New answers tagged geometry
1
C (gcc) (MinGW), 277 276 bytes
-1 byte thanks to @ceilingcat
Added -lm compiler flags to TiO for illustrative purposes, but MinGW does not need it.
Outputs a 3-shade PGM file to STDOUT.
q,i,c;main(s,v)char**v;{float u=atoi(v[1])/12.,d,r;q=u*15;for(*v=memset(calloc(q,q+1),2,i=q*q);i--;i[*v]-=d<6&d>5|r>6&r<7|r>8&r<9&&r&...
4
Python 3.8 (pre-release), 1533 bytes
def w(n,ll,ans):
global p,q
from math import sin,cos,pi,atan2
def y(s,e,f,a,b):
x,y=f(s),f(e)
g=lambda a,b,x:0<=(x-a)%2<=b-a
while e-s>1e-15:
m=(s+e)/2
z=f(m)
if x*z<=0:
e,y=m,z
else:
s,x=m,z
&#...
4
Google Sheets, 277 bytes
=ARRAYFORMULA((POW(COLUMN(A1:CV99)-5*A1,2)+POW(ROW(A1:CV99)-5*A1,2)<A1*A1)+(POW(COLUMN(A1:CV99)-5*A1,2)+POW(ROW(A1:CV99)-5*A1,2)>9/4*A1*A1)*(POW(COLUMN(A1:CV99)-5*A1,2)+POW(ROW(A1:CV99)-5*A1,2)<25*A1*A1)*IFERROR((MOD(3*ATAN2(COLUMN(A1:CV99)-5*A1,ROW(A1:CV99)-5*A1)/PI(),2)<1),0
Creates a grid of 1s and 0s which can be ...
2
Mathematica (Wolfram Language) 76 71 Bytes
This solution draws the radiation symbol using only the Disk[] primitive, which allows for arcs.
By default a Graphics object in Mathematica has a White background and objects are Black unless otherwise specified. So we first draw three black disk sectors, then a small white disk and finally a smaller inner black ...
1
Runic Enchantments, 216 bytes
D::i</~1-:0)?;{:}ak$2?X2 8?1]$#'~~\1?7 c?1$ '~~\?3 L
R:0)?/1-:̹{2,::}}-S{-2pS2p+'qA:{5,:}≮?/:}3,2*S:}S≯?/5,)?/]:̹{2,:}-S{-:̹'-A}'-A0)3*?P2?Z0]{{S:0)a*?:0(4*?~~10,'|A'aA{0)?S-:0≮4*?P2*+3*2P*%P)4*?'#2?' $]
Try it online!
Dear god. Why.
So much went wrong.
It makes me cry.
Several major impediments that ended ...
3
Mathematica, 93 88 85 76 74 70 bytes
RegionPlot[1.5<#<5&&Sin[3#2]>0||#<1&@@AbsArg[x+I y],{x,-5,5},{y,-5,5}]
-4 bytes thanks to Roman.
It is not clear from the description whether it is required to get rid of the outline and make it black on white, but I think the submission looks best and is shortest like this.
2
PostScript, 183 175 bytes
Code (compressed version):
20 20 scale 9 6 translate<</f{closepath fill}/c{0 360 arc f}/l{10 0 lineto -60 rotate}>>begin 0 0 5 c 1 setgray 3{0 0 moveto l l f} repeat 0 0 1.5 c 0 setgray 0 0 1 c showpage
Code (uncompressed version):
20 20 scale % over-all scale
9 6 translate % over-all shift
% ...
4
Desmos/Cartesian Plane, 103 97 86 84 bytes
(23 arithmetic and comparison operators)
(abs(mod(arctan(x,-y)/\pi*6,4)-2)-1)*max(0,182-abs(x^2+y^2-218))+max(0,16-x^2-y^2)>0
https://www.desmos.com/calculator/hadw7gayfn
1
OpenSCAD, 153 bytes
module a(){polygon([[0,0],[5,0],[5,10]]);}module b(){a();rotate(120)a();rotate(240)a();}circle();intersection(){difference(){circle(5);circle(1.5);}b();}
Uncompressed:
module triangle()
{
polygon(points=[[0,0],[5,0],[5,10]]);
}
module triangles() {
triangle();
rotate(120) triangle();
rotate(240) triangle();
}
module ...
2
Python 3.8, 129 124 122 120 119
def f(N):[print((abs(a:=(i%N+i//N*1j)*2/N-1-1j)<.2)+((a**3).imag<0<.3<abs(a)<1),end=(-i%N==1)*"\n")for i in range(N*N)]
Output (N=20):
00000000000000000000
00000000000000000000
00000100000000010000
00011100000000011100
00011110000000111100
00111111000001111110
01111111000001111111
01111111100011111111
...
2
Ruby, 91 100 bytes
->s{(w=-(s/=2)..s).map{|x|w.map{|r|a=x*x+r*=r;' O'[a<(b=s*s)/25||a>b/11&&a<b&&x*(r*4-a)<0?1:0]}*''}}
Try it online!
Accepting 1 parameter (size of the image).
Example output for s=40:
O O
OO OO
OOOOO OOOOO ...
14
SVG(HTML5), 128 126 124 120 bytes
(Thanks Wheat Wizard for saving 2 bytes by using .2 rather than 0.2)
(Drawing triangles from right to left saves 2 bytes)
(Thanks Grimmy for saving 2 bytes by using a modified path)
(And reduced another 2 bytes by using a different starting point on the path)
<svg viewbox=-2,-2,4,4>
<circle r=1 />
<...
4
Logo, 96 bytes
circle 16
rt 90
repeat 3[arc 80 60
arc 24 60
pu
repeat 2[fw 24
pd
fw 56
pu
backward 80
rt 60]pd]
Try it online! Draws just the border.
