Ways of geometrical multiplication












12












$begingroup$


There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).



Let $A,B$ be two integer points on the line $O1$:



enter image description here



Method 1




  1. Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).


enter image description here




  1. Draw a circle with radius $|OB|$ around $B$.


  2. Let $C$ be the (other) intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|OB|$ around $C$.


  2. Do this $a-1$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 2




  1. Construct a rectangle with side lengths $|OA|$, $|OB|$.


enter image description here




  1. Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).


enter image description here




  1. Draw a circle with radius $|O1|$ around $0$.


  2. Let $C$ be the intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|O1|$ around $C$.


  2. Do this $c$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 3




  1. Construct the line perpendicular to $O1$ through $O$.


  2. Construct the points $1'$ and $B'$.



enter image description here




  1. Draw the line $1'A$.


  2. Construct the parallel to $1'A$ through $B'$.


  3. The intersection point of this parallel with the line $O1$ is the product $A times B$.



enter image description here



Method 4




  1. Construct the perpendicular line to $O1$ through $O$.


  2. Construct the point $1'$.


  3. Construct the circle through $1'$, $A$ and $B$.


  4. The intersection point of this circle with the line $O1'$ is the product $A times B$.



enter image description here



Method 5



This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.




  1. Construct the unit parabola $(x,y)$ with $y = x^2$.


  2. Construct $B'$.


  3. Construct the line perpendicular to $O1$ through $A$.


  4. Construct the line perpendicular to $O1$ through $B'$.


  5. Draw the line through the intersection points of these two lines with the parabola.


  6. The intersection point of this line with the line $O1'$ is the product $A times B$.



enter image description here





For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).





Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):




  • Method 1: $sigma sim ab^2$


  • Method 2: $sigma sim ab$


  • Method 3: $sigma sim ab^2$


  • Method 4: $sigma sim a^2b^2$


  • Method 5: $sigma sim a^3b$



This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:




  • Method 1: $tau sim a$


  • Method 2: $tau sim ab$


  • Method 3: $tau sim 1$


  • Method 4: $tau sim 1$


  • Method 5: $tau sim 1$



From this point of view method 3 would be the most efficient.





Once again:




I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).






Those answers I managed to visualize I will add here:



Method 6 (due to Cia Pan)



enter image description here



Method 7 (due to celtschk)



enter image description here





Another list I'll try to keep up to date: Is $A times B$ defined by the intersection of a line or a circle (with $O1$ or $O1'$):




  • Method 1: circle


  • Method 2: circle


  • Method 3: line


  • Method 4: circle


  • Method 5: line


  • Method 6: line


  • Method 7: line











share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
    $endgroup$
    – CiaPan
    7 hours ago










  • $begingroup$
    @CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago












  • $begingroup$
    @CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago










  • $begingroup$
    @CiaPan So methods that count don't count?
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
    $endgroup$
    – CiaPan
    52 mins ago
















12












$begingroup$


There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).



Let $A,B$ be two integer points on the line $O1$:



enter image description here



Method 1




  1. Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).


enter image description here




  1. Draw a circle with radius $|OB|$ around $B$.


  2. Let $C$ be the (other) intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|OB|$ around $C$.


  2. Do this $a-1$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 2




  1. Construct a rectangle with side lengths $|OA|$, $|OB|$.


enter image description here




  1. Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).


enter image description here




  1. Draw a circle with radius $|O1|$ around $0$.


  2. Let $C$ be the intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|O1|$ around $C$.


  2. Do this $c$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 3




  1. Construct the line perpendicular to $O1$ through $O$.


  2. Construct the points $1'$ and $B'$.



enter image description here




  1. Draw the line $1'A$.


  2. Construct the parallel to $1'A$ through $B'$.


  3. The intersection point of this parallel with the line $O1$ is the product $A times B$.



enter image description here



Method 4




  1. Construct the perpendicular line to $O1$ through $O$.


  2. Construct the point $1'$.


  3. Construct the circle through $1'$, $A$ and $B$.


  4. The intersection point of this circle with the line $O1'$ is the product $A times B$.



enter image description here



Method 5



This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.




