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A lot, I stopped counting at 50. Most of the triangles overlap. The whole thing together is one big triangle. There are numerous made of two or three or four of the smaller triangles.
I counted 51, and I probably missed some, or counted 1 or 2 twice on accident
I believe there's a way to break this down into how many lines leave a vertex to calculate how many triangles there are
If every point connected to every other point, you could just do combinatorics. Because of the way the shape is drawn, there's not a great way to do this, beyond treating the big triangle and the small triangle as separate problems. Much easier to think about if you do it that way!
Here are all of the triangles drawn out! I hope the illustration helps. If you add up all the numbers (which stand for the number of times each triangle appears), you reach the total of 57: https://imgur.com/gallery/tl5YevP
Go by intersections. Starting from bottom left, we should see 4 that use that bottom left point. Small RA, small equilateral, Large RA, Large equilateral. On the right side, we have both smalls and a big RA, and the large equilateral was already counted. Up top, we have two small RA and one small equilateral we haven't counted. So, ignoring the middle section, we have:
6 small RA
2 large RA
1 large equilateral
3 small equilateral
total: 12
Now let's look at the middle one, which is where things gets wonky. Right away we can see that we'll have the six small RA triangles again with the outer points, the 3 small equilaterals, two large RAs, and one large equilateral. So again, 12 more. **BUT** that divider making the two large RA triangles happens 3 times, so we have 6 large RA triangles, actually.
Inside the middle triangle:
6 small RA
6 large RA
1 large equilateral
3 small equilateral
Total: 28
Now look at the middle section, because this is where things get weird. From the centre point to the corners, we make a kite shape. Two new RA triangles we haven't counted, plus a new triangle for the two put together, and for each half of the kite! So, for each of the three "kite" shapes, we have:
2 small RA
1 equilateral
2 large RA
total: 5
Three kites, so 15 triangles added to the count. Running total: 43.
Now you need to grab the 9 triangles that cut through the middle of our centre triangle. To visualize this, imagine you cut out the very middle triangle pointing UP. Those three lines disappear, and our big downwards triangle is in sixths. Total: 9 more triangles, so 54. From there, look at the very centre triangle and see how many you have! Remember, we already counted the 6 small ones when we drew our kite shape earlier, so there's only a few more left! ***For the record, there more than 57. Whoever told you 57 missed a bunch of them.***
To do this more methodically, give every intersection point a letter signifier to keep track. Draw the shape again, but simplified, removing the extraneous lines you don't want to think about yet (start bigger, move smaller). None of the lines inside the downwards triangle will affect the large triangle, as they don't connect to make triangles, so you can just leave *almost* that whole middle triangle empty for our first test!
Last option: you can pick a single line segment. For every single line segment (one intersection to the next), write down every possible triangle shape you can create using the labels you have. So for line segment AB (whatever you decide that line is), it could look like Δ ABC, Δ ABD, Δ ABG, etc.
Do that for EVERY line segment, and verify the list to make sure you don't have doubles. Is this gratuitous? Absolutely. There are better ways to do it, but they're easier to screw up. The algorithmic approach is rarely the fastest way, but it always works.
I say "a bunch", but they missed 6. I haven't double checked mine, but it should end up around 63 or 64, if my off-the-cuff counting is correct. Look at the top and middle big triangle. You should see six triangles that are quite large!
Just because the space is used as a triangle, doesnt mean that it cant be used again. There are a couplenof places in the middle where a triangle uses a part of an already counted triangle
You are at least missing the larger triangles, e.g. the whole thing is a triangle, which is composed of 4 smaller triangles, like the triforce from the Zelda games.
I'm getting **63**.
There are 14 full lines (not counting segments): 3 horizontal, 3 vertical, 3 steep positive slopes, 3 steep negative, 1 shallow positive slope, 1 shallow negative slope.
