Facebook puzzle

[Zag]

Beartalon commented on this in Facebook, with the wrong answer (at least, IMHO). I didn't want to air our dirty laundry of a GL'er getting one of these problems wrong, so I brought it over here to discuss.



Beary said 64. I'll reserve my response, but I get a higher number.

Comments?

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In a related note: I constantly get "friend suggestions" on Facebook for people I don't know, but I assume they must be GL'ers because we have 8 or 10 friends in common who are all GL'ers. Not everyone has a close an association between their real name and their screen name as MNO and I, so you have to speak up! If you're on Facebook, join the Grey Labyrinth group there and post what your GL name is so we have a clue!

[Trojan Horse]

I guess I'm an idiot, because I get 64 also.

- 24 that use the lower-left corner but not the lower-right
- 24 that use the lower-right corner but not the lower-left
- 16 that use both the lower-left and the lower-right corners

You're not counting the triangles that appear within the three capital A's, are you?

[Beartalon]

My original, definitely wrong, answer was 46. If 64 is wrong, then wordcross, who posted it on FB is wrong as well. Interesting to single me out.

Now I need to make a diagram to see if I confused myself with all those lines zig-Zagging around

[Zag]

Let's start just with the triangles with O as one corner.

4 have OA and OB sides
4 have OA and OC sides
4 have OA and OD sides
4 have OA and OE sides

4 have OB and OC sides
4 have OB and OD sides
4 have OB and OE sides

4 have OC and OD sides
4 have OC and OE sides

4 have OD and OE sides
___
40 so far

Count the same 40 with A as one corner.

Subtract the 4 that the two sets have in common (those with OA as one side and the top vertex in the middle).

76: final answer

[Zag]

[Beartalon]

There's more overlap than that.

Consider the outer-edge vertex above O, and call it G.

There are 4 triangles with sides OA and GA.
The smallest isoceles triangle OA is a duplicate as you state.
The smallest triangle in the angle AOC is also a duplicate. It's the second smallest triangle for angle OAG.
Similarly the other three triangles you can make with OAG are also duplicates.

All of the triangles you can make with base OA radiating from A are already accounted for in your first list of 40 radiating from O because they all contain the same base segment and intersection vertices.

The only unique triangles not in your original 40 come from triangles that do not include OA.

From G, name the next higher vertexes H, and I.

4 have AG and AH sides
4 have AH and AI sides
4 have AI and AE sides

4 have AG and AI sides
4 have AH and AE sides

4 have AG and AE sides

40 + 24 = 64

[Thok]

[raekuul]

Won't it be an odd number because of OAE?

How many unique three-letter combinations can be made with the letters A-L?

[Thok]

[raekuul]

If you say so...

Anyway, here's a diagram with all relevant vertices marked.



Here's an unshrunk, uncropped version

I still say it'll end up being an odd number, since we're disregarding duplicates.

Although... looking back, it's asking for how many triangles, not how many triangles that aren't mirror images.

[Thok]

[raekuul]

Hmm... still, how are we getting that from eighteen points to work with?

[Thok]

[raekuul]

But the 4x4 grid doesn't work in my mind because no matter which way you slice it there are two extra points to connect from. It's probably me misunderstanding how you're looking at this, but I'm still seeing you saying sixteen points when there are eighteen.

[Thok]

[raekuul]

Ah, that helps me to understand it a bit.

Still a little confused on the final result, though, since I can see triangles being missed by the grid approach. For example and using my grid as the source for point names, Triangle JMC.

Does anyone have a graphical way of explaining the grid approach?

[Thok]

[Zag]

[itisally]

Could someone point out for me the 16 matches. I see a symetrical image that starts each side with 4 triangles and then splits those 4 times, then 3 triangles split 4 times, then 2 triangles split 4 times then 1 set of 4 that is shared. it calculated like this
2[(4*4)+(4*3)+(4*2)]+4 = 76

I felt pretty confident because I counted the same by hand. Have to admit that time constrants held me up on listing all the letter combinations though.

I just want to know why I am wrong really.
_________________
I could agree with you, but then we would both be wrong.

[Thok]

[itisally*]

Ok I see now, thanks

[Zag]

Here's another puzzle from Facebook that I thought was interesting, though it's pretty easy once you do a little legwork.

I see these because I have 'liked' the Worldwide Center of Mathematics.

[raekuul]

The simple answer is that
because 2/9 < 1/4 < 1/3, 1/4 is in the Cantor Set
From the phrasing of the prompt 1/4 should always be in the Cantor Set. Since the Cantor Set is an ad infinitum procedure, however, the question is whether or not 1/4 can be proven to be outside of the Cantor Set given sufficient iterations.

How does one do that, again? Am I even on the right track?

[Zag]

So the Cantor set is what's left after your infinite iterations. Your job is to prove that, no mater how many iterations, the 1/4 point is still there. There are, of course, an infinite number of points that are still there, and 1/4 is, in fact, one of them. But can you prove it?

[raekuul]

That's the part that I don't know how to do. I was never good with sequences and infinitely-iterating processes. My guess is that it has something to do with the geometric something-or-other since it's going 1/3 -> 1/9 -> 1/27 for the step measurement, and that is convergent if I remember correctly.

[Thok]

Hints towards a solutions

Hint 1: Think what happens in base 3.

Hint 2: In base 3, 1/4=.020202020202..... (Proof left to the reader).

[raekuul]

Ahh... that helps me to understand that it is how things are

[Zag]

Really? I know the answer, and that didn't help me in the least.

I'm trying to think of a hint that doesn't just give it away, and I'm not coming up with anything. This is something of a giveaway, but: [Actually go through a couple of iterations and look at where the 1/4 point is on the continuous piece that it is part of.]

[Thok]

[bonanova]

What occurred to me is a simple word proof. The point 1/4 is 1/4 of the way from one end of the original 0-1 interval. Ditto for the 0-1/3 interval after the first cut. By induction for all remaining cuts.
_________________
_{Vidi, vici, veni.}

[Zag]

[bonanova]

[bonanova]

Sounds like "trust me." What initially seemed necessary was that either the fractional distance from one end remained constant or it followed a cycle. No pencil or paper handy. This morning it occurred to me that 3/12 is 1/4 of the way from 4/12, doh.
_________________
_{Vidi, vici, veni.}

[Thok]

[Chuck]

I get 64 in the original problem by numbering (changing the above lettering to numbering) the vertices and listing the vertices in each line of vertices,

[bonanova]

[Thok]

[bonanova]

Ah, I see. I at first thought the base-3 selector scheme identified [only] the interval end-points. But 0.25 dec is 0.020202 ... ternary. Now I'm thinking 0.25 eventually becomes an endpoint. I.e., all points in the set are endpoints of remaining intervals of length [measure] zero. Even tho we thought we were proving that .25 belonged to the set because it was an internal point, there can't be an internal point in any set that has measure zero.

So ... removing the closed middle third intervals would leave the final set empty.
_________________
_{Vidi, vici, veni.}

[Thok]