  1. Construct the unit parabola $(x,y)$ with $y = x^2$.


  2. Construct $B'$.


  3. Construct the line perpendicular to $O1$ through $A$.


  4. Construct the line perpendicular to $O1$ through $B'$.


  5. Draw the line through the intersection points of these two lines with the parabola.


  6. The intersection point of this line with the line $O1'$ is the product $A times B$.



enter image description here





For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).





Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):




  • Method 1: $sigma sim ab^2$


  • Method 2: $sigma sim ab$


  • Method 3: $sigma sim ab^2$


  • Method 4: $sigma sim a^2b^2$


  • Method 5: $sigma sim a^3b$



This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:




  • Method 1: $tau sim a$


  • Method 2: $tau sim ab$


  • Method 3: $tau sim 1$


  • Method 4: $tau sim 1$


  • Method 5: $tau sim 1$



From this point of view method 3 would be the most efficient.





Once again:




I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).






Those answers I managed to visualize I will add here:



Method 6 (due to Cia Pan)



enter image description here



Method 7 (due to celtschk)



enter image description here





Another list I'll try to keep up to date: Is $A times B$ defined by the intersection of a line or a circle (with $O1$ or $O1'$):




  • Method 1: circle


  • Method 2: circle


  • Method 3: line


  • Method 4: circle


  • Method 5: line


  • Method 6: line


  • Method 7: line











share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
    $endgroup$
    – CiaPan
    7 hours ago










  • $begingroup$
    @CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago












  • $begingroup$
    @CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago










  • $begingroup$
    @CiaPan So methods that count don't count?
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
    $endgroup$
    – CiaPan
    52 mins ago














12












12








12


4



$begingroup$


There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).



Let $A,B$ be two integer points on the line $O1$:



enter image description here



Method 1




  1. Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).


enter image description here




  1. Draw a circle with radius $|OB|$ around $B$.


  2. Let $C$ be the (other) intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|OB|$ around $C$.


  2. Do this $a-1$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 2




  1. Construct a rectangle with side lengths $|OA|$, $|OB|$.


enter image description here




  1. Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).


enter image description here




  1. Draw a circle with radius $|O1|$ around $0$.


  2. Let $C$ be the intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|O1|$ around $C$.


  2. Do this $c$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 3




  1. Construct the line perpendicular to $O1$ through $O$.


  2. Construct the points $1'$ and $B'$.



enter image description here




  1. Draw the line $1'A$.


  2. Construct the parallel to $1'A$ through $B'$.


  3. The intersection point of this parallel with the line $O1$ is the product $A times B$.



enter image description here



Method 4




  1. Construct the perpendicular line to $O1$ through $O$.


  2. Construct the point $1'$.


  3. Construct the circle through $1'$, $A$ and $B$.


  4. The intersection point of this circle with the line $O1'$ is the product $A times B$.



enter image description here



Method 5



This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.




  1. Construct the unit parabola $(x,y)$ with $y = x^2$.


  2. Construct $B'$.


  3. Construct the line perpendicular to $O1$ through $A$.


  4. Construct the line perpendicular to $O1$ through $B'$.


  5. Draw the line through the intersection points of these two lines with the parabola.


  6. The intersection point of this line with the line $O1'$ is the product $A times B$.



enter image description here





For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).





Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):




  • Method 1: $sigma sim ab^2$


  • Method 2: $sigma sim ab$


  • Method 3: $sigma sim ab^2$


  • Method 4: $sigma sim a^2b^2$


  • Method 5: $sigma sim a^3b$



This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:




  • Method 1: $tau sim a$


  • Method 2: $tau sim ab$


  • Method 3: $tau sim 1$


  • Method 4: $tau sim 1$


  • Method 5: $tau sim 1$



From this point of view method 3 would be the most efficient.





