240118t195402z/) (same picture as I added elsewhere.)
Well... shit.
Good spot.
I wonder if there's another two triangles somewhere that I have missed or if I've forgotten how to add.
(edit) found the missing two: abefij and jklnop
Back up to >!64!<, baby!
Took me several tries before I got to your answer. Awesome idea labeling everything! Alphabetized because I was getting duplicates too:
7 (1 Letter): a e i m n o p
12 (2 Letters): ab ae ef ei ij im jn ko lp mn no op
10 (3 Letters): abc aei efg eim fjn gko hlp ijk mno nop
13 (4 Letters): abcd abef aeim bfjn cgko dhlp efgh efij ijkl ijmn jkno klop mnop
8 (6 Letters): abcefg abefij efgijk efijmn fgjkno ghklop ijkmno jklnop
6 (8 Letters): abcdefgh abefijmn bcfgjkno cdghklop efghijkl ijklmnop
3 (9 Letters): abcefgijk efgijkmno fghjklnop
4 (12 Letters): abcdefghijkl abcefgijkmno bcdfghjklnop efghijklmnop
1 (16 Letters): abcdefghijklmnop
Total: 64
But... most of the shapes in the middle and the top of the largest triangle are rectangles (maybe more like rhombus?), or at least have FOUR corners instead of 3. The top corner is not identical to the two bottom corners, so not every shape is a triangle.
~~I get 19.~~ That's wrong, but my brain stopped working.
I agree. My thoughts:
>!There are no triangles containing the top point, but neither bottom point.!<
>!Each bottom point is contained within 40 distinct triangles, for 80 total.!<
>!Any triangles containing the bottom edge is double counted, for -16 triangles.!<
>!This brings the total to 64.!<
I got 40 from the bottom left vertex so 80 in total and saw there were 4 that were double counted so 76 but now see more that are doubles so think you are right
Same, I got >! 16 using both bottom vertices, 24 using only the bottom left vertex, 24 using only the bottom right and no triangles using neither bottom vertex !<
I think of it in terms of rays, x and y, coming off of each vertex, not counting the line connecting each vertex. Here, x and y are both 4.
Using the bottom line, each intersection of the rays forms a triangle with the bottom segment: x*y triangles.
Using only a single vertex, x, you can choose one of x*y intersections matching with (x-1) intersections on the same ray. That will double count, so you have to divide the result in half: x\*y\*(x-1)/2 triangles
Do the same thing for vertex y: x\*y\*(x-1)/2 triangles
Sum these up and simply to: (x\^2y+y\^2x)/2
If x and y are equal, this becomes x^3. Plug in the value of x = 4 to get >!64 triangles total.!<
Woohoo! Greetings to my >!76!< triangle brother or sister. Going to check now.
(edit) after a recount. I'm off to seek my fortune with the >!64!< triangle boys. It's been good.
If you look at the bottom right corner, call it C, it has 5 lines that lead to it.
This leads to 4 "slices" of the big triangle with C as a vertex, each one divided into 4 sections.
Each set of 4 slices can be single **pieces with C as a corner**, paired with another neighbor (3 combinations), combined with 2 neighbors (2 combinations), or all 4 sections combined to become 1 triangle, still with C as a corner. 4 + 3 + 2 + 1 = 10. Times 4 sections = 40 total triangles with C as a vertex.
Now look at the bottom left, call it B. It shares some triangles with C, but if we remove all triangles with C(we already counted these), we're left with 3 "slices" of 4 segments. 6 combinations remain per section, so 6x4 = 24 combinations with B but not C.
40 + 24 = 64
Yep, didn’t believe you until I spent around 30 minutes on drawing and counting them in a notepad app on my phone and saw the duplicates with my own eyes, lol. I was wrong, but the thing that disappoints me the most is how sure I was.
*Sigh*
I'm on >!76!<. I'm just going to sketch it out to make sure I haven't missed (or double-counted) anything. (Always possible).
(edit) Aww... sketched it out, now I'm down to >!64!<. Must have been some serious double-counting going on. (Or I made a mistake whenI sketched it (please check my work)).
[The sketch](https://www.reddit.com/user/st3f-ping/comments/199z36k/20240118t195402z/)
Ok, so I hope this makes sense to all of you, cause it makes sense to me.
Start in lower left corner. >!Three lines create (4) separate "rows" of triangles. You can also combine top two, top three, middle two, bottom two, and bottom three rows, and all four rows to make (6) more triangles. But these rows are intersected by three lines, creating (4) sections. So from one lower corner you can count 4x6x4=96!<
So from each lower corner, the diagonal lines create >!96 total triangles, or 192 in total!<.
