Draw the Ionising Radiation Hazard Symbol

Question

Draw the ionising-radiation-hazard-symbol in an arbitrary colour on a distinctly coloured background. The specific proportions were published in the June 27th 1974 issue of the Federal Register of the US Government.

Details

• As output, writing to a file (raster and vector formats are permitted) or displaying on the screen are both allowed.

• You can draw just the border or the filled shape.

• If you use raster images, you should take a parameter (or two) as input that lets you adjust the resolution of the output (e.g. width/height).

• The background must at least have the size of the bounding box of the symbol but may be larger.

• Outputting the unicode symbol ☢ is not sufficient.

• The exact ratios of the used distances are given in following diagram (which was originally from here):

Related: Draw the ☣ (Biohazard Symbol)

Answer 1

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 circle and finally drawing the central black circle.

Answer 2

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       % Sort. Gives [-n; -n+1; ... n]
t!      % Duplicate, transpose. Gives the row vector [-n -n+1 ... n]
Yy      % Hypotenuse, with implicit expansion. Gives a matrix containing radius
        % for all points in a grid from -n to -n
1M      % Push the two inputs of the last function, again
Z;      % atan2, with implicit expansion. Gives a matrix containing angle between
        % -pi and pi, where 0 is real positive semiaxis
3*YP/k  % Multiply by 3, divide by pi, round down. Gives an integer for each sextant
o       % Parity. This assigns 0 and 1 alternately to points in different sextants.
        % See this partial result as IMAGE 1 below.
y       % Duplicate from below: pushes the matrix of radii again
G.3*>   % Greater than 0.3*n?, element-wise. Gives 0 or 1 for each entry. This
        % realizes the "1.5R" circle from the specification. See IMAGE 2
*       % Multiply, element-wise. See IMAGE 3
y       % Duplicate from below: pushes the matrix of radii again
G<      % Less than n?, element-wise. Gives 0 or 1 for each entry. This realizes
        % the "5R" circle from the specification. See IMAGE 4
*       % Multiply, element-wise. See IMAGE 5
w       % Swap: pushes the original matrix of radii to the top
G5/<    % Mess than 0.2*n? This realizes the "R" circle in the spec. See IMAGE 6
=       % Equal?, element-wise. This combines the two matrices so far to produce
        % the final result, with 0 for foreground pixels and 1 for background pixels.
        % See IMAGE 7
3YG     % Show image. 0 is displayed as black, 1 as white

Intermediate results. 0 is shown as black, 1 as white:

• Image 1:

• Image 2:

• Image 3:

• Image 4:

• Image 5:

• Image 6:

• Image 7 (final result):

Answer 3

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, and draws the curved section's lines in a different order so that I can use an H instead of an L.

Answer 4

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 />
  <circle r=.3 fill=#FFF />
  <path d=m2,3.4V0h-4v3.4l4,-6.8h-4 fill=#FFF />
  <circle r=.2

<svg viewbox=-2,-2,4,4><circle r=1 /><circle r=.3 fill=#FFF /><path d=m2,3.4V0h-4v3.4l4,-6.8h-4 fill=#FFF /><circle r=.2

Rather than basing the SVG on wikipedia's vector image - which is drawn in a reasonable way, this solution draws:

1. A black large circle <circle r=1 />
2. A smaller white circle <circle r=.3 fill=#FFF />
3. Three white triangles as a single continuous path <path d=m2,3.4V0h-4v3.4l4,-6.8h-4 fill=#FFF />
4. And a small black circle <circle r=.2

Taking advantage of modern browsers forgiving interpretation which ignores unclosed tags (saved the final 9 bytes of /></svg> )

This is the result:

And this is the same code with the path in yellow to demonstrate what's going on:

Answer 5

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 effectively a PBM file without the header. Adding P1 L L to the start (where L is replaced by \$10n+1\$) would make it a valid PBM file. Ta

Explanation

Helper link 2

Dyadic link taking y coordinate as left and x as right argument; returns whether the angle is within the filled outer area

÷           | Divide (y by x)
 ÆṬ         | Arc-tangent
   ֯P      | Divide by pi
      Ḃ     | Mod 2
       ×3   | Times 3
         Ḟ  | Floor
          Ḃ | Mod 2

Helper link 1

Dyadic link taking y coordinate as left and x as right argument

æị                 | Convert to complex
  A                | Absolute (i.e. distance from origin)
   ©               | Copy to register
    >5,1.5         | Greater than 5,1.5 (vectorises, so will give two booleans)
          a        | Logical and with:
           ç       | - Result of helper link 2 applied to the input of this link
            ^/     | Reduce using exclusive or
              o  ¤ | Or following as a nilad:
               ®   | Value in register
                Ị  | Absolute of this <= 1

