This is a puzzle from one popular book called "The Man Who Counted: A Collection of Mathematical Adventures",author is Malba Tahan. How to arrange ten soldiers in five lines in such a way that each line contains four soldiers exactly?
Ten soldiers puzzle
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1http://mathoverflow.net/questions/74673/ten-soldiers-puzzle-closed – 2011-09-07
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0One could add another constraint and get the same solution: Each soldier stands in exactly two of the five lines. – 2011-09-07
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1Try this one: (1) Arrange ten soldiers into ten lines in such a way that each soldier is in three of the ten lines and each line contains three soldiers; (2) Do this in two different ways that are not incidence-isomorphic to each other. (Lots of people know an answer to #1. But #2 is also possible!) – 2011-09-07
4 Answers
Like this:
$\hskip1.7in$
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0Nevermind that any two points define a line... I've never heard this one before so just sayin'. – 2011-09-07
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0@Dan: The problem is using the word "line" in a colloquial sense; but regardless, collinear points all define the same line (by definition). So I'm afraid I don't understand your comment. – 2011-09-07
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1e.g., the line defined by the left-most and top-most points contains only those two points, and also there are more than five "lines". I suppose looking at it that way the problem is unsolvable. What is a more precise way to state it? Or am I being overly pedantic? – 2011-09-07
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0Ah, I see. It might be mildly pedantic :) but regardless it's still good to think about the issue. A precise way of stating the problem might be: Find 5 distinct lines in the plane such that the intersection of any of them with the union of the others contains exactly 4 points. – 2011-09-07
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6@Dan "How to arrange ten soldiers in such a way that there exist five lines each containing exactly four soldiers?" – 2011-09-07
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0Now how to prove the solution is unique? Any reason to jump out of the plane? – 2011-09-07
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0@Dan: I think the hard part of answering your question is figuring out, unique *up to what*? We could always adjust any subset of the lines by a small amount and still get a solution. We want to capture the idea that any pentagram should count as "the same" solution but I am not sure how to do that formally. – 2011-09-07
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0Zev, point taken! I was thinking of unique up to scaling in the plane. If it's not Euclidean then certainly all is lost! – 2011-09-07
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0Also I see any linear transformation works (not sure how to adjust any subset yet). What is the most interesting version of this problem? – 2011-09-07
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2You can do more than linear transformation. I think you can arrange the five outer vertices into any convex pentagon and you'll end up with a solution. Even more generally, perhaps *every* collection of five non-parallel lines gives a solution, as long as no three lines intersect at a single point. – 2011-09-07
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0[pentagram and golden ratio](http://mathworld.wolfram.com/Pentagram.html) – 2011-09-11
This is an alternative (sorry diagram is clunky)
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0,I am wondering whether we can find golden ratio in this configuration as it is found in pentagram – 2011-09-11
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1No - you can move the vertical line left or right, or rotate it. The central point of the other four lines can be moved left or right, even keeping a horizontal axis of symmetry. The angle at the arrowhead is not fixed either, All these change the ratios. So you can make the Golden Ratio happen, but it is not inherent in the arrangement. – 2011-09-11
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0[pentagram and golden ratio](http://mathworld.wolfram.com/Pentagram.html) – 2011-09-11
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1@pedja: The golden ratio is not inherent in the pentagrammatic solution to the puzzle either. It appears in what one might call a regular pentagram, but [a pentagram that is not regular](http://en.wikipedia.org/wiki/File:Haykal2.gif) still solves the puzzle despite having little to do with the golden ratio. – 2011-09-11
5 lines times 4 soldiers on a line equals 20 = two times 10 soldiers available. This suggests that every soldier belongs to two lines.
Draw $n$ lines on the plane such that no two are parallel, and no three intersect in one point.
You can always do that: if you already have $n-1$ lines then there are finite number of slopes of those lines and finite number of points of intersection -- choose a new slope not equal to any previous and draw the line with this slope not going through any previous points of intersection.
Each line contains exactly $n-1$ points of intersection with other lines.
There are $\frac{n(n-1)}{2}$ intersection points in total.
Now, if you put a soldier at every point of intersection, then there are $\frac{n(n-1)}{2}$ soldiers arranged in $n$ lines, each containing $n-1$ soldier. For $n=5$ you get the answer: any such configuration of 5 lines would work.
A more irregular looking solution.
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0This is the same as Mark Bennett's solution, just drawn with less symmetry. You could do the same with Zev Chonoles's pentagram: http://i.stack.imgur.com/h7rtU.png – 2012-11-10
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0@RahulNarain cool! I was gonna comment that my my picture might be essentially identical to one or more of the others and then I wasn't sure in what sense I mean't identical.. maybe in the sense of being the same graph or maybe in some topological sense..and then I was like ok this is taking too much time and I'll just let someone else point that out. :) So i'm guessing you mean as a graph? – 2012-11-11
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0Well, I don't know in what precise sense they are the same either. All I meant is that the figure has been rotated 90 degrees and the lines have been moved around a little, but the structure is the same. One could generate a unlimited number of answers that way. – 2012-11-11