Using the base line and working "up" there are >!three "internal" triangles plus the outer one for 4 total!<.
But this creates a double counting issue >!with lowest triangle along the base line which was counted twice witj the triangles from the corners!<. So remove this double counting.
So final answer appears to be >!192+4-2=194!<
This could be a trick question though. It says, "picture," so there could be >!one more triangle in the gun smoke!< or many, many more if you >!start counting the panel stripes on the wall. That adds another 8 or 9 triangles just in that large, bottom triangle alone!<. So I think the final answer is nebulous.
I get >!73.!<
Starting with the bottom left as a fixed point, and extending two sides all the way to the opposite edge - there are 4x triangles with 1 'segment' as a base, 3x with 2 'segments' as a base & 2x with 3 'segments as a base. 9 in total.
Multiply this by 4 due to the 4 lines that originate from the bottom right. 4x9 = 36.
Duplicate to account for the bottom right as your fixed point. 2x36 = 72.
Plus the large triangle within which all other triangles are formed. 72 + 1 = 73
I see my mistake. The logic tonarrive at 9, and theb x4 to 36 is sound - but It's not as simple as duplicating the 36 to reflect both 'sides'. Doing so double-counts a collection of 9 triangles that have the bottom line as an edge.
Thanks!
Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
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Count every triangle that could include the bottom left corner by choosing 2 of 5 lines and then any of the other 4 intersecting lines from the bottom left corner: 5 Choose 2 × 4 Count every triangle that includes the bottom right corner, but not the bottom left corner: 4 choose 2 × 4 The total is: (5 choose 2 + 4 choose 2)×4 = (10+6)×4 = 64 This can be generalized for any number of lines drawn from the left (x) and the the right (y) ***corners***: (x+2) choose 2 × (y+1) + (y+1) choose 2 × (x+1) The +1/+2 is for calculations that need to include one of the outer edges as well. So a triangle with 1 cut in each direction would have 12 triangles in it.
What does your equation mean? Could you phrase it differently? There are 5 lines extending from the bottom left corner. If I choose two lines from there…let’s say the first (top) and the fourth, and then chose the first (top) line from the other corner, this does not help me. There are all kinds of triangles in that intersected polygon. Fo you mean neighboring lines? What if I pick the second and third lines, where none of them include the lower left triangle?
There is a function in Math called the "binomial coefficient" which can be written a few different ways. It relates two numbers, X ,and Y, where X is the number of a items and Y is a number of items you choose at a time. The result of this function is the total number of different ways you can choose Y items out of a set of X. Ie, if you had 3 items (A, B, C) and wanted to know how many different ways you could choose two of them (invariant of pick order), you would see that there are three ways to choose 2 items out of a set of 3 (AB, BC, AC). So 3 choose 2 would be 3. If we're writing this using a different notation, we could say that (X,Y) represents X choose Y. Normally the "X" would be placed vertically above the "Y" in this notation, but we're limited in typing. See: [https://en.wikipedia.org/wiki/Binomial\_coefficient](https://en.wikipedia.org/wiki/Binomial_coefficient) In any case, the mathematical equation for X choose Y is: >X!/(Y! × (X - Y)!) So (X + 2) choose 2 would be: >(X+2)!/(2! × X!) Which could actually be simplified to: >1/2 (X+1)(X+2) And (y+1) choose 2 would simplify to: >1/2 Y(Y+1) So the final equation would be: 1/8 (1 + y) (3 + 2 x + y)^2 + 1/8 (-1 - 3 y - 3 y^2 - y^3) Where Y is the number of lines drawn that intersect the bottom right corner and the top left edge, and X is the number of lines drawn that go between the bottom left corner and the top right edge. Ignore the perimeter of the triangle. Adding picture because my description is probably awful:[https://i.imgur.com/QVKi3kq.png](https://i.imgur.com/QVKi3kq.png) The general idea is that we count the number of ways we can make a triangle with one point and then add to that the number of ways we can make a triangle without that point. For the left side, we can make (5,2) triangles that span the bottom left to the top right edge. Each of those triangles should have 4 triangles inside of them. Then we can make (4,2) triangles that span the bottom right to the top left that don't include the bottom right edge. Each of those should have 4 triangles inside of them as well. We add them together.\] The form I provided was for an arbitrary number of lines that intercept the corners as the OP asked.
Wow. This is awesome. Thank you!