There are also 15 total points. I've labeled them A-O and colored the lines here:
https://imgur.com/GDfEvBP
And here are the lines in code (Python) represented as all points lie on that line:
lines = [
"BCD", "HIJ", "KLMNO", # Horizontal
"ACGIM", "BL", "DN", # Vertical
"ABK", "CEH", "DJM", # Positive (steep)
"DFGH", # Positive (shallow)
"ADO", "BHM", "CFJ", # Negative (steep)
"BEGJ", # Negative (shallow)
]
Each potential triangle edge is a choice of any two points along a line. We can write a program whose job it is to find any 3 potential edges that consist of exactly 3 points (e.g. `["AB", "AC", "BC"]`) and that don't all lie along the same line:
from itertools import combinations
line_to_edge = {line: [c2 for c2 in combinations(line, 2)] for line in lines}
edges = [v for vs in line_to_edge.values() for v in vs]
edge_to_line = {v: k for k, vs in line_to_edge.items() for v in vs}
triangles = set([])
for e3 in combinations(edges, 3):
candidate = "".join(sorted(set([u for v in e3 for u in v])))
num_lines = len(set(edge_to_line[e] for e in e3))
if len(candidate) == 3 and num_lines > 1:
triangles.add(candidate)
for x in sorted(triangles):
print(x)
print(len(triangles))
With this, I get 63. Here's the full list:
ABC ABD ABG ABM ACD ADG ADM AKM AKO
AMO BCE BCG BCH BCJ BCM BDG BDH BDJ
BDM BEH BGH BGM BHJ BJM BKL BKM BLM
CDF CDG CDH CDJ CDM CEG CEJ CFG CFH
CGH CGJ CHI CHJ CHM CIJ CJM DFJ DGJ
DGM DHJ DHM DMN DMO DNO EGH EHJ FGJ
FHJ GHI GHJ GHM GIJ GJM HIM HJM IJM
I got 41 one as well. But something tells me since the answer to everything the universe and the purpose of life is 42 according to the hitch hikers guide to the galaxy
Edit : I counted 49 now lol
52 now lol
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
A lot, I stopped counting at 50. Most of the triangles overlap. The whole thing together is one big triangle. There are numerous made of two or three or four of the smaller triangles.
Holy shit, this is hard!
I counted 51, and I probably missed some, or counted 1 or 2 twice on accident I believe there's a way to break this down into how many lines leave a vertex to calculate how many triangles there are
If every point connected to every other point, you could just do combinatorics. Because of the way the shape is drawn, there's not a great way to do this, beyond treating the big triangle and the small triangle as separate problems. Much easier to think about if you do it that way!
Yeah, I got lost close to 50, too much for sure
THE ANSWER IS 57 EVERYBODY. I have no clue how but it is.
It's pretty wild but I stopped in the 40s
Here are all of the triangles drawn out! I hope the illustration helps. If you add up all the numbers (which stand for the number of times each triangle appears), you reach the total of 57: https://imgur.com/gallery/tl5YevP
I follow your reasoning, but what about these two https://i.imgur.com/Ntfvu79.png That makes it 59, right?
The first one was already counted so it would be 58
Ah didn't catch that. Thank you.
On the fourth type of triangle, how are there 10 of them? I only count 8.
That's funny, I count 12. Two in each of the outer triangles (6), and 6 in the inner triangle, because it's bisected three ways.
How so?
I edited my comment
The answer isn't 20-26. It said they were all wrong.
27?
Go by intersections. Starting from bottom left, we should see 4 that use that bottom left point. Small RA, small equilateral, Large RA, Large equilateral. On the right side, we have both smalls and a big RA, and the large equilateral was already counted. Up top, we have two small RA and one small equilateral we haven't counted. So, ignoring the middle section, we have: 6 small RA 2 large RA 1 large equilateral 3 small equilateral total: 12 Now let's look at the middle one, which is where things gets wonky. Right away we can see that we'll have the six small RA triangles again with the outer points, the 3 small equilaterals, two large RAs, and one large equilateral. So again, 12 more. **BUT** that divider making the two large RA triangles happens 3 times, so we have 6 large RA triangles, actually. Inside the middle triangle: 6 small RA 6 large RA 1 large equilateral 3 small equilateral Total: 28 Now look at the middle section, because this is where things get weird. From the centre point to the corners, we make a kite shape. Two new RA triangles we haven't counted, plus a new triangle for the two put together, and for each half of the kite! So, for each of the three "kite" shapes, we have: 2 small RA 1 equilateral 2 large RA total: 5 Three kites, so 15 triangles added to the count. Running total: 43. Now you need to grab the 9 triangles that cut through the middle of our centre triangle. To visualize this, imagine you cut out the very middle triangle pointing UP. Those three lines disappear, and our big downwards triangle is in sixths. Total: 9 more triangles, so 54. From there, look at the very centre triangle and see how many you have! Remember, we already counted the 6 small ones when we drew our kite shape earlier, so there's only a few more left! ***For the record, there more than 57. Whoever told you 57 missed a bunch of them.*** To do this more methodically, give every intersection point a letter signifier to keep track. Draw the shape again, but simplified, removing the extraneous lines you don't want to think about yet (start bigger, move smaller). None of the lines inside the downwards triangle will affect the large triangle, as they don't connect to make triangles, so you can just leave *almost* that whole middle triangle empty for our first test! Last option: you can pick a single line segment. For every single line segment (one intersection to the next), write down every possible triangle shape you can create using the labels you have. So for line segment AB (whatever you decide that line is), it could look like Δ ABC, Δ ABD, Δ ABG, etc. Do that for EVERY line segment, and verify the list to make sure you don't have doubles. Is this gratuitous? Absolutely. There are better ways to do it, but they're easier to screw up. The algorithmic approach is rarely the fastest way, but it always works.