Main link

×5         | Times 5
  ŒR       | Range from -this to +this
    ÷      | Divide by input
     µ     | Start a new monadic chain
      çþ   | Outer table using helper link 1 and:
        A  | - Absolute of the range as the right argument
         Z | Transpose rows and columns

Sample image for n=5

Answer 6

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 squared distance from the center:

$$r=x^2+y^2$$

We use \$\sqrt{r}\$ to quantize the distance from the center into 10 bins of width \$w\$:

$$\left\lfloor\frac{10\sqrt{r}}{w}\right\rfloor$$

We use \$r\$, \$x\$ and \$y\$ to determine in which kind of sector we are:

$$(4x^2>r)\text{ xor }(y<0)$$

See this graph!

We turn the above result into a drawing bit-mask, as summarized in the following schema:

Hence the value of the pixel at \$(x,y)\$:

(
  r = x * x + y * y,
  4 * x * x > r ^ y < 0 ?
    1019
  :
    3
)
>> r ** 0.5 / w * 10
& 1

Answer 7

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 coloured using a conditional format. It takes input in cell A1 as the value of R.

Here's a link.

Answer 8

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) />

Answer 9

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 up making this both (a) a nearly unreadable mess and (b) frustratingly difficult.

The first problem was the fact that Runic doesn't have atan2(x,y), so I had to heckin' work that out myself. And oh yeah, make sure its reasonably golfed, it had to be dropped into an existing program with no alterations, so in-line conditional logic was preferable.

The second problem was that I had a weird issue where I'd multiply the tangent value by 3, modulo to 2pi and get...nonsense. What even is that? Why is it not using angular values? Turned out I had a stray extra value at the bottom of the stack that was throwing the math off. Whoops.

THEN
AND THEN
The program would just suddenly die after the first four bytes of the outer wedges. Oh my, what fun we had. Lets simplify, shall we? Lets just isolate the inputs at that point and... oh. It works fine.

I.

Uh.

Hmm.

When in doubt, check the source code. Can you spot the difference?

Yep. Integer values are totes allowed to divide by zero (and return infinity) and doubles aren't. Oh and that's not even true in the original Unity C# code (which I use as a graphical debugger) as it was an intentional design decision to terminate on div-by-zero.

Wot. Also, conversion to float instead of double for no reason. Just sittin' there.

(╯°Д°)╯︵┻━┻

Explanation

I need a drink first.

That's better.

D::i<                     Set up the stack, we need a current position (X,Y) and total size. Init all to read value, counting down is easier than counting up.
                          The D here, and the R below it, constitute the main loop entry
R:0)?\1-                  Compare current X against zero, else decrement
     \~1-:0)?;{:}ak$      if zero, decrement Y. If Y is zero, terminate, else return to the D on the first line (there's a lot of skip-over-logic)
:̹ {2,::}}-S{-             dupe the entire stack, then translate origin by 1/2 width
2pS2p+'qA                 square root (x*x+y*y)
:{5,:}≮?\:}3,2*S:}S≯?\    compare that against 1/20th the width and 1.5/20th the width (R and 1.5R)
  ]$#'~~L       $ '~~L    Print # or <space> for the inner circle (and return to D) skip-bypasses omitted
5,)?/                     Compare against 1/10th the width (5R), print a space and return to D 

]:̹ {2,:}-S{-:̹ '-A}'-A0)3*?P2?Z0]{{S:0)a*?:0(4*?~~10,'|A'aA{0)?S-    Dupe the entire stack and compute atan2(x,y)

:0≮4*?P2*+                       Convert negative values to values greater than pi up to 2 pi
          3*2P*%P)4*?'#2?' $]    Multiply by 3, modulo 2pi, compare against pi. Less: print # else print space. Return to R.

Be wary of improperly stacking unicode combining characters. The little ̹ makes the command apply to the stack-of-stacks and ̸ is boolean-not; eg. ≮ is "not less than (aka greater-equals)." Looks like it costs bytes, but because of the upper line, there's a byte savings in that line being shorter, which pays for the +1 byte cost for an extended character in the lower line (typically assuming that the lower line could be ?!/ instead of ?/).

Answer 10

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!

Answer 11

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;m[*v+(r+m)*q+c]=d<R*5&&d<=R|d>=1.5*R&asin(r/d)+1.5707>(r>0?5:1)*0.5235?:2)d=hypot(c,r);printf("P5 %d %d 2 %s",q,q,*v);}

Try it online!

