You will start with only two elements: Points \$A\$ and \$B\$ such that \$A \neq B\$. These points occupy a plane that is infinite in all directions.

At any step in the process you may do any of the three following actions:

  1. Draw a line that passes through two points.

  2. Draw a circle centered at one point such that another point lies on the circle.

  3. Add a new point where two objects (lines and circles) intersect.

Your goal is to create 5 points such that they form the vertices of a regular pentagon (a convex polygon with 5 sides equal in length) using as few circles as possible. You may, of course, have other points but 5 of them must for a regular pentagon. You do not have to draw the edges of the pentagon for your scoring.


When comparing two answers the one that draws fewer circles is better. In the case of a tie in circles the answer that draws the fewest lines is better. In the case of a tie in both circles and lines the answer that adds the fewest points is better.


While the rules list is exhaustive and details everything you can do, this list is not. Just because I don't say you can't do something does not mean you can.

  • You cannot create "arbitrary" objects. Some constructions you will find will do things like add a point at an "arbitrary" location and work from there. You cannot add new points at locations other than intersections.

  • You cannot copy a radius. Some constructions will involve taking a compass setting it to a radius between two points and then picking it up and drawing a circle elsewhere. You cannot do this.

  • You cannot perform limiting processes. All constructions must take a finite number of steps. It is not good enough to approach the answer asymptotically.

  • You cannot draw an arc or part of a circle in order to avoid counting it as a circle in your scoring. If you want to visually use arcs when showing or explaining your answer because they take up less space go ahead but they count as a circle for scoring.


You can think through the problem on GeoGebra. Just go over to the shapes tab. The three rules are equivalent to the point, line and circle with center tools.

Burden of Proof

This is standard but I would like to reiterate. If there is a question as to whether a particular answer is valid the burden of proof is on the answerer to show that their answer is valid rather than the public to show that the answer is not.

What is this doing on my Code-Golf site?!

This is a form of similar to albeit in a bit of a weird programming language. There is currently a +22/-0 consensus on the meta that this sort of thing is allowed.

  • 12
    \$\begingroup\$ This is like the game I have on my phone called Euclidea. \$\endgroup\$
    – mbomb007
    Commented Jul 13, 2019 at 16:40
  • \$\begingroup\$ closely related: codegolf.stackexchange.com/q/38653/15599 \$\endgroup\$ Commented Jul 13, 2019 at 17:03
  • 6
    \$\begingroup\$ Next time you should ask people to draw a heptagon, which would be slightly more challenging:) \$\endgroup\$
    – flawr
    Commented Jul 13, 2019 at 18:09
  • 3
    \$\begingroup\$ It's the regular 17-gon which is constructible using ruler and compasses. I can give you a heptagon but it won't necessarily be regular! \$\endgroup\$
    – Rosie F
    Commented Jul 14, 2019 at 18:25
  • 1
    \$\begingroup\$ Heptagon (7 sides) is not possible with only ruler and compass. Mathologer covered it. \$\endgroup\$ Commented Jul 14, 2019 at 19:00

3 Answers 3


2 circles, 13 lines, 17 points


Try it on GeoGebra

  • Let circle(A, B) intersect circle(B, A) at C and D.
  • Let AB intersect circle(A, B) again at E.
  • Let AB intersect circle(B, A) again at F.
  • Let AD intersect circle(A, B) again at G.
  • Let AD intersect CF at H.
  • Let BG intersect DF at I.
  • Let HI intersect circle(A, B) at J and K.
  • Let BG intersect EJ at L.
  • Let BJ intersect EG at M.
  • Let BG intersect EK at N.
  • Let BK intersect EG at O.
  • Let LM intersect circle(A, B) at P and S.
  • Let NO intersect circle(A, B) at Q and R.

Then EPQRS is a regular pentagon.

Why it works

Let BE intersect GJ at T, and let BE intersect GK at U. The complete quadrilateral BEGJ shows that T is the polar of LM, which is the intersection of the tangents at P and S. Similarly, the complete quadrilateral BEGK shows that U is the polar of NO, which is the intersection of the tangents at Q and R.

Let FG intersect HI at V. The diagonals DV and GI of the complete quadrilateral DGVI intersect FH at harmonic conjugates with respect to F and H; since the first is at ∞, the second is the midpoint C of FH, which is to say that C, D, V are collinear.

Let CG intersect HI at W.


