# Equivalence of different ways of geometrical multiplication

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 $$O1$$:

## Method 1

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

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

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

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

2. Do this $$a-1$$ times.

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

## Method 2

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

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

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

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

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

2. Do this $$c$$ times.

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

## Method 3

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

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

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 $$A \times B$$.

## 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 $$A \times B$$.

## 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 $$A \times B$$.

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).

## Method 8 (due to Accumulation)

• 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 So methods that count don't count? – Acccumulation Mar 20 '19 at 21:37
• 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
• (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
• According to the Mohr–Mascheroni theorem (Wikipedia), 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

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 $$OK = OA\cdot OB.$$

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.

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 $$A/(1/B)$$.

• I added your construction to my big list. – Hans-Peter Stricker Mar 21 '19 at 8:28
• Isn't it the Method 3? – CiaPan Mar 21 '19 at 12:28