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Topic: 6 Matches (Read 6775 times) |
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PSesulka
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Take 6 matches out of a match box. Arrange them in a way that you get 6 equilateral triangles.
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SWF
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Re: 6 Matches
« Reply #1 on: Dec 10th, 2002, 6:53pm » |
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Is it too many if they are arranged to give 8 equilateral triangles?
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PSesulka
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Re: 6 Matches
« Reply #2 on: Dec 10th, 2002, 8:44pm » |
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It is supposed to be 6 Equilateral triangles. Are you sure you are using 6 matches? And are you sure they are equilateral? If they are, could you post the answer because I didn't know it was possible to get 8.
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towr
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Re: 6 Matches
« Reply #3 on: Dec 11th, 2002, 12:24am » |
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I'd guess making a david's star works..
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\/ \/
/\ /\
/__\____/__\
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\/
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| « Last Edit: Dec 11th, 2002, 12:25am by towr » |
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aero man
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That is what I was thinking towr, and it answers both SgtAcid and SWF, SgtAcid just didn't notice the two large equilateral triangles formed by three matches each.
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James Fingas
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Re: 6 Matches
« Reply #5 on: Dec 11th, 2002, 12:07pm » |
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I suppose if a David's Star is okay, we might also consider deforming the star to get only six triangles (exercise left to the reader). That is, supposing we want only six triangles...
Edited: I thought I had ten, but I didn't  You'd think I'd be able to count by now.
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| « Last Edit: Dec 11th, 2002, 12:30pm by James Fingas » |
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towr
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Re: 6 Matches
« Reply #6 on: Dec 11th, 2002, 1:41pm » |
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I hadn't thought I had 8 myself, I was just answering the original problem.. But I see how you could count the large traingles as well..
The problem could be considered ambigue. Equilateral could have the scope over all triangles, meaning all sides of all triangles need to be the same. Or every single triangle could be equilateral, meaning the sides of one triangle are all the same, but they can differ from the sides of another triangle..
I'd say the latter is actually the more probable interpretation if you only look at this ambiguity.
The former is more natural in the context of th 'real world', and thus more probable in that sense. (That's also way most people would say a checkerboard has 64 squares, rather than 8*8+7*7+6*6+5*5+4*4+3*3+2*2+1*1)
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| « Last Edit: Dec 11th, 2002, 1:42pm by towr » |
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Garzahd
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Re: 6 Matches
« Reply #7 on: Dec 11th, 2002, 3:17pm » |
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I heard one matchstick puzzle before that went as follows:
Start with 3 matches in an equilateral triangle.
Add 3 more matches such that there are 4 equilateral triangles formed.
Which is, of course, full of ambiguity: Must it be exactly 4? Is it okay if there are extra pieces of matchstick hanging off that don't form a triangle? Must all triangles be congruent? Must the target shape be two-dimensional? (The author's intended answer was a tetrahedron.)
It's a fairly interesting problem if you constrain it to exactly 4 triangles with no hanging bits (and not necessarily congruent triangles).
One solution follows hidden, but I'm pretty sure that others exist:
________
\ . /\
\____/__\
\ / \
\/______\
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SWF
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Re: 6 Matches
« Reply #8 on: Dec 11th, 2002, 8:06pm » |
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I was also thinking of the hexagram/Star of David for 8 triangles, but here is a way to reduce it to exactly 6 (spoiler):
Code:
\ \ /\ /
\ \/ \/
\ /\ /\
\/ \/ \
/\ /\ \
/ \/ \ \
/ /\ \ \
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Also here is another solution Garzahad's problem (add 3 matches to an equilateral triangle of matches to form 4 triangles) (spoiler):
Add the 3 matches such that it forms a tetrahedron.
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towr
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Re: 6 Matches
« Reply #9 on: Dec 11th, 2002, 11:42pm » |
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another way to get exactly six: ________
\ /\ /
\/ \/
/\ /\
/__\/__\
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| « Last Edit: Dec 11th, 2002, 11:42pm by towr » |
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PSesulka
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Re: 6 Matches
« Reply #10 on: Dec 12th, 2002, 4:57am » |
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Isn't a triangle supposed to have three sides?