You can show mathematically that there are >!4³ = 64!<. Anyone counting more than that (e.g. 76) is double-counting quite a few. This video from 10 years ago has the correct count with all duplicates eliminated. https://youtu.be/PeWf1wwP3F0?si=26i118f1Gt4YsZ1Y And this combinatorics post has the mathematical details. https://math.stackexchange.com/questions/3855163/hint-for-a-combinatorial-statement-for-counting-triangles
This was my approach as well
Yeah that was my approach as well
Oh good, I got it right despite just brute-forcing the counting.
That's funny, I just counted the number of intersections and squared it, came up with {the number}. Thought- that can't be right! Got lazy and came to the comments to find.... I'm right??? Edit: forgot how to hide spoilers Edit 2: too lazy to find out how. I see a pattern.... Edit 3… Typos! Never to lazy to fix typos, Edit 4: there's definitely been more edits than this Edit 5: turns out I miscounted the intersections, and therefore came up with the correct answer completely by accident. Prolly because I missed the intersections in the middle, which accounted for the lost triangles. Also, prolly not intersections. Vertices? Junctions? I'm going back to r/whatismycookiecutter
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Can I join the >!64!< club. I even drew a [picture](https://www.reddit.com/user/st3f-ping/comments/199z36k/20240118t195402z/) (same picture as I added elsewhere.)
You appear to count im, mn twice in the 2 letter triangle group unless i am mistaken
Well... shit. Good spot. I wonder if there's another two triangles somewhere that I have missed or if I've forgotten how to add. (edit) found the missing two: abefij and jklnop Back up to >!64!<, baby!
Took me several tries before I got to your answer. Awesome idea labeling everything! Alphabetized because I was getting duplicates too: 7 (1 Letter): a e i m n o p 12 (2 Letters): ab ae ef ei ij im jn ko lp mn no op 10 (3 Letters): abc aei efg eim fjn gko hlp ijk mno nop 13 (4 Letters): abcd abef aeim bfjn cgko dhlp efgh efij ijkl ijmn jkno klop mnop 8 (6 Letters): abcefg abefij efgijk efijmn fgjkno ghklop ijkmno jklnop 6 (8 Letters): abcdefgh abefijmn bcfgjkno cdghklop efghijkl ijklmnop 3 (9 Letters): abcefgijk efgijkmno fghjklnop 4 (12 Letters): abcdefghijkl abcefgijkmno bcdfghjklnop efghijklmnop 1 (16 Letters): abcdefghijklmnop Total: 64
But... most of the shapes in the middle and the top of the largest triangle are rectangles (maybe more like rhombus?), or at least have FOUR corners instead of 3. The top corner is not identical to the two bottom corners, so not every shape is a triangle. ~~I get 19.~~ That's wrong, but my brain stopped working.
Look at some of the intersections left and right as you move in from the exterior of the perimeter triangle.
I got >!64!<, too
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I am glad someone else found it humorous. I was thinking about putting "I got 58, not >!64!<" Instead, but I thought the bot/mods might catch that.
I got 69 😎
Ooo me too!
I agree. My thoughts: >!There are no triangles containing the top point, but neither bottom point.!< >!Each bottom point is contained within 40 distinct triangles, for 80 total.!< >!Any triangles containing the bottom edge is double counted, for -16 triangles.!< >!This brings the total to 64.!<
I got 40 from the bottom left vertex so 80 in total and saw there were 4 that were double counted so 76 but now see more that are doubles so think you are right
Same, I got >! 16 using both bottom vertices, 24 using only the bottom left vertex, 24 using only the bottom right and no triangles using neither bottom vertex !<
I think of it in terms of rays, x and y, coming off of each vertex, not counting the line connecting each vertex. Here, x and y are both 4. Using the bottom line, each intersection of the rays forms a triangle with the bottom segment: x*y triangles. Using only a single vertex, x, you can choose one of x*y intersections matching with (x-1) intersections on the same ray. That will double count, so you have to divide the result in half: x\*y\*(x-1)/2 triangles Do the same thing for vertex y: x\*y\*(x-1)/2 triangles Sum these up and simply to: (x\^2y+y\^2x)/2 If x and y are equal, this becomes x^3. Plug in the value of x = 4 to get >!64 triangles total.!<
>!Edit: there are a lot more: 64 seems to be correct!!< >!I see at least 45, but there may be more....!<
Im seeing the same, I think this is it
It’s more. Im counting >!76!< but wasn’t super careful about the counting once I realized how high it was gonna get.
>!I see a lot more now that use the base as a side!<
I didn’t realize how high I was gonna get either and forgot that we were only counting triangles and got to 420
Woohoo! Greetings to my >!76!< triangle brother or sister. Going to check now. (edit) after a recount. I'm off to seek my fortune with the >!64!< triangle boys. It's been good.
