8

In a Geometry problem appears the figure inscribed in a circle, having outline consisting of $8$ ~semicircles, see figure on the left. The radius of each semicircle is $1$.

enter image description here

By doing some calculations, it's easy to realize that the radius of the circle is $\sqrt{10}$

For the solution of the problem, it is interesting to observe the figure on the right.

\begin{center}

\begin{tikzpicture}

\draw (1,1) circle [radius=3.16];

\draw (0.5,0) arc(0:-180:1);

\draw (1.5,0) arc(0:180:1);

\draw (0,1.5) arc(-90:-270:1); 

\end{tikzpicture}

\end{center}
3
  • John Kormylo, the radius is $\sqrt{10}$. Feb 29, 2020 at 14:44
  • Assuming the tangent occurs at 60 degrees, (2+\sqrt(3)/2)^2+1.5^2 is not 10. A test draw seems to fit this. Feb 29, 2020 at 15:11
  • I'm sorry, but I can form a right triangle, inscribed in the circle, whose catetos measure 2 and 6 respectively. Hypotenuse is the diameter of the circle. Feb 29, 2020 at 15:18

6 Answers 6

4

And a version in Metapost, relying on the fact that the tangent point on the small circle must be the same point on the circumcircle, and using some nice colours.

enter image description here

prologues := 3;
outputtemplate := "%j%c.eps";

input colorbrewer-cmyk

beginfig(1);

    path base, edge, propeller, circumcircle, square;

    base = halfcircle shifted 1/2 right;
    edge = (base & reverse base rotated 180) shifted up scaled 89;
    propeller = for i=0 upto 3: edge rotated 90i .. endfor cycle;
    circumcircle = fullcircle scaled 2 abs(point 1/45 angle 
                    1/2[point 0 of edge, point 4 of edge] of edge);
    square = for i=0 upto 3: point 9i of propeller -- endfor cycle;

    picture P[];
    P1 = image(
        fill circumcircle withcolor Blues 8 5;
        fill propeller withcolor Blues 8 4;
        draw propeller;
    );
    P2 = image(
        fill square withcolor Blues 8 5;
        fill propeller withcolor Blues 8 4;
        clip currentpicture to square;
    );
    P3 = image(
        fill propeller withcolor Blues 8 2;
        draw P2;
        draw propeller;
    );
    draw P1;
    draw P3 shifted 300 right;

endfig;
end.
10

Not an answer, but should settle the issue of the radius.

derivation

2
  • John Kormylo - I recognize I made a mistake. In fact, the radius can't be $sqrt{10}$. It was really nice to see you! Mar 1, 2020 at 10:59
  • Very kind John...I have removed my question on the rocket. Always thank you.
    – Sebastiano
    Oct 17, 2020 at 8:12
9

You don't need to know the radius of the big circle

\documentclass{standalone} 
\usepackage[dvipsnames,svgnames]{xcolor}
\usepackage{tkz-euclide}

\begin{document} 

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B,2/2/O,3/4/X,4/1/Y,1/0/Z,
              0/3/W,3/0/R,4/3/S,1/4/T,0/1/U}
\tkzDefSquare(A,B)\tkzGetPoints{C}{D}
\tkzInterLC(O,X)(X,C) \tkzGetSecondPoint{F}
% or \tkzDefPointWith[colinear normed=at X,K=1](O,X) \tkzGetPoint{F}
\begin{scope}
  \tkzFillCircle[fill=MidnightBlue](O,F)
  \tkzFillPolygon[purple!40](A,...,D)
  \tkzClipPolygon(A,...,D)
  \foreach \c/\t in {S/C,R/B,U/A,T/D}
  {\tkzFillCircle[MidnightBlue](\c,\t)}
\end{scope}
\foreach \c/\t in {X/C,Y/B,Z/A,W/D}
{\tkzFillCircle[purple!40](\c,\t)}
  \foreach \c/\t in {S/C,R/B,U/A,T/D}
  {\tkzFillCircle[MidnightBlue](\c,\t)}
\end{tikzpicture}
\end{document}

enter image description here

9

With the radius r=1+sqrt(5) (easy to get!).

enter image description here

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\def\cc{(2,0) arc(270:90:1) arc(0:180:1) arc(0:-180:1) arc(90:270:1) arc(90:-90:1) arc(180:360:1) arc(180:0:1) arc(-90:90:1)--cycle;}
\begin{tikzpicture}
\pgfmathsetmacro{\r}{1+sqrt(5)}
\fill[teal] (0,0) circle(\r);
\draw[fill=white] \cc;
\end{tikzpicture}
\hspace*{1cm}
\begin{tikzpicture}
\fill[green!50] (-2,-2) rectangle (2,2);
\begin{scope}
\clip (-2,-2) rectangle (2,2);
\fill[purple!50] \cc;
\end{scope}
\draw \cc;
\end{tikzpicture}
\end{document}

Update. Now I realise that using fit library is not suitable for this situation. In fact, smallest-circle problem is more complicated than direct calculating the above radius.