Answer 12

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}

Answer 13

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.

Answer 14

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

Answer 15

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 disk. Unlike @Mario Carneiro's solution (which is a nice use of RegionPlot), this comes without axes and without the artifacts produced by the inner workings of RegionPlot.

Saved a byte by letting R=2

Graphics@{(d={0,0}~Disk~##&)[10,{#,#+1}Pi/3]&/@{0,2,4},{White,d@3},d@2}

Answer 16

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.

Answer 17

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

% define some short-named procedures for later use
<<
  /f { closepath fill }
  /c { 0 360 arc f }  % filled circle (x, y, radius are taken from stack)
  /l { 10 0 lineto
       -60 rotate }   % long line + rotate by 60°
>> begin

0 0 5 c               % big black circle
1 setgray             % set white color
3 {
  0 0 moveto l l f    % white triangle + rotate by 120°
} repeat
0 0 1.5 c             % white circle
0 setgray             % set black color
0 0 1 c               % small black circle
showpage

Result (as animation to see how it is drawn):

Answer 18

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
01111111000001111111
01111111011101111111
00000000011100000000
00000000011100000000
00000000000000000000
00000000010100000000
00000000111110000000
00000000111110000000
00000001111111000000
00000011111111100000
00000011111111100000
00000011111111100000

The idea is to represent each point in the plane by a complex number \$x\$. Let the radius of the innermost circle be \$2\$. Then \$x\$ is inside the inner circle if \$|x|<2\$.

The outer parts are a bit more involved. To lie in one of the outer parts, we have to have \$3=1.5\cdot 2< |x| < 5\cdot 2 = 10\$ so that the distance is neither too small nor too large. We also have an angle restriction: the polar angle of \$x\$ has to lie between \$60^\circ\$ and \$120^\circ\$, when taken mod \$120^\circ\$. But if we instead look at \$x^3\$, the angle is multiplied by three, so that the polar angle of \$x^3\$ has to lie between \$180^\circ\$ and \$360^\circ\$ (mod \$360^\circ\$). This is just equivalent to \$\Im(x^3)<0\$.

To put it all together, we have to have \$|x|<2\$ or both of \$3<|x|<10\$ and \$\Im(x^3)<0\$ for \$x\$ to be inside the image. As both conditions are mutually exclusive, their truth values can just be added together.

Answer 19

TikZ, 93 bytes

The last comma ; at the end of the code can be missed.

\foreach\i in{0,120,240}\fill[rotate=\i](60:3)arc(60:0:3)--(0:10)arc(0:60:10);\fill circle(2)

TikZ, 164 bytes

\documentclass[tikz]{standalone}\begin{document}\tikz{\foreach\i in{0,120,240}\fill[rotate=\i](60:3)arc(60:0:3)--(0:10)arc(0:60:10);\fill circle(2)}\end{document}

Asymptote 109 bytes

// Run on http://asymptote.ualberta.ca/
for(int i=0;i<3;++i)fill(rotate(i*120)*(arc((0,0),3,60,0)--arc((0,0),10,0,60)--cycle));fill(circle((0,0),2));

Asymptote 117 bytes

// Run on http://asymptote.ualberta.ca/
for(int i: new int[]{0,120,240})fill(rotate(i)*(arc((0,0),3,60,0)--arc((0,0),10,0,60)--cycle));fill(circle((0,0),2));

Answer 20

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 between point and the center.
The point is black if either r<r0 where r0 is 1/5 distance from center to a side
or if 5*r0<r<1.5*r0 (see the diagram in the question) and the angle from center to the point (computed with math.atan2) is in selected ranges. 1.5 and 7/6 there were obtained almost empirically, but could be reasoned too.
The other (most) part is a wrap-in.

Answer 21

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.

Answer 22

Desmos, 3+70 64=73 67 bytes

After seeing Will Chen's answer and seeing that the online link no longer seemed to work, I figured I'd start at it from scratch, and ended up with 11 fewer bytes.

r<2
max(0,mod(-sign(x)arctan(y/x),120)-60)(x^2+y^2-9)(x^2+y^2-100)<0

View it online

Edit: realized max is unnecessary for the latter two functions. It's required for the first function to zero out a lot of areas, but this might be able to be improved. This makes this the shortest answer in a non-golfing language, beating out Mathematica, with some potential for improvement.

This originally used a bunch of piecewise conditionals, which simply evaluate to 1 if true and undefined if false, making them handy for this purpose. However, by just multiplying them through and comparing to 0, we can check for polarity in a shorter way. The reason we do <0 at the end instead of >0 like you may expect is that the last block, checking if r>10, needs to evaluate to false (negative). The only other notable thing is the polarity check in the first part because arctan(y/x) doesn't necessarily work as well for angle-finding as you might hope.

