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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?

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One could add another constraint and get the same solution: Each soldier stands in exactly two of the five lines. –  Michael Hardy Sep 7 '11 at 14:00
Try 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!) –  Michael Hardy Sep 7 '11 at 14:01

4 Answers 4

up vote 18 down vote accepted

Like this:

$\hskip1.7in$ enter image description here

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Nevermind that any two points define a line... I've never heard this one before so just sayin'. –  Dan Brumleve Sep 7 '11 at 5:26
@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. –  Zev Chonoles Sep 7 '11 at 5:30
e.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? –  Dan Brumleve Sep 7 '11 at 5:34
@Dan "How to arrange ten soldiers in such a way that there exist five lines each containing exactly four soldiers?" –  ShreevatsaR Sep 7 '11 at 5:39
You 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. –  Tanner Swett Sep 7 '11 at 17:30

A more irregular looking solution.

enter image description here

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This is the same as Mark Bennett's solution, just drawn with less symmetry. You could do the same with Zev Chonoles's pentagram: i.stack.imgur.com/h7rtU.png –  Rahul Nov 10 '12 at 23:05
@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? –  Amatya Nov 11 '12 at 1:14
Well, 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. –  Rahul Nov 11 '12 at 19:17

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.

  1. Each line contains exactly $n-1$ points of intersection with other lines.

  2. 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.

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Five lines ten points

This is an alternative (sorry diagram is clunky)

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,I am wondering whether we can find golden ratio in this configuration as it is found in pentagram –  pedja Sep 11 '11 at 4:52
No - 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. –  Mark Bennet Sep 11 '11 at 5:17
pentagram and golden ratio –  pedja Sep 11 '11 at 13:07
@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 still solves the puzzle despite having little to do with the golden ratio. –  Rahul Sep 11 '11 at 15:28

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