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towr
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Re: 6 Matches
« Reply #11 on: Dec 12th, 2002, 5:41am » |
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yes, but that doesn't mean it can't overlap with another triangle, or share sides (or part thereof)
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fenomas
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Re: 6 Matches
« Reply #12 on: Dec 16th, 2002, 7:30pm » |
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It's interesting to extend this problem thus: using six matchsticks (of the same length and unbent), for how many integers N can you make exactly N equilateral triangles? (Naturally, shared sides can count for several triangles, and the triangles need not be congruent.)
N=1,2 are trivial. N=3,4 aren't hard, and I think N=6,8 have been posted. I've found solutions for N=5,7 if you allow overhanging matches (whose ends don't align to another match). Can anyone find "cleaner" solutions for 5 or 7 (with no overhangs)?
Further, if 8 is the maximum possible, can anyone think of a reasonably intuitive proof why? I can't, so far, so maybe nine is possible? (Note that if you assume that very small triangles are okay, then you can generalize to infinite length matches. Thus, the more general case would be to find the maximum number of equilateral triangles that can be created by X straight coplanar lines...)
Any good ideas?
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James Fingas
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Re: 6 Matches
« Reply #13 on: Dec 17th, 2002, 10:56am » |
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fenomas,
I have a pretty good idea of why 8 is the maximum number. The reason is pretty simple: to make an equilateral triangle, you use three different directions of matches (0, 120, and 240 degrees). There is no point putting matches at any other angles, because that won't help you make more equilateral triangles. Incidentally, this also includes non-planar angles (so keeping the matches planar is always best).
Now consider this: you need at least one match in each direction (to get any triangles at all). Furthermore, assuming that overhanging ends is okay, then we can just consider all the possible number of matches going in each direction:
4,1,1 (by symmetry, it doesn't matter which direction has the 4 matches)
3,2,1 (ditto)
2,2,2 (this gives the maximum number of triangles)
There are no other combinations, and the actual arrangement of the matches is immaterial, since your best strategy is to put them very close together, allowing as many match intersections as possible. It turns out that for these three cases, the number of triangles you make is the product of the numbers of matches in the three directions. (I haven't yet convinced myself that that result applies when you have more than six matches)
Now, if you don't want the maximum possible number of triangles, then the situation is more complicated ...
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fenomas
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Re: 6 Matches
« Reply #14 on: Dec 17th, 2002, 6:19pm » |
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James-
I had been thinking along those lines too, and I think this can be generalized for n lines. Consider:
As you point out, to make equillateral triangles, every line we place should be coplanar, and oriented at 0, 120, or 240 degrees from horizontal. Call these orientations A, B, and C. To maximize the number of triangles formed, there should be no point where an A line, a B line, and a C meet at the same point.
Now consider- if the lines are infinite, then any three of them (of different orientations) will form a triangle. Thus, the total number of triangles formed is basically the combination of the A, B, and C lines. If you place your lines in the 3,2,1 formation, you have these lines:
a1, a2, a3
b1, b2
c1
So your set of triangles is the combination of those elements:
{ a1b1c1, a1b2c1, a2b1c1, a2b2c1, a3b1c1, a3b2c1 }
This explains why the number of triangles formed is the product of the number of lines placed in each orientation.
Now, to go one step further, consider the case where lines of each orientation meet at a single point (this is the only way, with infinite lines, to get a non-maximum number of triangles, other than by superimposing lines or putting them at orientations other than a,b,c). Then, all that actually happens is that one triangle dissappears. To prove it, move one of the three lines an infinitessimal distance, and the figure is unchanged except that a tiny triangle is created where the intersection point was.
So, for N infinite lines, where A, B, and C are the numbers of them placed at 0, 120, and -120 degrees w.r.t to the horizontal, and X places where lines of all orientations meet at the same point, you will create (A*B*C)-X triangles. Since A+B+C = N, it should be easy to show that A*B*C is maximized when each A,B,C is as close to N/3 as possible.
Does that sound right?
