Please read the Cops post for all details.

Points are scored on the basis of how long a challenge remains unsolved before you solve. You may solve any number of challenges. You cannot solve your own challenges. Points scored by robbers and by cops are independent, and you can participate in both posts separately.

  • $\begingroup$ With all due respect, why is this posted separately? The challenge I'm familiar with at PCG (codegolf.stackexchange.com/questions/54807/…) is a single post for both Cops and Robbers $\endgroup$ – question_asker Jan 13 '16 at 13:25
  • $\begingroup$ @question_asker I just saw this post and decided to write two. codegolf.stackexchange.com/q/40932/40929 If there is any reason to combine them, I'll do so. $\endgroup$ – ghosts_in_the_code Jan 13 '16 at 14:17
  • $\begingroup$ So that people don't have to look at two separate threads to do the same thing? $\endgroup$ – question_asker Jan 13 '16 at 14:48
  • $\begingroup$ I'm voting to close this question as off-topic because it doesn't contain a question - please edit the question to include a question text. $\endgroup$ – user20 Jan 14 '16 at 20:14
  • $\begingroup$ @Emrakul I'm shutting down the challenge (because f" gave a simple theorem that can convert any compass and straightedge construction to a compass only construction). Please delete (or at least close) both the posts. $\endgroup$ – ghosts_in_the_code Jan 15 '16 at 8:01

From the Cop's challenge thread:

OK, I've got an easy one to start with. Given a line segment radius R, find a point that is exactly R away from one endpoint of the segment, and lies on the same vector. Essentially, double the line segment.


complete the circle that the segment is on. next, place the points of the compass on each end of the segment. complete a circle using one point as the center. Create 6 circles using the radius of the new circle, centered on the end of the first segment intersection through the center of the new circle, and adding new circles around the first at the intersection of each new circle and the original.


enter image description here

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  • $\begingroup$ that lies on the same vector Meaning to say, the three points are collinear. I know the answer, it's quite easy, but your answer does not work. $\endgroup$ – ghosts_in_the_code Jan 13 '16 at 14:55
  • $\begingroup$ ah, you're right. missed that part. $\endgroup$ – dfperry Jan 13 '16 at 14:55
  • $\begingroup$ Yeah, this was meant to be a straight line, so at least slightly more involved than your solution. $\endgroup$ – Cain Jan 13 '16 at 14:55
  • $\begingroup$ @Cain updated with what should be the correct answer $\endgroup$ – dfperry Jan 13 '16 at 15:10
  • $\begingroup$ I meant for the starting segment to also be straight, but this answer solves for that case as well. Nice job. $\endgroup$ – Cain Jan 13 '16 at 16:08

Given a line segment radius R, find a point that is exactly R away from one endpoint of the segment, and lies on the same vector.

Draw two circles of radius $R$ about the two ends of the line segement $a$ and $b$. Call the circles $A$ and $B$. They meet at two new points $c$ and $d$. $c$ and $d$ are $\sqrt{3}R$ from each other. Draw a circle of radius $\sqrt{3}R$ from $c$ and for good measure $d$. These circles will both intersect $A$ at the same point $f$. That point $f$ meets all requirements.

This works because you effectively make a right triangle with the following properties:

  1. The right angle is located at the midpoint between $a$ and $b$ (call it $e$). This is not a marked point and is, therefore, invisible. It is the intersection between $ab$ and $cd$.
  2. Since two circles are produced with the same radius about $c$ and $d$, the intersection will rest on a line that bisects $cd$. As $cd$ was produced by bisecting $ab$ these are on the same vector.
  3. We know $ce$ equals $\sqrt{3}/2$ as it is the height of the equilateral triangle $abc$. The hypoteneus is $\sqrt{3}$ in length as it was from $cd$ which is twice $ce$. $ef$ should be $1.5$ and $cd^2=ef^2+ce^2=\sqrt{3}^2=3=(\sqrt{3}/2)^2+1.5^2=3/4+9/4=3$.
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