Is the set of paths between any two points moving only in units on the plane countable or uncountable?












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$begingroup$


Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?



I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?



    I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?



      I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.










      share|cite|improve this question











      $endgroup$




      Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?



      I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.







      geometry elementary-set-theory






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      edited 2 hours ago









      Andrés E. Caicedo

      65.6k8159250




      65.6k8159250










      asked 2 hours ago









      Knight98Knight98

      253




      253






















          5 Answers
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          $begingroup$

          Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$



            Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.



            The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .



            And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.






            share|cite|improve this answer











            $endgroup$





















              2












              $begingroup$

              Go one unit from a at an angle of t to c.

              Go in unit steps along ca until one is less than a unit away from b to a point p.

              If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.



              As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                I believe it is uncountable also, and here is my thought process:



                For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.



                Then if $vneq w$, we have $alpha_vneqalpha_w$.



                But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                  $endgroup$
                  – coffeemath
                  2 hours ago










                • $begingroup$
                  I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                  $endgroup$
                  – Chris Custer
                  1 hour ago










                • $begingroup$
                  Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                  $endgroup$
                  – coffeemath
                  23 mins ago










                • $begingroup$
                  @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                  $endgroup$
                  – Chris Custer
                  19 mins ago



















                0












                $begingroup$

                Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.



                If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.






                share|cite|improve this answer









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                  $begingroup$

                  Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.






                      share|cite|improve this answer









                      $endgroup$



                      Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      Henning MakholmHenning Makholm

                      241k17308547




                      241k17308547























                          2












                          $begingroup$

                          Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$



                          Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.



                          The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .



                          And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.






                          share|cite|improve this answer











                          $endgroup$


















                            2












                            $begingroup$

                            Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$



                            Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.



                            The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .



                            And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.






                            share|cite|improve this answer











                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$



                              Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.



                              The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .



                              And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.






                              share|cite|improve this answer











                              $endgroup$



                              Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$



                              Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.



                              The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .



                              And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 hours ago

























                              answered 2 hours ago









                              DanielWainfleetDanielWainfleet

                              35.3k31648




                              35.3k31648























                                  2












                                  $begingroup$

                                  Go one unit from a at an angle of t to c.

                                  Go in unit steps along ca until one is less than a unit away from b to a point p.

                                  If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.



                                  As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    Go one unit from a at an angle of t to c.

                                    Go in unit steps along ca until one is less than a unit away from b to a point p.

                                    If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.



                                    As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Go one unit from a at an angle of t to c.

                                      Go in unit steps along ca until one is less than a unit away from b to a point p.

                                      If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.



                                      As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Go one unit from a at an angle of t to c.

                                      Go in unit steps along ca until one is less than a unit away from b to a point p.

                                      If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.



                                      As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 1 hour ago









                                      William ElliotWilliam Elliot

                                      8,4022720




                                      8,4022720























                                          0












                                          $begingroup$

                                          I believe it is uncountable also, and here is my thought process:



                                          For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.



                                          Then if $vneq w$, we have $alpha_vneqalpha_w$.



                                          But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                            $endgroup$
                                            – coffeemath
                                            2 hours ago










                                          • $begingroup$
                                            I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                            $endgroup$
                                            – Chris Custer
                                            1 hour ago










                                          • $begingroup$
                                            Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                            $endgroup$
                                            – coffeemath
                                            23 mins ago










                                          • $begingroup$
                                            @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                            $endgroup$
                                            – Chris Custer
                                            19 mins ago
















                                          0












                                          $begingroup$

                                          I believe it is uncountable also, and here is my thought process:



                                          For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.



                                          Then if $vneq w$, we have $alpha_vneqalpha_w$.



                                          But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                            $endgroup$
                                            – coffeemath
                                            2 hours ago










                                          • $begingroup$
                                            I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                            $endgroup$
                                            – Chris Custer
                                            1 hour ago










                                          • $begingroup$
                                            Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                            $endgroup$
                                            – coffeemath
                                            23 mins ago










                                          • $begingroup$
                                            @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                            $endgroup$
                                            – Chris Custer
                                            19 mins ago














                                          0












                                          0








                                          0





                                          $begingroup$

                                          I believe it is uncountable also, and here is my thought process:



                                          For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.



                                          Then if $vneq w$, we have $alpha_vneqalpha_w$.



                                          But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.






                                          share|cite|improve this answer









                                          $endgroup$



                                          I believe it is uncountable also, and here is my thought process:



                                          For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.



                                          Then if $vneq w$, we have $alpha_vneqalpha_w$.



                                          But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered 2 hours ago









                                          Chris CusterChris Custer

                                          14.1k3827




                                          14.1k3827












                                          • $begingroup$
                                            It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                            $endgroup$
                                            – coffeemath
                                            2 hours ago










                                          • $begingroup$
                                            I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                            $endgroup$
                                            – Chris Custer
                                            1 hour ago










                                          • $begingroup$
                                            Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                            $endgroup$
                                            – coffeemath
                                            23 mins ago










                                          • $begingroup$
                                            @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                            $endgroup$
                                            – Chris Custer
                                            19 mins ago


















                                          • $begingroup$
                                            It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                            $endgroup$
                                            – coffeemath
                                            2 hours ago










                                          • $begingroup$
                                            I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                            $endgroup$
                                            – Chris Custer
                                            1 hour ago










                                          • $begingroup$
                                            Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                            $endgroup$
                                            – coffeemath
                                            23 mins ago










                                          • $begingroup$
                                            @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                            $endgroup$
                                            – Chris Custer
                                            19 mins ago
















                                          $begingroup$
                                          It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                          $endgroup$
                                          – coffeemath
                                          2 hours ago




                                          $begingroup$
                                          It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
                                          $endgroup$
                                          – coffeemath
                                          2 hours ago












                                          $begingroup$
                                          I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                          $endgroup$
                                          – Chris Custer
                                          1 hour ago




                                          $begingroup$
                                          I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
                                          $endgroup$
                                          – Chris Custer
                                          1 hour ago












                                          $begingroup$
                                          Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                          $endgroup$
                                          – coffeemath
                                          23 mins ago




                                          $begingroup$
                                          Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
                                          $endgroup$
                                          – coffeemath
                                          23 mins ago












                                          $begingroup$
                                          @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                          $endgroup$
                                          – Chris Custer
                                          19 mins ago




                                          $begingroup$
                                          @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
                                          $endgroup$
                                          – Chris Custer
                                          19 mins ago











                                          0












                                          $begingroup$

                                          Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.



                                          If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.



                                            If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.



                                              If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.



                                              If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 2 hours ago









                                              John DoumaJohn Douma

                                              5,56211319




                                              5,56211319






























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