I've drawn two random points on a two dimensional plane. If the distance between these two points is one unit, can a line of π units be drawn using a compass and straight edge?
Or, is there a proof that shows this is impossible?
[Math] Can a line of π lengths be drawn with a compass and straight edge
geometric-constructionpi
Related Solutions
Your solution fails because you give up a move creating the center point of your bounding circle. To complete the challenge in eight moves, you need to use one of the preexisting intersections as your center point, and then rely on the fact that extending a line is not considered a move to create the fourth vertex of the square.
The following diagram illustrates these steps, where step 4b is the non-move extension of the line:
The following solution to the problem uses the ruler only and is based on Desargues's theorem.
Let $R=AB\cap A'B'$ be the point out from the sheet and $P$ a point on the sheet. It's required to draw the line $PR$.
Let $O=AA'\cap BB'$, and choose a line $c\supset O$. Let $C=c\cap AP$ and $C'=c\cap A'P$. The triangles $ABC$ and $A'B'C'$ are homologous since $O=AA'\cap BB'\cap CC'$, thus $R=AB\cap A'B'$, $P=AC\cap A'C'$ and $Q=BC\cap B'C'$ lies on a line. Consequently, $PQ$ is the required line.
Best Answer
No (to your first question). A constructible number must be algebraic. However $\pi$ is not an algebraic number ($\pi$ is a transcendental number). Also of interest with regards to this is the famous "squaring the circle" problem.