Once again:




I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).






Those answers I managed to visualize I will add here:



Method 6 (due to Cia Pan)



enter image description here



Method 7 (due to celtschk)



enter image description here





Another list I'll try to keep up to date: Is $A times B$ defined by the intersection of a line or a circle (with $O1$ or $O1'$):




  • Method 1: circle


  • Method 2: circle


  • Method 3: line


  • Method 4: circle


  • Method 5: line


  • Method 6: line


  • Method 7: line











share|cite|improve this question











$endgroup$




There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).



Let $A,B$ be two integer points on the line $O1$:



enter image description here



Method 1




  1. Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).


enter image description here




  1. Draw a circle with radius $|OB|$ around $B$.


  2. Let $C$ be the (other) intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|OB|$ around $C$.


  2. Do this $a-1$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 2




  1. Construct a rectangle with side lengths $|OA|$, $|OB|$.


enter image description here




  1. Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).


enter image description here




  1. Draw a circle with radius $|O1|$ around $0$.


  2. Let $C$ be the intersection point of this circle with the line $O1$.



enter image description here




  1. Draw a circle with radius $|O1|$ around $C$.


  2. Do this $c$ times.


  3. The last intersection point $C$ is the product $A times B$.



enter image description here



Method 3




  1. Construct the line perpendicular to $O1$ through $O$.


  2. Construct the points $1'$ and $B'$.



enter image description here




  1. Draw the line $1'A$.


  2. Construct the parallel to $1'A$ through $B'$.


  3. The intersection point of this parallel with the line $O1$ is the product $A times B$.



enter image description here



Method 4




  1. Construct the perpendicular line to $O1$ through $O$.


  2. Construct the point $1'$.


  3. Construct the circle through $1'$, $A$ and $B$.


  4. The intersection point of this circle with the line $O1'$ is the product $A times B$.



enter image description here



Method 5



This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.




  1. Construct the unit parabola $(x,y)$ with $y = x^2$.


  2. Construct $B'$.


  3. Construct the line perpendicular to $O1$ through $A$.


  4. Construct the line perpendicular to $O1$ through $B'$.


  5. Draw the line through the intersection points of these two lines with the parabola.


  6. The intersection point of this line with the line $O1'$ is the product $A times B$.



enter image description here





For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).





Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):




  • Method 1: $sigma sim ab^2$


  • Method 2: $sigma sim ab$


  • Method 3: $sigma sim ab^2$


  • Method 4: $sigma sim a^2b^2$


  • Method 5: $sigma sim a^3b$



This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:




  • Method 1: $tau sim a$


  • Method 2: $tau sim ab$


  • Method 3: $tau sim 1$


  • Method 4: $tau sim 1$


  • Method 5: $tau sim 1$



From this point of view method 3 would be the most efficient.





Once again:




I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).






Those answers I managed to visualize I will add here:



Method 6 (due to Cia Pan)



enter image description here



Method 7 (due to celtschk)



enter image description here





Another list I'll try to keep up to date: Is $A times B$ defined by the intersection of a line or a circle (with $O1$ or $O1'$):




  • Method 1: circle


  • Method 2: circle


  • Method 3: line


  • Method 4: circle


  • Method 5: line


  • Method 6: line


  • Method 7: line








euclidean-geometry arithmetic conic-sections big-list geometric-construction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







Hans-Peter Stricker

















asked 7 hours ago









Hans-Peter StrickerHans-Peter Stricker

6,56443995




6,56443995








  • 2




    $begingroup$
    IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
    $endgroup$
    – CiaPan
    7 hours ago










  • $begingroup$
    @CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago












  • $begingroup$
    @CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago










  • $begingroup$
    @CiaPan So methods that count don't count?
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
    $endgroup$
    – CiaPan
    52 mins ago














  • 2




    $begingroup$
    IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
    $endgroup$
    – CiaPan
    7 hours ago