11
JavaScript (ES7), 118 113 98 93 bytes
Saved 15 bytes thanks to @tsh!
Takes a parameter \$w\$ as input and draws an ASCII art of width \$2w-1\$ and height \$2w\$, using 0 and 1 for the 'pixels'.
f=(x,w=y=x)=>y+w?(--x+w?(r=x*x+y*y,4*x*x>r^y<0?1019:3)>>r**.5/w*10&1:(x=w,y--,`
`))+f(x,w):''
Try it online!
How?
We compute the ...
1
GLSL / Shadertoy, 181 bytes
void mainImage(out vec4 o,vec2 c)
{
vec2 p=(c-iResolution.xy/2.)/iResolution.y;
float k=length(p),h=acos(-1.)/3.;
o=step(0.,min(k-.1,max(k-.5,.15-k)*step(mod(atan(p.y,p.x),h+h),h)))-o;
}
Copy and paste this code into the editor then press the play button to run the shader.
The shader uses the signed distance of the ...
6
Haskell (code.world dialect), 108 86 bytes
drawingOf(c(2)&colored(c(3),white)&0%1&2%3&4%5)
c=solidCircle
a%b=sector(a*60,b*60,10)
Run on code.world!
33
BBC BASIC, 200 bytes
f=500
g=1/SQR(3)
CIRCLE FILLf,f,f
GCOL15
MOVE-100,f
MOVEf,f
PLOT85,f*(1-g),0
MOVE1100,f
MOVEf,f
PLOT85,f*(1+g),0
MOVEf*(1-g),2*f
MOVEf,f
PLOT85,f*(1+g),2*f
CIRCLE FILLf,f,150
GCOL0
CIRCLE FILLf,f,100
Somehow outgolfing Python, this works by drawing a large black circle, filling in the white triangular sections, drawing a central white ...
7
C (gcc) (MinGW), 245 244 242 236 223 221 bytes
-6 bytes thanks to ceilingcat.
Added -lm on TiO so that it works there. MinGW does not require it.
Outputs a 3-shade PGM file to STDOUT. Takes width (height is the same) of image as a command line argument.
main(q,v,r,c,m)char**v;{*v=calloc(q+1,q=atoi(v[1]));float R=q/10.1,d;for(r=m=q/2;r--+m;)for(c=m;c--+m;...
14
Jelly, 42 bytes
÷ÆṬ÷ØPḂ×3ḞḂ
æịA©>5,1.5aç^/o®Ị¤
×5ŒR÷µçþAZY
Try it online!
A full program that takes an integer n that determines the radius of the central circle and implicitly outputs to STDOUT a \$10n+1\$ square with the desired image encoded as 1 for black and 0 for white.
Per @flawr, this is a permitted output format. Note also that this is ...
4
TikZ, 170 bytes
\documentclass[tikz]{standalone}\begin{document}\tikz{\fill circle(10);\foreach\a in{0,120,240}{\fill[white]circle(3)--(\a-60:13)--(\a:13);}\fill circle(2)}\end{document}
4
Red, 237 218 bytes
Red[needs: 'View]
g: func[n][a: compose[arc(s: n / 2.0 * 50x50)(s)60 60 closed
rotate 120(s)]view compose/deep[base(s * 2)draw[(f: 'fill-pen)black(a)(a)(a)
pen gray line-width(0.1 * s/1)(f)black circle(s)(s/1 / 4.0)]]]
Takes a parameter n - the width/height in multiples of 50.
3
python 3.8, 220
from PIL import Image as I
from math import*
d=300
h=d//2
r0=d/10
g=range(-h,h)
i=I.frombytes('L',(d,d),bytes([255,0][1.5<(r:=(x*x+y*y)**.5)/r0<5 and(atan2(x,y)/pi+7/6)*1.5%1.>.5 or r<r0]for y in g for x in g))
i.show()
The idea behind is very simple: we count (x,y) from the center of image and r=(x*x+y*y)**.5 is the distance ...
17
MATL, 46 44 41 bytes
:t_0vSt!Yy1MZ;3*YP/koyG.3*>*yG<*wG5/<=3YG
Given an input n, the image has width 2*n+1 in pixels.
Try it at MATL Online!
Explanation
: % Implicit input n. Range [1 ... n]
t_ % Duplicate, negate. Gives [-1 ... -n]
0 % Push 0
v % Concatenate stack as a column vector: [1; ...; n; -1; ...; -n; 0]
S % ...
14
SVG(HTML5), 175 bytes
<svg viewBox=-24,-24,48,48><circle r=4 /><path id=b d=M3,-5.2A6,6,0,0,1,6,0H20A20,20,0,0,0,10,-17.32z /><use href=#b transform=rotate(120) /><use href=#b transform=rotate(240)
Originally based on @Arnauld's answer, but removes unnecessary characters, scales the numbers by 80% to make them golfier, ...
7
HTML / SVG, 201 189 bytes
A revamped version of this file from Wikimedia Commons.
<svg viewBox=-30,-30,60,60><circle r=5 /><path id=b d=M7.5,0A7.5,7.5,0,0,0,3.7,-6.5L12.5,-21.6A25,25,0,0,1,25,0z /><use href=#b transform=rotate(120) /><use href=#b transform=rotate(240) />
0
Jelly, 23 bytes
×ḤÆS÷ÆS²¥_ÆTİ×ɗS
ØP÷ç÷4
Try it online!
A full program that takes the number of sides as its first argument and the list of folds as its right. Returns/prints the area of the folded polygon.
Loosely based on the formula in @Arnauld’s answer.
Top 50 recent answers are included
