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Construct a pentagon avoiding compass use
Construct n-gons with a ruler and compassLittle Chandler is sad. Draw him a cloud to cheer him upConstruct n-gons with a ruler and compassCircle packing in a rectangleRandomize points on a discCircular BluesFinding Exclusive Area in Circle IntersectionsText on a circleBorders of overlapping circlesI really wanted a rhombus, but all I got was this stupid rectangleGolf the smallest circle!
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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:
Draw a line that passes through two points.
Draw a circle centered at one point such that another point lies on the circle.
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 atomic-code-golf similar to proof-golf 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.
math geometry atomic-code-golf
$endgroup$
add a comment |
$begingroup$
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:
Draw a line that passes through two points.
Draw a circle centered at one point such that another point lies on the circle.
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 atomic-code-golf similar to proof-golf 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.
math geometry atomic-code-golf
$endgroup$
2
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
1
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago
add a comment |
$begingroup$
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:
Draw a line that passes through two points.
Draw a circle centered at one point such that another point lies on the circle.
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 atomic-code-golf similar to proof-golf 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.
math geometry atomic-code-golf
$endgroup$
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:
Draw a line that passes through two points.
Draw a circle centered at one point such that another point lies on the circle.
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 atomic-code-golf similar to proof-golf 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.
math geometry atomic-code-golf
math geometry atomic-code-golf
edited 7 hours ago
Sriotchilism O'Zaic
asked 10 hours ago
Sriotchilism O'ZaicSriotchilism O'Zaic
36.7k10 gold badges164 silver badges375 bronze badges
36.7k10 gold badges164 silver badges375 bronze badges
2
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
1
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago
add a comment |
2
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
1
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago
2
2
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
1
1
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
76 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found here.
$endgroup$
add a comment |
$begingroup$
4 circles, 4 lines, 9 points
- Let circle(A, B) intersect circle(B, A) at C.
- Let AB intersect circle(A, B) again at D.
- Let circle(D, B) intersect circle(B, A) farthest from C at E.
- Let AB intersect CE at F.
- Let circle(D, F) intersect circle(A, B) at G, H.
- Let circle(D, F) intersect AB again at I.
- Let GI intersect circle(A, B) again at J.
- Let HI intersect circle(A, B) again at K.
Then BGJKH is a regular pentagon.
$endgroup$
add a comment |
$begingroup$
4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem.
- Draw $mathrmCircle(A,B)$
- Draw $overlineAB$
- Mark the intersection of $mathrmCircle(A,B)$ and $overlineAB$ as $C$
- Draw $mathrmCircle(B,C)$
- Draw $mathrmCircle(C,B)$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $D$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineAB$ as $E$
- Draw $overlineDC$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineDC$ as $F$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $G$
- Draw $overlineBG$
- Mark the intersection of $overlineBG$ and $overlineEF$ as $H$
- Draw $overlineHC$
- Mark the intersection of $overlineHC$ and $mathrmCircle(C,B)$ as $I$
- Draw $overlineIA$
- Mark the intersection of $overlineIA$ and $mathrmCircle(A,B)$ as $J$
- Draw $mathrmCirlce(I,J)$
- Mark the intersection of $mathrmCircle(I,J)$ and $overlineHC$ as $L$
- Mark the intersections of $mathrmCircle(I,J)$ and $mathrmCircle(C,B)$ as $M$ and $K$.
- Draw $overlineML$
- Draw $overlineKL$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineML$ as $N$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineHC$ as $O$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineKL$ as $P$
$MKPON$ is a regular pentagon.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
76 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found here.
$endgroup$
add a comment |
$begingroup$
76 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found here.
$endgroup$
add a comment |
$begingroup$
76 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found here.
$endgroup$
76 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found here.
edited 7 hours ago
answered 8 hours ago
flawrflawr
28.5k6 gold badges74 silver badges199 bronze badges
28.5k6 gold badges74 silver badges199 bronze badges
add a comment |
add a comment |
$begingroup$
4 circles, 4 lines, 9 points
- Let circle(A, B) intersect circle(B, A) at C.