Another way without calculation is using library through after finding out the tangent point as follows.

enter image description here

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{through}
\begin{document}
\begin{tikzpicture}
\path (0,0)--(1,2)--([turn]0:1) coordinate (M);
\node[circle through=(M),draw,fill=cyan] at (0,0) {};
\draw[fill=white] (2,0) arc(270:90:1) arc(0:180:1) arc(0:-180:1) arc(90:270:1) arc(90:-90:1) arc(180:360:1) arc(180:0:1) arc(-90:90:1)--cycle;
\end{tikzpicture}
\end{document}

Is it reasonable that the library through is just for circle through?!

enter image description here

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\path (0,0)--(1,2)--([turn]0:1) coordinate (M);
\draw[fill=violet!50] (0,0) 
let \p1=(M) in circle({veclen(\x1,\y1)});
\draw[fill=white] (2,0) arc(270:90:1) arc(0:180:1) arc(0:-180:1) arc(90:270:1) arc(90:-90:1) arc(180:360:1) arc(180:0:1) arc(-90:90:1)--cycle;
\end{tikzpicture}
\end{document} 
6
  • 1
    You seem not to use fit, do you?
    – user194703
    Feb 29, 2020 at 16:26
  • @Schrödinger'scat ah, yes! something like this \usetikzlibrary{fit} \begin{tikzpicture}[draw,circle,inner sep=0,outer sep=0] \path[nodes={minimum size=2cm}] (1,2) node (A) {} (-2,1) node (B) {} (-1,-2) node (C) {} (2,-1) node (D) {}; \node[fill=teal,fit=(A) (B) (C) (D)] {}; \draw[fill=white] \cc; \end{tikzpicture} I will update later!
    – Black Mild
    Feb 29, 2020 at 16:46
  • 1
    I added some comment. fit library is not suitable here!
    – Black Mild
    Feb 29, 2020 at 20:28
  • 1
    I added another way using through library, and without calculation.
    – Black Mild
    Mar 1, 2020 at 17:14
  • 1
    @BlackMild Good idea.I don't have turn with tkz-euclide but possible with the same idea \tkzDefPointWith[colinear normed=at X,K=1](O,X) \tkzGetPoint{F} Mar 1, 2020 at 19:44
7

enter image description here

I think The radius of big circle is \sqrt{5}+1. Here XY=4(two small circle diameter= 2*(1+1)) YZ=2. so XZ=\sqrt{4^2+2^2}=\sqrt{20}, so radius of big circle would be (\sqrt{20}+2)/2=\sqrt{5}+1. I used @Schrödinger's cat code but used radius=\sqrt{5}+1. and here is the figure.

enter image description here

6

I agree with John Kormylo, i.e. I also do not get a radius of sqrt(10), but maybe I do not understand the construction. The sqrt(5)+1 is from this answer.

\documentclass[tikz,border=3mm]{standalone}
\begin{document}
\begin{tikzpicture}[arcs/.style={insert path={foreach \X [count=\Y] in {180,90,0,-90} 
   {arc[start angle=\X,end angle=\X+180,radius=1]
   arc[start angle=\X,end angle=\X-180,radius=1] coordinate (#1\Y)}}}]
 \begin{scope}[local bounding box=arcs]
  \draw[fill=blue!60,even odd rule] (0,0) [arcs]
  (arcs.center) circle[radius={sqrt(5)+1}];  
 \end{scope} 
 \begin{scope}[local bounding box=arcs2,xshift=7.5cm]
  \draw (0,0) [arcs=p] (p1) rectangle (p3);  
  \clip[postaction={fill=purple}](p1) rectangle (p3);  
  \draw[fill=green!50,even odd rule] (0,0) 
  [arcs=p] (p1) rectangle (p3);
 \end{scope} 
\end{tikzpicture}
\end{document}
4
  • You beat me by seconds. Feb 29, 2020 at 15:24
  • @JohnKormylo Sorry to hear that!
    – user194703
    Feb 29, 2020 at 15:25
  • Thank you. But I keep thinking that the measure of the radius of the circle is $sqrt{10}$. I can't put here the figure that makes clear the value of the radius Feb 29, 2020 at 15:39
  • @BeneditoFreire All I know is that if I uses sqrt(10) above the circle is too small. I think John Kormylo is right.
    – user194703
    Feb 29, 2020 at 15:42

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