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Based on The square and the compass

This is a new kind of challenge proposed on Meta Puzzling SE. Any discussion about the general type of puzzle (rather than this particular one) can be done there.

The robbers' post is given here.

You have a compass and a pencil but no scale/straightedge. You need to decide upon a construction task that can be achieved accurately in a finite number of moves. The following moves are valid.

  • Make the compass radius equal to the distance between two already marked points.
  • Draw a circle with any marked point as centre.
  • Use any intersection of arcs/circles as a marked point.
  • Select a continuous region (either an arc or a 2D region) and mark an approximate point. For example, I could draw an arc and then mark a point that is approximately (but not exactly) at the centre.

For example, a task could be "find the midpoint of a line segment" or "mark 5 points that form a regular pentagon"

Your challenge will be scored by the amount of time it remains unsolved. At the end of this month (31 Jan), you must post the solution(s) to any unsolved challenges. You can make up to 3 posts, and your total score will be the sum of the individual scores.

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closed as off-topic by Aza Jan 15 '16 at 10:19

  • This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What if someone deliberately posts a problem that is not solvable? $\endgroup$ – G-man Jan 13 '16 at 14:55
  • $\begingroup$ No points could be awarded to the 'cop'. If by the end of the month (when scoring is done), the OP cannot post a viable solution to the problem, he cannot score points from it. $\endgroup$ – Tim Couwelier Jan 13 '16 at 15:09
  • $\begingroup$ The Mohr-Mascheroni theorem states that any construction possible with compass and straightedge is possible with compass alone. This challenge reduces to finding a compass-and-straightedge construction and then converting it, e.g. by following the constructions in this paper. $\endgroup$ – f'' Jan 13 '16 at 16:06
  • $\begingroup$ @f" Oh, ok. That was disappointing. Now only I understood this theorem. And it does seem easy to convert a construction. So I'm thinking of just shutting down the challenge. Could you please post the comment as an answer for me to accept? $\endgroup$ – ghosts_in_the_code Jan 13 '16 at 17:01
  • $\begingroup$ For extra props, use Geogebra to make a test bed along the lines of Euclid the Game. $\endgroup$ – Peter Taylor Jan 13 '16 at 22:06
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OK, I've got an easy one to start with. Given a straight line segment radius R, find a point that is exactly R away from one endpoint of the segment, and is colinear with the two endpoints. Essentially, double the line segment.

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Might as well do another easy one:

Define two points $a$ and $b$. Generate two more points $c$ and $d$ such that $abcd$ is a perfect square.

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  • $\begingroup$ To stress this: $a$ and $b$ are already on the sheet in front of you (aka the starting point). They cannot be a byproduct of you drawing the square. $\endgroup$ – kaine Jan 13 '16 at 16:50

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