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James Fingas
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Re: 6 Matches
« Reply #15 on: Dec 18th, 2002, 8:46am » |
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fenomas,
That sounds good. I'm wasn't completely convinced about the three-lines-meeting only removing a single triangle, but maybe we can look at it this way:
If three lines meet at a point, that affects our ability to use "any combination of 3 lines of different direction" to make a triangle. Specifically, if we pick three lines that meet at a point, they don't make a triangle. Any two of those lines can still make a triangle with a third line of the third direction, since nothing is unusual (the two lines just meet at a point normally). I think I'm convinced now.
I would also like to add that before you start counting the triangles created from N lines, you should remove all lines that lie on top of another line, since they don't contribute anything.
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BNC
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on Dec 11th, 2002, 3:17pm, Garzahd wrote:I heard one matchstick puzzle before that went as follows:
Start with 3 matches in an equilateral triangle.
Add 3 more matches such that there are 4 equilateral triangles formed.
Which is, of course, full of ambiguity: Must it be exactly 4? Is it okay if there are extra pieces of matchstick hanging off that don't form a triangle? Must all triangles be congruent? Must the target shape be two-dimensional? (The author's intended answer was a tetrahedron.)
It's a fairly interesting problem if you constrain it to exactly 4 triangles with no hanging bits (and not necessarily congruent triangles).
One solution follows hidden, but I'm pretty sure that others exist:
________
\ . /\
\____/__\
\ / \
\/______\
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How 'bout forming a 3D pyramid?
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fenomas
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Re: 6 Matches
« Reply #17 on: Dec 19th, 2002, 6:21am » |
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on Dec 18th, 2002, 8:46am, James Fingas wrote:| I would also like to add that before you start counting the triangles created from N lines, you should remove all lines that lie on top of another line, since they don't contribute anything. |
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Yep.. like I said, the only ways to get a non-maximum number of triangles are lines that all meet at a point, or superimposed lines, though I should have mentioned that out-of-plane lines or lines that aren't oriented to 0,120, or 240 degrees also qualify. But superimposed lines are trivially redundant, and off-angle lines just represent a parallel problem-- if there are lines on 5, 125, and 245 degrees, for example, they're a separate issue. Out of plane lines are the most interesting.. but beyond me. 
So I've been wondering about coplanar lines at any angle, and what happens if you expand to non-equilateral triangles. If the lines are infinite, I think similar logic applies- except that all lines can make triangles with each other, so the number of triangles formed is just nC3, with n lines. The exceptions are: mutually parallel lines, and points where more than two lines meet.
At a point where more than m lines meet (m>2), I believe the number of triangles removed from the total is mC3. To justify this, visualize a figure with such a point. Nudging each of the intersecting lines infinitessimally shouldn't affect the rest of the figure, so disregard all lines in the figure that don't intersect at the point. Move the remaining lines about, and you have mC3 triangles (if what I said earlier is right), and that's how many triangles are removed from the total because of the intersection.
For mutually parallel lines, I'm not as sure.. I guess that really, two parallel lines are kind of redundant, as they make the same set of triangles with all the other lines. So if you have x lines, none of them parallel, that are coplanar with y mutually parallel lines, then if there are no points where more than two lines intersect, I guess you have ((X+1)C3 * y)triangles made? Does that sound right?
So my point is: Consider N randomly placed (but not superimposed), infinitely long coplanar lines, where there are P sets of mutually parallal lines, the first set containing P1 lines, the second P2, lines ... and the last with Pp lines. Also suppose that there are X points where more than two lines intersect, with X1 lines intersecting at the first such point, X2 at the second, ..... and Xx at the last. Then the total number of triangles formed should be: (lord I wish there was a fancy equation dealy that would let me write nested subscripts..)
(N+P-P1-P2-...Pp)C3 * P1 * P2 * .... PP - X1C3 - X2C3 - .... XxC3
Or possibly I'm biting off more than I can chew with this. Any of that sound legal? All I can say at this point is that it seems to satisfy boundary conditions, such as all lines parallel, or all lines intersecting at the same point. Also, this degrades nicely to the case where you only consider equilateral triangles, because P=3, and P1+P2+P3=N, so (ignoring multiline intersections), the number of triangles is
(N+3-(P1+P2+P3))C3 * P1*P2*P3
= (3)C3 * P1*P2*P3
= P1*P2*P3
which is what we found before.
Of course, I've been drinking tonight  so who can say?
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