  • $begingroup$
    @CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago












  • $begingroup$
    @CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago










  • $begingroup$
    @CiaPan So methods that count don't count?
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
    $endgroup$
    – CiaPan
    52 mins ago








2




2




$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
7 hours ago




$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
7 hours ago












$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
7 hours ago






$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
7 hours ago














$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
6 hours ago




$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
6 hours ago












$begingroup$
@CiaPan So methods that count don't count?
$endgroup$
– Acccumulation
2 hours ago




$begingroup$
@CiaPan So methods that count don't count?
$endgroup$
– Acccumulation
2 hours ago












$begingroup$
When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
$endgroup$
– CiaPan
52 mins ago




$begingroup$
When you say 'count it – let $n$ be the numer – do something $n$ times' you introduce some counter $n$ and some variable 'iteration number' running from 1 through $n$. In my feeling this goes beyond classic constructions and I would translate it into purely geometric actions. For example in Method 1: construct a chain of consecutive copies of the segment $O1$ along the line, until you reach $A$; at each constructed endpoint construct a copy of $OB$ perpendicular to the line; construct a chain of consecutive copies of... (to be continued)
$endgroup$
– CiaPan
52 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$


  1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

  2. Construct the perpendicular at $O$.

  3. Construct the semicircle on the diameter $A'B$.

  4. Find $H$ at the intersection of the semicircle and the perpendicular.
    $(OH)^2 = OA'cdot OB = OAcdot OB$.

  5. Draw line $1H$ and construct a perpendicular to it through $H$.

  6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I'll try to visualize this (or do you have a diagram at hand)?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago










  • $begingroup$
    Done (have a look at my edited question).
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago



















3












$begingroup$

The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).




  1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

  2. Select on $g$ an arbitrary point $P$ other than the origin.

  3. Draw a line through $1$ and $P$.

  4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

  5. Draw a line through $P$ and $B$.

  6. Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.


Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).



With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have added your construction to my list (see above). Thanks a lot!
    $endgroup$
    – Hans-Peter Stricker
    5 hours ago



















0












$begingroup$

If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.



Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155691%2fways-of-geometrical-multiplication%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$


    1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

    2. Construct the perpendicular at $O$.

    3. Construct the semicircle on the diameter $A'B$.

    4. Find $H$ at the intersection of the semicircle and the perpendicular.
      $(OH)^2 = OA'cdot OB = OAcdot OB$.

    5. Draw line $1H$ and construct a perpendicular to it through $H$.

    6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! I'll try to visualize this (or do you have a diagram at hand)?
      $endgroup$
      – Hans-Peter Stricker
      7 hours ago










    • $begingroup$
      Done (have a look at my edited question).
      $endgroup$
      – Hans-Peter Stricker
      6 hours ago
















    5












    $begingroup$


    1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

    2. Construct the perpendicular at $O$.

    3. Construct the semicircle on the diameter $A'B$.

    4. Find $H$ at the intersection of the semicircle and the perpendicular.
      $(OH)^2 = OA'cdot OB = OAcdot OB$.

    5. Draw line $1H$ and construct a perpendicular to it through $H$.

    6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! I'll try to visualize this (or do you have a diagram at hand)?
      $endgroup$
      – Hans-Peter Stricker
      7 hours ago










    • $begingroup$
      Done (have a look at my edited question).
      $endgroup$
      – Hans-Peter Stricker
      6 hours ago














