In the book Geometry: Euclid and beyond, the exercise 2.20 says:
Using a ruler and rusty compass, given a line $l$ and given a segment $AB$ more than one inch long, construct one of the points $C$ which the circle of center $A$ and radius $AB$ meets $l$.(Rusty compass's radius is 1 inch)
But I don't think it can be done use ruler and rusty compass. As the picture below shown, we need to get point $C$, use ruler and rusty compass, we can easily twice long $BA$ to point $B'$ and construct perpendicular $BG,B'H$, but in order to get length of $GC$, we need to solve a quadratic equation and need to use square root operation, but some results before said that we can only get the length in $\mathbf{Q}$ use ruler and rusty compass. I am very confused about that. Is this exercise solvable?
Best Answer
Apologies, but I didnt like MathManiacs answer. To say the least its confusingly written, and fails to carry a construction through to the end, and makes assumptions about what is provided. Some vital details seem to be brushed over. I attempted to follow his construction and failed; I dont know why. I couldnt see what direction he was going and the instructions themselves dont make a whole lot of sense to me when you actually try to carry it out. Perhaps its my failure but I thought Id give a different perspective, filling some gaps. I of course dont mean any disrespect to the community but I didnt feel as though this 3 year old question was ever truly answered. I only came across this question in an Ask Jeeves search because I particularly love geometry.
Here is my own construction. Hopefully you can follow along with the animated GIF image. This isn't a proof; it's just a method of construction.
We have an arbitrary line m, and an arbitrary line segment AB defining the center, A, and the radius AB of a circle. No circle is drawn there because circle A(B) is of any size and you only have a rusty compass, which is defined for you in the upper-right corner. We wish to find the points of intersection on line m of where circle A(B) intersects the line.
If X1 and X2 dont exist then neither do the intersections I1 and I2, for obvious reasons.
I have no citation for this construction. It is based on the notion of shrinking down the scale, so that the circle A(B) becomes A(r), and line m becomes a new line that can be directly intersected with A(r) (producing the X intersections). The two I points - the intersections of interest - are projections of the two X points, away from the point A and onto the line m again at appropriate scale. Circle A(r) is to circle A(B) what the horizontal line passing through H is to the line m. The bulk of the construction is scaling line m down in proportion.
Creating a Parallel Line
To draw a parallel line using a straightedge and a single circle, the Wikipedia has an article entitled Poncelet-Steiner Theorem which contains an animated GIF depicting the construction. It requires 5 intermediary lines to be drawn for each parallel line to be made, and 1 circle (of any size) for each line from which a parallel is to be made.
The Case of Parallel AB
In the case that line segment AB is parallel to line m, a separate construction could be implemented to rotate AB into a new non-parallel position. You then restart this line-circle intersection construction using a new line segment AB'.
That said, the construction for this case is actually easier than just explained. You dont need to rotate AB at all. I wont show you the modification, but you can swap the roles of lines m and n in the original construction and adapt the construction appropriately. This can be done fairly readily and with no additional cost.
I absolutely love geometric constructions and in particular restricted constructions. Send me more. The aforementioned Poncelet-Steiner theorem, and of course the Mohr-Mascheroni Theorem, are fascinating restricted constructions in their own right.