geometric-constructiongeometry
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?
EUCLIDEA IOS PROBLEM 2.8 / ANDROID 5.7 | 5E SOLUTION REQUIRED
Given Two parallel lines ${a}$ (for 'above') and ${b}$ ('below'), separated by a distance $2D$.
Goal Construct line ${m}$ (for 'middle') parallel to and equidistant from both given lines, i.e., separated by a distance $D$ from both $a$ and $b$.
Constraints Only an unmarked straightedge and a non-rusty collapsible compass (i.e., a compass that, while on paper, can have any radius but which cannot maintain said radius when not on paper) can be used. Use only from the construction steps below a number of steps that together require just five elementary steps; the fifth step must construct the goal line itself.
- Construct a point: 0 elementary steps (E).
- Mark the intersection of two curves with a point: 0E.
- Construct a new line (*or line segment or ray): 1E.
- Extend a given line segment (*or ray): 1E.
- Construct a circle (non-rusty collapsible compass): 1E.
- Construct the perpendicular bisector of a line segment: 3E.
- Construct a new line perpendicular to an old line: 3E.
Hints from Euclidea (including Twitter support) A known 5E solution employs five elementary steps in the order: circle, circle, line, line, line. The first circle is twice the radius of the second one. Homothety is used in the known solution.
SOLUTION Please refer to the figure below. Apologies for not having labels on this figure.
Construct two arbitrary points $O$ and $P$ on $b$ separated by a distance $OP > D$ (just eyeball it) [OE running total].
Construct circle $P(O)$ centered on point $P$ and with radius $OP$ [1E running total].
Construct $Q(O)$ [2E running total].
Construct $\overleftrightarrow{OL}$ [3E running total].
Construct $\overleftrightarrow{OR}$ [4E running total].
Construct the desired line $m = \overleftrightarrow{M_{L}M_{R}}$, shown in yellow in the figure below [5E final total].
PROOF OF THE SOLUTION
Given that $OP$ is chosen such that $OQ$ is larger than the gap $2D$ between the two give lines, there exist exactly two points $L$ and $R$ of intersection between $Q(O)$ and $a$.
Homothety is then used in this construction. First, we define a homothetic point ${O}$ on one of the lines, with $b$ used here without loss of generality. $Q(O)$ is constructed to have twice the radius of $P(O)$. Hence, taking $O$ as the homothetic point, also known as the center of dilation, we can view $P(O)$ as the half-fold 'dilation' of $Q(O)$. This means that every line through $O$ intersects $P(O)$ and $Q(O)$ at corresponding points, with the point on $P(O)$ half as far from $O$ as its corresponding point on $Q(O)$ is.
More specifically, the point $M_L$ on $P(O)$ that corresponds to the point $L$ on $P(O)$ can be found by intersecting $\overleftrightarrow{OL}$ and $P(O)$. Point $M_R$ can also be found analogously using $R$.
Homothety preserves parallelism. Therefore, line $m := \overleftrightarrow{M_LM_R}$ is parallel to both $a$ and $b$, as desired.
Given that the homothety involved here involves a half-fold dilation of $Q(O)$ to $P(O)$, we know that the gap between $a$ and $m$ is also half of the gap between $a$ and $b$. This is equivalent to saying that $m$ is equidistant from both $a$ and $b$, as desired. $\blacksquare$
Best Answer
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:
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!