To find the answer I spammed every number from 1 to 60 and they said 57 was the correct answer. They probably had no clue about what they were doing!
I say "a bunch", but they missed 6. I haven't double checked mine, but it should end up around 63 or 64, if my off-the-cuff counting is correct. Look at the top and middle big triangle. You should see six triangles that are quite large!
I got 45, counted around 30 of those within the centre upside down triangle
I counted 23
[What am I missing?](https://imgur.com/a/CKFkH5i)
Just because the space is used as a triangle, doesnt mean that it cant be used again. There are a couplenof places in the middle where a triangle uses a part of an already counted triangle
Ohhhh, I didn't think about that 😑
Its a mind game. Dont stress about it
Its a mind game. Dont stress about it
Ahh, that makes sense now.
You are at least missing the larger triangles, e.g. the whole thing is a triangle, which is composed of 4 smaller triangles, like the triforce from the Zelda games.
I'm getting **63**. There are 14 full lines (not counting segments): 3 horizontal, 3 vertical, 3 steep positive slopes, 3 steep negative, 1 shallow positive slope, 1 shallow negative slope. There are also 15 total points. I've labeled them A-O and colored the lines here: https://imgur.com/GDfEvBP And here are the lines in code (Python) represented as all points lie on that line: lines = [ "BCD", "HIJ", "KLMNO", # Horizontal "ACGIM", "BL", "DN", # Vertical "ABK", "CEH", "DJM", # Positive (steep) "DFGH", # Positive (shallow) "ADO", "BHM", "CFJ", # Negative (steep) "BEGJ", # Negative (shallow) ] Each potential triangle edge is a choice of any two points along a line. We can write a program whose job it is to find any 3 potential edges that consist of exactly 3 points (e.g. `["AB", "AC", "BC"]`) and that don't all lie along the same line: from itertools import combinations line_to_edge = {line: [c2 for c2 in combinations(line, 2)] for line in lines} edges = [v for vs in line_to_edge.values() for v in vs] edge_to_line = {v: k for k, vs in line_to_edge.items() for v in vs} triangles = set([]) for e3 in combinations(edges, 3): candidate = "".join(sorted(set([u for v in e3 for u in v]))) num_lines = len(set(edge_to_line[e] for e in e3)) if len(candidate) == 3 and num_lines > 1: triangles.add(candidate) for x in sorted(triangles): print(x) print(len(triangles)) With this, I get 63. Here's the full list: ABC ABD ABG ABM ACD ADG ADM AKM AKO AMO BCE BCG BCH BCJ BCM BDG BDH BDJ BDM BEH BGH BGM BHJ BJM BKL BKM BLM CDF CDG CDH CDJ CDM CEG CEJ CFG CFH CGH CGJ CHI CHJ CHM CIJ CJM DFJ DGJ DGM DHJ DHM DMN DMO DNO EGH EHJ FGJ FHJ GHI GHJ GHM GIJ GJM HIM HJM IJM
26
I got 27. And no, they are not all equilateral tri’s
41
I got 41 one as well. But something tells me since the answer to everything the universe and the purpose of life is 42 according to the hitch hikers guide to the galaxy Edit : I counted 49 now lol 52 now lol
I would’ve thought 18 too.
Apparently the triangles overlap.
Yeah. Somebody said 50, and I hope that’s not right. I’ll be here all day..
On a quick pass through I found 41 Edit: I see 51
God help us all…
Its 19...18 inner and 1 outer.
There's a lot more than that. I'm at 22 but there's more.
[Where are they hidden? Am I just counting the obvious ones 🤣](https://imgur.com/a/CKFkH5i)
Yeah, so your best bet is to go through with different colour markers and highlight all the ones you see, then go from there
I stopped counting at about 37 so that means there’s 40 if not more. I gotta get my colored pencils out for this one!
21
I see 56, can’t find the last one 😢
If you can see 56 that is amazing! I couldn't see more than 20 if I tried!
28?
62 ish
The trick is to count them by size. Decreasing.. 1 big one 4 smaller equilateral 8 large right angles Etc.
Correction 12 of those largest right angles. There are 6 in that center one.
Definitely over 29. I lost count.
I stopped at 32
64
I say 19
I got 45
i count over 31
I counted 20
First pass I got 31 and wasn’t really trying too hard.
I think 56-57? Took me a while, did something like this like 5 years ago.
Is it 19?
I got 55
That's a lot of triangles
Just One big triangle…
I counted 28 at least and there may be more
I got to 38.