- Let AB intersect circle(A, B) again at D.
- Let circle(D, B) intersect circle(B, A) farthest from C at E.
- Let AB intersect CE at F.
- Let circle(D, F) intersect circle(A, B) at G, H.
- Let circle(D, F) intersect AB again at I.
- Let GI intersect circle(A, B) again at J.
- Let HI intersect circle(A, B) again at K.
Then BGJKH is a regular pentagon.
$endgroup$
add a comment |
$begingroup$
4 circles, 4 lines, 9 points
- Let circle(A, B) intersect circle(B, A) at C.
- Let AB intersect circle(A, B) again at D.
- Let circle(D, B) intersect circle(B, A) farthest from C at E.
- Let AB intersect CE at F.
- Let circle(D, F) intersect circle(A, B) at G, H.
- Let circle(D, F) intersect AB again at I.
- Let GI intersect circle(A, B) again at J.
- Let HI intersect circle(A, B) again at K.
Then BGJKH is a regular pentagon.
$endgroup$
add a comment |
$begingroup$
4 circles, 4 lines, 9 points
- Let circle(A, B) intersect circle(B, A) at C.
- Let AB intersect circle(A, B) again at D.
- Let circle(D, B) intersect circle(B, A) farthest from C at E.
- Let AB intersect CE at F.
- Let circle(D, F) intersect circle(A, B) at G, H.
- Let circle(D, F) intersect AB again at I.
- Let GI intersect circle(A, B) again at J.
- Let HI intersect circle(A, B) again at K.
Then BGJKH is a regular pentagon.
$endgroup$
4 circles, 4 lines, 9 points
- Let circle(A, B) intersect circle(B, A) at C.
- Let AB intersect circle(A, B) again at D.
- Let circle(D, B) intersect circle(B, A) farthest from C at E.
- Let AB intersect CE at F.
- Let circle(D, F) intersect circle(A, B) at G, H.
- Let circle(D, F) intersect AB again at I.
- Let GI intersect circle(A, B) again at J.
- Let HI intersect circle(A, B) again at K.
Then BGJKH is a regular pentagon.
edited 6 hours ago
answered 7 hours ago
Anders KaseorgAnders Kaseorg
27k2 gold badges47 silver badges96 bronze badges
27k2 gold badges47 silver badges96 bronze badges
add a comment |
add a comment |
$begingroup$
4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem.
- Draw $mathrmCircle(A,B)$
- Draw $overlineAB$
- Mark the intersection of $mathrmCircle(A,B)$ and $overlineAB$ as $C$
- Draw $mathrmCircle(B,C)$
- Draw $mathrmCircle(C,B)$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $D$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineAB$ as $E$
- Draw $overlineDC$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineDC$ as $F$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $G$
- Draw $overlineBG$
- Mark the intersection of $overlineBG$ and $overlineEF$ as $H$
- Draw $overlineHC$
- Mark the intersection of $overlineHC$ and $mathrmCircle(C,B)$ as $I$
- Draw $overlineIA$
- Mark the intersection of $overlineIA$ and $mathrmCircle(A,B)$ as $J$
- Draw $mathrmCirlce(I,J)$
- Mark the intersection of $mathrmCircle(I,J)$ and $overlineHC$ as $L$
- Mark the intersections of $mathrmCircle(I,J)$ and $mathrmCircle(C,B)$ as $M$ and $K$.
- Draw $overlineML$
- Draw $overlineKL$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineML$ as $N$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineHC$ as $O$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineKL$ as $P$
$MKPON$ is a regular pentagon.
$endgroup$
add a comment |
$begingroup$
4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem.