Answer 23

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       
      OOOOOO                 OOOOOO      
     OOOOOOOO               OOOOOOOO     
     OOOOOOOOO             OOOOOOOOO     
    OOOOOOOOOO             OOOOOOOOOO    
   OOOOOOOOOOOO           OOOOOOOOOOOO   
   OOOOOOOOOOOO           OOOOOOOOOOOO   
  OOOOOOOOOOOOOO         OOOOOOOOOOOOOO  
  OOOOOOOOOOOOOO         OOOOOOOOOOOOOO  
 OOOOOOOOOOOOOOOO       OOOOOOOOOOOOOOOO 
 OOOOOOOOOOOOOOOO       OOOOOOOOOOOOOOOO 
 OOOOOOOOOOOOOOO         OOOOOOOOOOOOOOO 
 OOOOOOOOOOOOOO   OOOOO   OOOOOOOOOOOOOO 
 OOOOOOOOOOOOOO  OOOOOOO  OOOOOOOOOOOOOO 
 OOOOOOOOOOOOOO  OOOOOOO  OOOOOOOOOOOOOO 
                 OOOOOOO                 
                 OOOOOOO                 
                 OOOOOOO                 
                  OOOOO                  


                 OOO OOO                 
                OOOOOOOOO                
                OOOOOOOOO                
               OOOOOOOOOOO               
               OOOOOOOOOOO               
              OOOOOOOOOOOOO              
              OOOOOOOOOOOOO              
             OOOOOOOOOOOOOOO             
            OOOOOOOOOOOOOOOOO            
            OOOOOOOOOOOOOOOOO            
           OOOOOOOOOOOOOOOOOOO           
           OOOOOOOOOOOOOOOOOOO           
            OOOOOOOOOOOOOOOOO            
              OOOOOOOOOOOOO

Answer 24

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 radiation()
{
    circle($fn = 100);
    intersection() {
        difference() {
            circle(5, $fn = 100);
            circle(1.5, $fn = 100);
        }
        triangles();
    }
}

radiation();

Notice the extra $fn = 100 in the circles, which are used to make them higher-resolution polygon. We could add $fs=.1; at the beginning of the file, which would also improve the resolution.

Answer 25

SVG(HTML5), 114 bytes

<svg viewBox=-1,-1,2,2><circle r=.65 fill=none stroke=#000 stroke-width=.7 stroke-dasharray=0,.68,0 /><circle r=.2

An alternative solution to the other SVG answers. Uses the stroke dash property to create the 'wings'.

The value of the .68 should be equal to 0.65 / 3 * PI, or 0.680678408... to be more accurate.

If incorrect rotation is allowed, this can be reduced to 110 bytes:

<svg viewBox=-1,-1,2,2><circle r=.65 fill=none stroke=#000 stroke-width=.7 stroke-dasharray=.68 /><circle r=.2

Answer 26

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 shapes to be able to use the union and intersection operations to generate the symbol.

Answer 27

TI-BASIC (TI-84+), 1042 bytes

"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→Str1
AxesOff
15→N
4→dim(∟A
For(I,1,60N
sub(Str1,I,1
inString("0123456789ABCDEF",Ans)-1→D
For(B,4,1,-1
D≥2(B-1→∟A(B
If ∟A(B
D-2^(B-1→D
End
For(E,1,4
If ∟A(5-E
Pxl-On(iPart(I/N),E+4remainder(I,N
End
End

Oh man, this one was rough. The TI series of calculators do have shading functionality between functions, but that just don't cut it.
Furthermore, I was limited to 60x60 pixels, since drawing on the 62nd row or lower would result in ERR: ARGUMENT error being thrown and 60 fits the radii ratios mentioned in the challenge perfectly: $$ R = 12 \\ 1.5R = 18 \\ 5R = 60$$

So, I opted to convert a string of hexadecimal characters to pixels.

