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How to draw this figure using Tikz?


LaTeX equivalent of ConTeXt buffersHow can I put a coloured outline around fraction lines?Rotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?Numerical conditional within tikz keys?Why do I get an extra white page before my TikZ picture?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to draw a square and its diagonals with arrows?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









4


















I would like to draw this figure, where two circles are tangents to a semicircle, as in the following figure:



Two circles tangents to a semicircle



I can partially draw:



documentclass[12pt]article

usepackagetikz

begindocument

begincenter

begintikzpicture[scale=0.3]

draw [ultra thick] (0,0) arc (0:180:6);

draw [ultra thick] (-12,0)--(0,0);

node at (-6,0) $bullet$;

node at (-12,0) $bullet$;

node at (0,0) $bullet$;

endtikzpicture

endcenter

enddocument


Semicircle without the two tangents circles










share|improve this question

































    4


















    I would like to draw this figure, where two circles are tangents to a semicircle, as in the following figure:



    Two circles tangents to a semicircle



    I can partially draw:



    documentclass[12pt]article

    usepackagetikz

    begindocument

    begincenter

    begintikzpicture[scale=0.3]

    draw [ultra thick] (0,0) arc (0:180:6);

    draw [ultra thick] (-12,0)--(0,0);

    node at (-6,0) $bullet$;

    node at (-12,0) $bullet$;

    node at (0,0) $bullet$;

    endtikzpicture

    endcenter

    enddocument


    Semicircle without the two tangents circles










    share|improve this question





























      4













      4









      4


      2






      I would like to draw this figure, where two circles are tangents to a semicircle, as in the following figure:



      Two circles tangents to a semicircle



      I can partially draw:



      documentclass[12pt]article

      usepackagetikz

      begindocument

      begincenter

      begintikzpicture[scale=0.3]

      draw [ultra thick] (0,0) arc (0:180:6);

      draw [ultra thick] (-12,0)--(0,0);

      node at (-6,0) $bullet$;

      node at (-12,0) $bullet$;

      node at (0,0) $bullet$;

      endtikzpicture

      endcenter

      enddocument


      Semicircle without the two tangents circles










      share|improve this question
















      I would like to draw this figure, where two circles are tangents to a semicircle, as in the following figure:



      Two circles tangents to a semicircle



      I can partially draw:



      documentclass[12pt]article

      usepackagetikz

      begindocument

      begincenter

      begintikzpicture[scale=0.3]

      draw [ultra thick] (0,0) arc (0:180:6);

      draw [ultra thick] (-12,0)--(0,0);

      node at (-6,0) $bullet$;

      node at (-12,0) $bullet$;

      node at (0,0) $bullet$;

      endtikzpicture

      endcenter

      enddocument


      Semicircle without the two tangents circles







      tikz-pgf






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Sep 19 at 21:17









      Peter Grill

      190k28 gold badges459 silver badges790 bronze badges




      190k28 gold badges459 silver badges790 bronze badges










      asked Sep 19 at 21:15









      Benedito FreireBenedito Freire

      2971 silver badge7 bronze badges




      2971 silver badge7 bronze badges























          1 Answer
          1






          active

          oldest

          votes


















          13



















          documentclass[12pt]article
          usepackagetikz
          begindocument
          begincenter
          begintikzpicture[scale=0.3,declare
          function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          endcenter
          enddocument


          enter image description here



          Here is a possible derivation.



          documentclass[12pt,fleqn]article
          usepackagetikz
          begindocument
          Call the large radius $R$ and the small radius $a$. The condition that the
          circles ``touch'' means that the slopes of the circles at the touching point
          coincide. This implies that the circles touch at points with coinciding polar
          angle $alpha$ (see figure~reffig:Computation). Therefore,
          [ a+a,sinalpha~=~R,sinalpha]
          and thus
          [ alpha~=~arcsinfracaR-a;.]
          This means that the centers of the circles are at
          $(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

          beginfigure[!h]
          centering
          begintikzpicture[scale=0.8,declare function=R=6;a=1;]

