# Construct a pentagon avoiding compass use

## Rules

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.

## 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.

## Anti-Rules

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 thinks 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.

## Tools

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.

• This is like the game I have on my phone called Euclidea. – mbomb007 Jul 13 at 16:40
• closely related: codegolf.stackexchange.com/q/38653/15599 – Level River St Jul 13 at 17:03
• Next time you should ask people to draw a heptagon, which would be slightly more challenging:) – flawr Jul 13 at 18:09
• 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! – Rosie F Jul 14 at 18:25
• Heptagon (7 sides) is not possible with only ruler and compass. Mathologer covered it. – Draco18s Jul 14 at 19:00

# 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.

• Please explain each step of your algorithm in words and symbols. – Rosie F Jul 14 at 18:08
• @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. – Wheat Wizard Jul 14 at 19:08
• @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. – Anders Kaseorg Jul 15 at 0:10
• 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? – Minethlos Jul 15 at 15:04
• @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. – Anders Kaseorg Jul 15 at 22:38

# 7 6 circles, 3 lines

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

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