The pixels drawn to the screen will look like this:

The very long string you see on the first line of the program looks like this when formatted:

000000000000000
000000000000000
000000000000000
000000000000000
000300000000000
000780000006000
001FC000000F800
003FC000001FC00
007FE000001FE00
00FFE000003FF00
01FFF000003FF80
03FFF000007FFC0
03FFF80000FFFC0
07FFFC0000FFFE0
0FFFFC0001FFFF0
0FFFFE0001FFFF0
1FFFFE0003FFFF8
1FFFFF0007FFFF8
3FFFFF8007FFFFC
3FFFFF800FFFFFC
3FFFFFC00FFFFFC
7FFFFFC01FFFFFE
7FFFFFC01FFFFFE
7FFFFF801FFFFFE
7FFFFF0F0FFFFFE
FFFFFE3FC7FFFFF
FFFFFE7FE7FFFFF
FFFFFC7FE3FFFFF
FFFFFCFFF3FFFFF
FFFFFCFFF3FFFFF
000000FFF3FFFFF
000000FFF000000
0000007FE000000
0000007FE000000
0000003FC000000
0000000F0000000
000000000000000
000000604000000
000000FFE000000
000001FFE000000
000001FFF000000
000003FFF800000
000007FFF800000
000007FFFC00000
00000FFFFC00000
00000FFFFE00000
00001FFFFF00000
00003FFFFF00000
00003FFFFF80000
00007FFFFF80000
00007FFFFFC0000
0000FFFFFFC0000
0001FFFFFFE0000
0001FFFFFFF0000
0003FFFFFFF0000
0001FFFFFFF8000
0000FFFFFFF0000
00003FFFFFC0000
000007FFFE00000
0000007FE000000

Each hexadecimal digit encodes to a series of 4 pixels drawn to the screen. For each bit set in the digit, a pixel will be drawn.

Explanation:

"000 ... 000→Str1                       ;set the digit string to convert
AxesOff                                 ;get rid of the axes on the graph since
                                        ;  it'll be in the way
15→N                                    ;set N to 15 in order to save a few bytes
4→dim(∟A                                ;set the length of the list A to 4 elements
For(I,1,60N                             ;loop 900 times.  There are 60 lines to draw
                                        ;  and 15 digits per line to convert
sub(Str1,I,1                            ;get the current hexadecimal digit
inString("0123456789ABCDEF",Ans)-1→D    ;convert it to base 10, store the result in D
For(B,4,1,-1                            ;convert number to its binary representation
D≥2(B-1→∟A(B
If ∟A(B
D-2^(B-1→D
End
For(E,1,4                               ;loop over each bit in said representation
If ∟A(5-E                               ;and only draw the pixel if this bit is set
                                        ; (bit order is reversed)
Pxl-On(iPart(I/N),E+4remainder(I,N      ;pixel row (Y) is I/N, pixel column (X) is
                                        ; 4*(I%N)+E
End
End

---

Note: TI-BASIC is a tokenized language. Character count does not equal byte count.

Answer 28

T-SQL, 185 bytes

DECLARE @ GEOMETRY='POINT(6 6)',@b GEOMETRY='POLYGON((12 6,9 0,3 12,9 12,3 0,0 6,12 6))'SELECT @.STBuffer(5).STDifference(@b.MakeValid().STUnion(@.STBuffer(1.5))).STUnion(@.STBuffer(1))

I shifted the center from (0,0) to (6,6) to reduce the number of negative numbers needed; this saves a byte or two.

This code isn't quite long enough to benefit from the REPLACE() trick that was needed for the Biohazard symbol.

This was done using SQL geo-spatial features introduced in SQL 2008.

Visual of output, with annotated code:

For more T-SQL drawing, see my Biohazard symbol and my Easter Bunny.

Answer 29

Desmos, 60 bytes

w=hsv(0,0,1)
r<5\{[0,3,6]40<\theta<[1,3,5]60\}
xx+yy<9/4
r<1

Try it on Desmos

This is the version without scaling, now the shortest Desmos answer.

Desmos, 178 161 153 126 119 72 68 67 bytes

R=1
w=hsv(0,0,1)
r<5R\{[0,3,6]40<\theta<[1,3,5]60\}
xx+yy<9RR/4
r<R

Try it on Desmos

-12 bytes thanks to @Steffan

-27 bytes by switching to degrees and extracting the "arms" to a function

-55 bytes and fix thanks to @Aiden Chow

-1 byte thanks to emanresu A

By far the longest Desmos answer yet, this is now the second shortest answer. Oh, and mine is scalable, just move the \$R\$ slider.

Answer 30

Processing, 172 bytes

noStroke();background(255);fill(0);for(float i=-60;i<=300;i+=120)arc(50,50,100,100,radians(i),radians(i+60),PIE);fill(255);circle(50,50,30);fill(0);circle(50,50,20);

Un-minified code:

noStroke();
background(255);
fill(0);
for(float i=-60;i<=300;i+=120)arc(50,50,100,100,radians(i),radians(i+60),PIE);
fill(255);
circle(50,50,30);
fill(0);
circle(50,50,20);

Output