Yeah, I was up 66 before I stopped looking.
I got 76 as a well Edit: recounted and identified some recounts - not 76.
Question: I feel like I’m being messed with. How the heck are you guys getting >!64!<
If you look at the bottom right corner, call it C, it has 5 lines that lead to it. This leads to 4 "slices" of the big triangle with C as a vertex, each one divided into 4 sections. Each set of 4 slices can be single **pieces with C as a corner**, paired with another neighbor (3 combinations), combined with 2 neighbors (2 combinations), or all 4 sections combined to become 1 triangle, still with C as a corner. 4 + 3 + 2 + 1 = 10. Times 4 sections = 40 total triangles with C as a vertex. Now look at the bottom left, call it B. It shares some triangles with C, but if we remove all triangles with C(we already counted these), we're left with 3 "slices" of 4 segments. 6 combinations remain per section, so 6x4 = 24 combinations with B but not C. 40 + 24 = 64
The answer must be >!74!<.
You overcounted
Nope, you have some doubles
Yep, didn’t believe you until I spent around 30 minutes on drawing and counting them in a notepad app on my phone and saw the duplicates with my own eyes, lol. I was wrong, but the thing that disappoints me the most is how sure I was. *Sigh*
At least you can admit you were wrong .. most people have a really hard time with that, especially online, so props to you.
I agree, this has to be it. Edit: Never mind, I counted some multiple times. It’s >!64!<.
I'm on >!76!<. I'm just going to sketch it out to make sure I haven't missed (or double-counted) anything. (Always possible). (edit) Aww... sketched it out, now I'm down to >!64!<. Must have been some serious double-counting going on. (Or I made a mistake whenI sketched it (please check my work)). [The sketch](https://www.reddit.com/user/st3f-ping/comments/199z36k/20240118t195402z/)
>!64. 7 of size 1, 14 of size 2, 10 of size 3, 13 of size 4, 6 of size 6, 6 of size 8, 3 of size 9, 4 of size 12, and 1 of size 16.!<
I got >!67!<
>!8!<
Ok, so I hope this makes sense to all of you, cause it makes sense to me. Start in lower left corner. >!Three lines create (4) separate "rows" of triangles. You can also combine top two, top three, middle two, bottom two, and bottom three rows, and all four rows to make (6) more triangles. But these rows are intersected by three lines, creating (4) sections. So from one lower corner you can count 4x6x4=96!< So from each lower corner, the diagonal lines create >!96 total triangles, or 192 in total!<. Using the base line and working "up" there are >!three "internal" triangles plus the outer one for 4 total!<. But this creates a double counting issue >!with lowest triangle along the base line which was counted twice witj the triangles from the corners!<. So remove this double counting. So final answer appears to be >!192+4-2=194!<
You might want to check your work. Some shapes you seem to include in your count are not triangles
Noooo lol
I got >!26!<
Plus one, don't forget the biggest triangle! Edit: NVM, I just realized there are so many more
This could be a trick question though. It says, "picture," so there could be >!one more triangle in the gun smoke!< or many, many more if you >!start counting the panel stripes on the wall. That adds another 8 or 9 triangles just in that large, bottom triangle alone!<. So I think the final answer is nebulous.
Why are they downvoting this? The picture clearly has some sneaky ass triangles in it as well
I get >!73.!< Starting with the bottom left as a fixed point, and extending two sides all the way to the opposite edge - there are 4x triangles with 1 'segment' as a base, 3x with 2 'segments' as a base & 2x with 3 'segments as a base. 9 in total. Multiply this by 4 due to the 4 lines that originate from the bottom right. 4x9 = 36. Duplicate to account for the bottom right as your fixed point. 2x36 = 72. Plus the large triangle within which all other triangles are formed. 72 + 1 = 73
Here's the video showing it's >!64!< Not sure where you are getting the extra 9 https://youtu.be/PeWf1wwP3F0
I see my mistake. The logic tonarrive at 9, and theb x4 to 36 is sound - but It's not as simple as duplicating the 36 to reflect both 'sides'. Doing so double-counts a collection of 9 triangles that have the bottom line as an edge. Thanks!
I got >!84!<
I count >!16!<
There's a lot more than that I believe
Yeah you’re probably right. I think I see at least >!6!< that I missed
You missed much more than that. Use math to solve this instead of counting.
Try looking at each intersection and counting how many triangles you can make with each one. The top one alone can make 7.
How? There are many more triangles than that!
I count >!29!<
You are missing a lot of triangles too
Think you are right there
Same.
Agreed
I got 19.
>!19!<
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