How to construct a line with only a short ruler

Question

Suppose I want to draw a line between the points A and B

but I only have a ruler that covers only something between a fifth and a quarter of the distance between the two points.

Also available a compass but with the same limitation. (maximum diameter of a circle is somewhere between a fifth and a quarter of the distance between the two points)

How can I draw a straight line beteen A and B?

Answer 1

You want to draw a line form $A$ to $B$. With the following method you can find the middle point of this line, $M$. Repeating this method wou will be able to find points in the desired line as close as you want for your initial points, and this is enough to draw the line.

With a ruler and a compass you can draw lines of the length that you want. First of all we will draw a line passing from A and enough close to B. In order to do this do the following:

1. Draw two lines starting at A more or less close to B, say $l_1$ and $l_2$, and let $\alpha$ be the angle between them.
2. If $B$ is located between the lines, OK. If not, draw a fan of lines forming an angle $\alpha$ with the previous ones until you get two lines for which $B$ is located betwwen them.
3. Now we have two lines starting at $A$ and with $B$ located "inside". Let $l_3$ be the bisectrix.
4. Repeat the step 3 enough times choosing each time lines such that the point $B$ is "inside".

After this process we get a line $l$ passing through $A$ as close as we want to $B$. Since $B$ and $l$ are enough close we can draw with our small ruler and our small compass an orthogonal line to $l$ passing through $B$. So we have a right triangle with vertex on $A$ and $B$ and edges $l$ and $r$. Transporting angles from $B$ to $A$ you can draw a rectangle with vertex $A$, $E$, $B$ and $F$. This rectangle have the property that its height (length of the segment $AE$) is enough small and its diagonal is $AB$ is the desired line.

Now, using additional lines we are able to find the middle points of the long sides. Let $E_1$ be the middle point of $AE$ and $F_1$ the middle point of $BF$.

Since $E_1$ and $F_1$ are enugh close one from the other, we can determine the point $M$ in the middle of them, and this point belongs to the desired line $AB$.

Iterating this algorithm you can draw the line.

I hope that the you can understand my poor explanation...sorry about my english!

Answer 2

Start at point A. Draw a short line in the approximate direction of B. Slide the ruler along the line (half way along will do) and extend the line. Repeat the process until you are quite close to B. Use the compass to step along this line, marking off the points P1, P2, etc that are evenly spaced along the line. When you have a point Pn that is just a little ways beyond B (but within striking distance of B), then stop extending the line. Call this point C. C should also be within striking distance of Pn-1. If it isn't, construct points between all the Pn to give a finer spacing.

Construct a triangle with vertices C, B and Pn-1. Using your measurements from that triangle, use your compass to construct a triangle with vertices at A, P1 and D, such that AD=CB, DP1=CPn.

Construct another triangle with vertices at A, D and E, such that AE = BC.

A, C, B and E are the four corners of a parallelogram. You have the start of the side EB - just extend the line ED towards B.

once you have the parallelogram constructed, you can pace along ED with the compass to consruct a grid of congruent triangles - the framework, if you like, that builds the bridge from A to B.

You should be able to find the centre of the bridge very easily. You now have a point halfway between A and B. Repeat to find the Quarter points. Repeat to find the eighth points, which you can join with your ruler.

Answer 3

How about constructing a quadratic grid first using your ruler and compass:

and then use that as a coordinate system to find points on the straight line between $A$ and $B$ via a linear function.