Now for the fun part. Line FUBAT is perspective about G to line VKIHJ, which is perspective about D to circle CKDGJ, which is perspective about C to line HKVWJ, which is perspective about G to line AUF∞T. Composing these four perspecitivities yields a projectivity FUBAT ⌅ AUF∞T. Since a one-dimensional projectivity is determined by three points, T and U are determined as the two fixed points of FBA ⌅ AF∞.

Assigning coordinates with A = 0, B = −1, F = −2, this projectivity is defined by x ↦ 4/x + 2, and its fixed points T = 1 + √5 = sec(2π/5) and U = 1 − √5 = −sec(2π/10), exactly as required to make EPQRS a regular pentagon.

  • 11
    \$\begingroup\$ Please explain each step of your algorithm in words and symbols. \$\endgroup\$
    – Rosie F
    Commented Jul 14, 2019 at 18:08
  • 2
    \$\begingroup\$ @Servaes This answer could use some explanation, but I can tell you that the third line is fine, it is a perpendicular bisector but it is defined in terms of two preexisting points rather than as a perpendicular bisector. Same goes for the fourth one. \$\endgroup\$
    – Wheat Wizard
    Commented Jul 14, 2019 at 19:08
  • 2
    \$\begingroup\$ @RosieF Sorry about that, the labels were annoying to add with the way I had been producing the pictures. I redid this in GeoGebra with labelled points and added instructions and a link to the interactive app where you can play with the construction. \$\endgroup\$ Commented Jul 15, 2019 at 0:10
  • 2
    \$\begingroup\$ Looks like a neat solution, but do you care to explain why the result is a regular pentagon? I.e. why EP = PQ = QR = RS = SE? \$\endgroup\$
    – Minethlos
    Commented Jul 15, 2019 at 15:04
  • 2
    \$\begingroup\$ @Minethlos It took a while to come up with a nice proof but I finally found one that I’m happy with. Be warned that it requires a fair amount of background in projective geometry. \$\endgroup\$ Commented Jul 15, 2019 at 22:38

7 6 circles, 3 lines

This is a classical pentagon construction, a proof of its correctness can be found here.

enter image description here


4 circles, 7 lines

Since it has been beaten I thought I would just post my original solution to the problem. This solution is modified from the method given by Dixon in Mathographics, a proof of correctness for that method can be found here.

  • Draw \$\mathrm{Circle}(A,B)\$
  • Draw \$\overline{AB}\$
  • Mark the intersection of \$\mathrm{Circle}(A,B)\$ and \$\overline{AB}\$ as \$C\$
  • Draw \$\mathrm{Circle}(B,C)\$
  • Draw \$\mathrm{Circle}(C,B)\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\mathrm{Circle}(B,C)\$ as \$D\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{AB}\$ as \$E\$
  • Draw \$\overline{DC}\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{DC}\$ as \$F\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\mathrm{Circle}(B,C)\$ as \$G\$
  • Draw \$\overline{BG}\$
  • Mark the intersection of \$\overline{BG}\$ and \$\overline{EF}\$ as \$H\$
  • Draw \$\overline{HC}\$
  • Mark the intersection of \$\overline{HC}\$ and \$\mathrm{Circle}(C,B)\$ as \$I\$
  • Draw \$\overline{IA}\$
  • Mark the intersection of \$\overline{IA}\$ and \$\mathrm{Circle}(A,B)\$ as \$J\$
  • Draw \$\mathrm{Cirlce}(I,J)\$
  • Mark the intersection of \$\mathrm{Circle}(I,J)\$ and \$\overline{HC}\$ as \$L\$
  • Mark the intersections of \$\mathrm{Circle}(I,J)\$ and \$\mathrm{Circle}(C,B)\$ as \$M\$ and \$K\$.
  • Draw \$\overline{ML}\$
  • Draw \$\overline{KL}\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{ML}\$ as \$N\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{HC}\$ as \$O\$
  • Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{KL}\$ as \$P\$

\$MKPON\$ is a regular pentagon.


  • 1
    \$\begingroup\$ This is marvellous! Some of your construction resembles Dixon's method, but your method cleverly avoids bisecting anything or constructing a perpendicular. \$\endgroup\$
    – Rosie F
    Commented Jul 14, 2019 at 18:17
  • \$\begingroup\$ @RosieF It is modified from Dixon's method, I probably should have mentioned that. \$\endgroup\$
    – Wheat Wizard
    Commented Jul 14, 2019 at 18:49

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