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Ways of geometrical multiplication


Construct circle tangent to two lineshyperbolic geometry (and circle ) construction problemConstruct a circle with straight edge and compass with some given conditions.Finding tangents to a circle with a straightedgeInscribing square in circle in just seven compass-and-straightedge stepsfinding the center of a circle (elementary geometry)Algebra - Construct circle with radius AB around distinct point CConstruct a perpendicular to a given line from a given (external) point, using a compass only onceDoes this mean that one can construct the cube root of two in three dimensions?Approximation of the quadratic formula with straightedge and compass













6












$begingroup$


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










share|cite|improve this question











$endgroup$











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















6












$begingroup$


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










share|cite|improve this question











$endgroup$











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













6












6








6


1



$begingroup$


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










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







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 34 mins ago







Hans-Peter Stricker

















asked 2 hours ago









Hans-Peter StrickerHans-Peter Stricker

6,53443994




6,53443994











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










2 Answers
2






active

oldest

votes


















4












$begingroup$

  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
    1 hour ago










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


















2












$begingroup$

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
    25 mins ago










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2 Answers
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$begingroup$

  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
    1 hour ago










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















4












$begingroup$

  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
    1 hour ago










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













4












4








4





$begingroup$

  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 1 hour 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
    1 hour ago










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
















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










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















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




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












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




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











2












$begingroup$

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
    25 mins ago















2












$begingroup$

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
    25 mins ago













2












2








2





$begingroup$

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 1 hour ago

























answered 1 hour ago









celtschkceltschk

30.3k755101




30.3k755101











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
















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















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




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

















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