- Draw $mathrmCircle(A,B)$
- Draw $overlineAB$
- Mark the intersection of $mathrmCircle(A,B)$ and $overlineAB$ as $C$
- Draw $mathrmCircle(B,C)$
- Draw $mathrmCircle(C,B)$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $D$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineAB$ as $E$
- Draw $overlineDC$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineDC$ as $F$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $G$
- Draw $overlineBG$
- Mark the intersection of $overlineBG$ and $overlineEF$ as $H$
- Draw $overlineHC$
- Mark the intersection of $overlineHC$ and $mathrmCircle(C,B)$ as $I$
- Draw $overlineIA$
- Mark the intersection of $overlineIA$ and $mathrmCircle(A,B)$ as $J$
- Draw $mathrmCirlce(I,J)$
- Mark the intersection of $mathrmCircle(I,J)$ and $overlineHC$ as $L$
- Mark the intersections of $mathrmCircle(I,J)$ and $mathrmCircle(C,B)$ as $M$ and $K$.
- Draw $overlineML$
- Draw $overlineKL$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineML$ as $N$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineHC$ as $O$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineKL$ as $P$
$MKPON$ is a regular pentagon.
$endgroup$
add a comment |
$begingroup$
4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem.
- Draw $mathrmCircle(A,B)$
- Draw $overlineAB$
- Mark the intersection of $mathrmCircle(A,B)$ and $overlineAB$ as $C$
- Draw $mathrmCircle(B,C)$
- Draw $mathrmCircle(C,B)$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $D$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineAB$ as $E$
- Draw $overlineDC$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineDC$ as $F$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $G$
- Draw $overlineBG$
- Mark the intersection of $overlineBG$ and $overlineEF$ as $H$
- Draw $overlineHC$
- Mark the intersection of $overlineHC$ and $mathrmCircle(C,B)$ as $I$
- Draw $overlineIA$
- Mark the intersection of $overlineIA$ and $mathrmCircle(A,B)$ as $J$
- Draw $mathrmCirlce(I,J)$
- Mark the intersection of $mathrmCircle(I,J)$ and $overlineHC$ as $L$
- Mark the intersections of $mathrmCircle(I,J)$ and $mathrmCircle(C,B)$ as $M$ and $K$.
- Draw $overlineML$
- Draw $overlineKL$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineML$ as $N$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineHC$ as $O$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineKL$ as $P$
$MKPON$ is a regular pentagon.
$endgroup$
4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem.
- Draw $mathrmCircle(A,B)$
- Draw $overlineAB$
- Mark the intersection of $mathrmCircle(A,B)$ and $overlineAB$ as $C$
- Draw $mathrmCircle(B,C)$
- Draw $mathrmCircle(C,B)$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $D$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineAB$ as $E$
- Draw $overlineDC$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineDC$ as $F$
- Mark the intersection of $mathrmCircle(C,B)$ and $mathrmCircle(B,C)$ as $G$
- Draw $overlineBG$
- Mark the intersection of $overlineBG$ and $overlineEF$ as $H$
- Draw $overlineHC$
- Mark the intersection of $overlineHC$ and $mathrmCircle(C,B)$ as $I$
- Draw $overlineIA$
- Mark the intersection of $overlineIA$ and $mathrmCircle(A,B)$ as $J$
- Draw $mathrmCirlce(I,J)$
- Mark the intersection of $mathrmCircle(I,J)$ and $overlineHC$ as $L$
- Mark the intersections of $mathrmCircle(I,J)$ and $mathrmCircle(C,B)$ as $M$ and $K$.
- Draw $overlineML$
- Draw $overlineKL$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineML$ as $N$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineHC$ as $O$
- Mark the intersection of $mathrmCircle(C,B)$ and $overlineKL$ as $P$
$MKPON$ is a regular pentagon.
answered 6 hours ago
Sriotchilism O'ZaicSriotchilism O'Zaic
36.7k10 gold badges164 silver badges375 bronze badges
36.7k10 gold badges164 silver badges375 bronze badges
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2
$begingroup$
This is like the game I have on my phone called Euclidea.
$endgroup$
– mbomb007
10 hours ago
$begingroup$
closely related: codegolf.stackexchange.com/q/38653/15599
$endgroup$
– Level River St
9 hours ago
$begingroup$
"setting it two a radius" a likely typo
$endgroup$
– Grzegorz Oledzki
9 hours ago
1
$begingroup$
Next time you should ask people to draw a heptagon, which would be slightly more challenging:)
$endgroup$
– flawr
8 hours ago