Geometry From Hell












5












$begingroup$


You’re locked in a room with nothing but a pencil, a math compass, and paper. You do not have a straightedge. Your captors have informed you that you cannot leave until you construct (the endpoints of) a line segment of length $sqrt{7}$.



What do you do?






mathematics geometry construction







share|improve this question











$endgroup$








  • 2




    $begingroup$
    Is this a puzzle you created, and do you know of a solution?
    $endgroup$
    – Hugh
    23 hours ago






  • 9




    $begingroup$
    Presumably you have access to something of length 1?
    $endgroup$
    – Dr Xorile
    22 hours ago






  • 2




    $begingroup$
    I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
    $endgroup$
    – Bass
    17 hours ago






  • 1




    $begingroup$
    Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
    $endgroup$
    – Chris Cudmore
    13 hours ago
















5












$begingroup$


You’re locked in a room with nothing but a pencil, a math compass, and paper. You do not have a straightedge. Your captors have informed you that you cannot leave until you construct (the endpoints of) a line segment of length $sqrt{7}$.



What do you do?






mathematics geometry construction







share|improve this question











$endgroup$








  • 2




    $begingroup$
    Is this a puzzle you created, and do you know of a solution?
    $endgroup$
    – Hugh
    23 hours ago






  • 9




    $begingroup$
    Presumably you have access to something of length 1?
    $endgroup$
    – Dr Xorile
    22 hours ago






  • 2




    $begingroup$
    I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
    $endgroup$
    – Bass
    17 hours ago






  • 1




    $begingroup$
    Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
    $endgroup$
    – Chris Cudmore
    13 hours ago














5












5








5





$begingroup$


You’re locked in a room with nothing but a pencil, a math compass, and paper. You do not have a straightedge. Your captors have informed you that you cannot leave until you construct (the endpoints of) a line segment of length $sqrt{7}$.



What do you do?






mathematics geometry construction







share|improve this question











$endgroup$




You’re locked in a room with nothing but a pencil, a math compass, and paper. You do not have a straightedge. Your captors have informed you that you cannot leave until you construct (the endpoints of) a line segment of length $sqrt{7}$.



What do you do?






mathematics geometry construction




mathematics geometry construction






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 23 hours ago









Hugh

2,26811127




2,26811127










asked 23 hours ago









Dirge of DreamsDirge of Dreams

41317




41317








  • 2




    $begingroup$
    Is this a puzzle you created, and do you know of a solution?
    $endgroup$
    – Hugh
    23 hours ago






  • 9




    $begingroup$
    Presumably you have access to something of length 1?
    $endgroup$
    – Dr Xorile
    22 hours ago






  • 2




    $begingroup$
    I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
    $endgroup$
    – Bass
    17 hours ago






  • 1




    $begingroup$
    Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
    $endgroup$
    – Chris Cudmore
    13 hours ago














  • 2




    $begingroup$
    Is this a puzzle you created, and do you know of a solution?
    $endgroup$
    – Hugh
    23 hours ago






  • 9




    $begingroup$
    Presumably you have access to something of length 1?
    $endgroup$
    – Dr Xorile
    22 hours ago






  • 2




    $begingroup$
    I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
    $endgroup$
    – Bass
    17 hours ago






  • 1




    $begingroup$
    Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
    $endgroup$
    – Chris Cudmore
    13 hours ago








2




2




$begingroup$
Is this a puzzle you created, and do you know of a solution?
$endgroup$
– Hugh
23 hours ago




$begingroup$
Is this a puzzle you created, and do you know of a solution?
$endgroup$
– Hugh
23 hours ago




9




9




$begingroup$
Presumably you have access to something of length 1?
$endgroup$
– Dr Xorile
22 hours ago




$begingroup$
Presumably you have access to something of length 1?
$endgroup$
– Dr Xorile
22 hours ago




2




2




$begingroup$
I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
$endgroup$
– Bass
17 hours ago




$begingroup$
I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds.
$endgroup$
– Bass
17 hours ago




1




1




$begingroup$
Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
$endgroup$
– Chris Cudmore
13 hours ago




$begingroup$
Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit?
$endgroup$
– Chris Cudmore
13 hours ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Suppose you can set your pair of compasses to length 1. Then




Then draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.




Then we know that the length of the line that would join those two points is




$sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.




The simplest construction would us:




Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $sqrt{7}$. construction







share|improve this answer











$endgroup$





















    4












    $begingroup$

    Edited: added a drawing for the first step.

    Edited again: Dr Xorile very clever solution eliminates the second step.




    You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

    enter image description here

    The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml



    In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $sqrt{7}$







    share|improve this answer











    $endgroup$













    • $begingroup$
      Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
      $endgroup$
      – ppgdev
      21 hours ago











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    7












    $begingroup$

    Suppose you can set your pair of compasses to length 1. Then




    Then draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.




    Then we know that the length of the line that would join those two points is




    $sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.




    The simplest construction would us:




    Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $sqrt{7}$. construction







    share|improve this answer











    $endgroup$


















      7












      $begingroup$

      Suppose you can set your pair of compasses to length 1. Then




      Then draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.




      Then we know that the length of the line that would join those two points is




      $sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.




      The simplest construction would us:




      Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $sqrt{7}$. construction







      share|improve this answer











      $endgroup$
















        7












        7








        7





        $begingroup$

        Suppose you can set your pair of compasses to length 1. Then




        Then draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.




        Then we know that the length of the line that would join those two points is




        $sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.




        The simplest construction would us:




        Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $sqrt{7}$. construction







        share|improve this answer











        $endgroup$



        Suppose you can set your pair of compasses to length 1. Then




        Then draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.




        Then we know that the length of the line that would join those two points is




        $sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.




        The simplest construction would us:




        Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $sqrt{7}$. construction








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 9 hours ago

























        answered 22 hours ago









        Dr XorileDr Xorile

        13.5k32572




        13.5k32572























            4












            $begingroup$

            Edited: added a drawing for the first step.

            Edited again: Dr Xorile very clever solution eliminates the second step.




            You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

            enter image description here

            The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml



            In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $sqrt{7}$







            share|improve this answer











            $endgroup$













            • $begingroup$
              Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
              $endgroup$
              – ppgdev
              21 hours ago
















            4












            $begingroup$

            Edited: added a drawing for the first step.

            Edited again: Dr Xorile very clever solution eliminates the second step.




            You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

            enter image description here

            The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml



            In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $sqrt{7}$







            share|improve this answer











            $endgroup$













            • $begingroup$
              Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
              $endgroup$
              – ppgdev
              21 hours ago














            4












            4








            4





            $begingroup$

            Edited: added a drawing for the first step.

            Edited again: Dr Xorile very clever solution eliminates the second step.




            You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

            enter image description here

            The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml



            In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $sqrt{7}$







            share|improve this answer











            $endgroup$



            Edited: added a drawing for the first step.

            Edited again: Dr Xorile very clever solution eliminates the second step.




            You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

            enter image description here

            The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml



            In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $sqrt{7}$








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 21 hours ago

























            answered 22 hours ago









            ppgdevppgdev

            1215




            1215












            • $begingroup$
              Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
              $endgroup$
              – ppgdev
              21 hours ago


















            • $begingroup$
              Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
              $endgroup$
              – ppgdev
              21 hours ago
















            $begingroup$
            Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
            $endgroup$
            – ppgdev
            21 hours ago




            $begingroup$
            Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution!
            $endgroup$
            – ppgdev
            21 hours ago


















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