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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 $O1$:

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Method 1

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

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

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

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

  2. Do this $a-1$ times.

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

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Method 2

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

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

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

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

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

  2. Do this $c$ times.

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

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Method 3

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

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

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  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 $A \times B$.

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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 $A \times B$.

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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) $A \times B$.


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)

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Method 7 (due to celtschk)

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Method 8 (due to Accumulation)

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Hans-Peter Stricker
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    IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical. – CiaPan Mar 20 '19 at 17:06
  • @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? – Hans-Peter Stricker Mar 20 '19 at 17:10
  • @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. – Hans-Peter Stricker Mar 20 '19 at 17:31
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    @CiaPan So methods that count don't count? – Acccumulation Mar 20 '19 at 21:37
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    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) – CiaPan Mar 20 '19 at 23:38
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    (cont.) ...all those segments along the line; the final endpoint is the point sought. This way we iterate over some set of concrete objects (segments or their endpoints). We can say 'I proces _this_ one, and now _this_ one... And I processed all of them so here is the result.' as opposite to 'I do one construction, and the second one... And I _remember_ I was to make five of them, so now I'm done.' which involves some criterion not visible in the drawing. – CiaPan Mar 20 '19 at 23:38
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    According to the [Mohr–Mascheroni theorem (Wikipedia)](https://en.wikipedia.org/wiki/Mohr%E2%80%93Mascheroni_theorem), any point classicaly constructible by a straightedge and a compass can also be constructed by a compass alone. Applying this would turn all points in your recent list to "circle". :) – CiaPan Mar 21 '19 at 12:17
  • @CiaPan: I'll drop it. – Hans-Peter Stricker Mar 21 '19 at 12:55
  • There's another geometrical way to multiply that is not a compass+straightedge construction in the sense you're after since it requires 3 dimensions, but I find it quite beautiful so I can't resist to mention it. I described it in [this question](https://math.stackexchange.com/questions/2210189/geometric-notion-of-addition-for-the-real-projective-line) (the first image). It requires constructing two lines of slope $a$ and $b$ first, its output is a line of slope $a \cdot b$. – pregunton Mar 22 '19 at 09:04
  • This question should be retitled "Different ways of geometrical multiplication", or even "Equivalence of different ways of geometrical multiplication". Since that's what it's looking for. It's not just looking for a jumbled list without any discussion of equivalence (or "interrrelationship"). My edit on that was rejected; one of you can fix it if you care. – smci Mar 27 '19 at 08:05
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    @smci: I: agree and changed the title. – Hans-Peter Stricker Mar 27 '19 at 08:45

3 Answers3

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  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. $(OH)^2 = OA'\cdot OB = OA\cdot OB$.
  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 $(OH)^2 = 1\cdot OK,$ hence $OK = OA\cdot OB.$
CiaPan
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4

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 $A\times B$.

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.

celtschk
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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 $A/(1/B)$.

Acccumulation
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