    5












    5








    5





    $begingroup$


    1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

    2. Construct the perpendicular at $O$.

    3. Construct the semicircle on the diameter $A'B$.

    4. Find $H$ at the intersection of the semicircle and the perpendicular.
      $(OH)^2 = OA'cdot OB = OAcdot OB$.

    5. Draw line $1H$ and construct a perpendicular to it through $H$.

    6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$






    share|cite|improve this answer









    $endgroup$




    1. Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

    2. Construct the perpendicular at $O$.

    3. Construct the semicircle on the diameter $A'B$.

    4. Find $H$ at the intersection of the semicircle and the perpendicular.
      $(OH)^2 = OA'cdot OB = OAcdot OB$.

    5. Draw line $1H$ and construct a perpendicular to it through $H$.

    6. Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    CiaPanCiaPan

    10.1k11247




    10.1k11247












    • $begingroup$
      Thanks! I'll try to visualize this (or do you have a diagram at hand)?
      $endgroup$
      – Hans-Peter Stricker
      7 hours ago










    • $begingroup$
      Done (have a look at my edited question).
      $endgroup$
      – Hans-Peter Stricker
      6 hours ago


















    • $begingroup$
      Thanks! I'll try to visualize this (or do you have a diagram at hand)?
      $endgroup$
      – Hans-Peter Stricker
      7 hours ago










    • $begingroup$
      Done (have a look at my edited question).
      $endgroup$
      – Hans-Peter Stricker
      6 hours ago
















    $begingroup$
    Thanks! I'll try to visualize this (or do you have a diagram at hand)?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago




    $begingroup$
    Thanks! I'll try to visualize this (or do you have a diagram at hand)?
    $endgroup$
    – Hans-Peter Stricker
    7 hours ago












    $begingroup$
    Done (have a look at my edited question).
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago




    $begingroup$
    Done (have a look at my edited question).
    $endgroup$
    – Hans-Peter Stricker
    6 hours ago











    3












    $begingroup$

    The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).




    1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

    2. Select on $g$ an arbitrary point $P$ other than the origin.

    3. Draw a line through $1$ and $P$.

    4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

    5. Draw a line through $P$ and $B$.

    6. Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.


    Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).



    With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have added your construction to my list (see above). Thanks a lot!
      $endgroup$
      – Hans-Peter Stricker
      5 hours ago
















    3












    $begingroup$

    The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).




    1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

    2. Select on $g$ an arbitrary point $P$ other than the origin.

    3. Draw a line through $1$ and $P$.

    4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

    5. Draw a line through $P$ and $B$.

    6. Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.


    Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).



    With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have added your construction to my list (see above). Thanks a lot!
      $endgroup$
      – Hans-Peter Stricker
      5 hours ago














    3












    3








    3





    $begingroup$

    The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).




    1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

    2. Select on $g$ an arbitrary point $P$ other than the origin.

    3. Draw a line through $1$ and $P$.

    4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

    5. Draw a line through $P$ and $B$.

    6. Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.


    Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).



    With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.






    share|cite|improve this answer











    $endgroup$



    The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).




    1. Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

    2. Select on $g$ an arbitrary point $P$ other than the origin.

    3. Draw a line through $1$ and $P$.

    4. Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

    5. Draw a line through $P$ and $B$.

    6. Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.


    Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).



    With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 6 hours ago

























    answered 6 hours ago









    celtschkceltschk

    30.3k755101




    30.3k755101












    • $begingroup$
      I have added your construction to my list (see above). Thanks a lot!
      $endgroup$
      – Hans-Peter Stricker
      5 hours ago


















    • $begingroup$
      I have added your construction to my list (see above). Thanks a lot!
      $endgroup$
      – Hans-Peter Stricker
      5 hours ago
















    $begingroup$
    I have added your construction to my list (see above). Thanks a lot!
    $endgroup$
    – Hans-Peter Stricker
    5 hours ago




    $begingroup$
    I have added your construction to my list (see above). Thanks a lot!
    $endgroup$
    – Hans-Peter Stricker
    5 hours ago











    0












    $begingroup$

    If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.



    Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.



      Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.



        Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.






        share|cite|improve this answer









        $endgroup$



        If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$.



        Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        AcccumulationAcccumulation

        7,1252619




        7,1252619






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155691%2fways-of-geometrical-multiplication%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            宮崎県

            濃尾地震

            シテ島