          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          node at (-R,0) $bullet$;
          node at (R,0) $bullet$;
          node at (0,0) $bullet$;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[red] (0,0) -- (180-myalpha:R)
          (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a)
          -- ++ (180-myalpha:2)
          ((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
          node[midway,fill=white]$a$;
          draw (0,0) -- (40:R) node[midway,fill=white]$R$;
          draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
          ((R-a)*cos(myalpha),a) circle[radius=a];
          endtikzpicture
          captionComputation of the center of the circle.
          labelfig:Computation
          endfigure
          enddocument


          enter image description here



          Of course, you can vary a.



          documentclass[tikz,border=3mm]standalone
          begindocument
          foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
          begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
          path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer



























          • Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

            – Benedito Freire
            Sep 19 at 21:57











          • @BeneditoFreire I added the derivation.

            – Schrödinger's cat
            Sep 19 at 22:01











          • Schrödinger's cat - Show!!!!

            – Benedito Freire
            Sep 19 at 22:44











          • Schrödinger's cat - It looks really good. Show!!!!

            – Benedito Freire
            Sep 19 at 22:47












          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          13



















          documentclass[12pt]article
          usepackagetikz
          begindocument
          begincenter
          begintikzpicture[scale=0.3,declare
          function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          endcenter
          enddocument


          enter image description here



          Here is a possible derivation.



          documentclass[12pt,fleqn]article
          usepackagetikz
          begindocument
          Call the large radius $R$ and the small radius $a$. The condition that the
          circles ``touch'' means that the slopes of the circles at the touching point
          coincide. This implies that the circles touch at points with coinciding polar
          angle $alpha$ (see figure~reffig:Computation). Therefore,
          [ a+a,sinalpha~=~R,sinalpha]
          and thus
          [ alpha~=~arcsinfracaR-a;.]
          This means that the centers of the circles are at
          $(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

          beginfigure[!h]
          centering
          begintikzpicture[scale=0.8,declare function=R=6;a=1;]

          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          node at (-R,0) $bullet$;
          node at (R,0) $bullet$;
          node at (0,0) $bullet$;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[red] (0,0) -- (180-myalpha:R)
          (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a)
          -- ++ (180-myalpha:2)
          ((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
          node[midway,fill=white]$a$;
          draw (0,0) -- (40:R) node[midway,fill=white]$R$;
          draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
          ((R-a)*cos(myalpha),a) circle[radius=a];
          endtikzpicture
          captionComputation of the center of the circle.
          labelfig:Computation
          endfigure
          enddocument


          enter image description here



          Of course, you can vary a.



          documentclass[tikz,border=3mm]standalone
          begindocument
          foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
          begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
          path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer



























          • Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

            – Benedito Freire
            Sep 19 at 21:57











          • @BeneditoFreire I added the derivation.

            – Schrödinger's cat
            Sep 19 at 22:01











          • Schrödinger's cat - Show!!!!

            – Benedito Freire
            Sep 19 at 22:44











          • Schrödinger's cat - It looks really good. Show!!!!

            – Benedito Freire
            Sep 19 at 22:47















          13



















          documentclass[12pt]article
          usepackagetikz
          begindocument
          begincenter
          begintikzpicture[scale=0.3,declare
          function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          endcenter
          enddocument


          enter image description here



          Here is a possible derivation.



          documentclass[12pt,fleqn]article
          usepackagetikz
          begindocument
          Call the large radius $R$ and the small radius $a$. The condition that the
          circles ``touch'' means that the slopes of the circles at the touching point
          coincide. This implies that the circles touch at points with coinciding polar
          angle $alpha$ (see figure~reffig:Computation). Therefore,
          [ a+a,sinalpha~=~R,sinalpha]
          and thus
          [ alpha~=~arcsinfracaR-a;.]
          This means that the centers of the circles are at
          $(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

          beginfigure[!h]
          centering
          begintikzpicture[scale=0.8,declare function=R=6;a=1;]

          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          node at (-R,0) $bullet$;
          node at (R,0) $bullet$;
          node at (0,0) $bullet$;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[red] (0,0) -- (180-myalpha:R)
          (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a)
          -- ++ (180-myalpha:2)
          ((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
          node[midway,fill=white]$a$;
          draw (0,0) -- (40:R) node[midway,fill=white]$R$;
          draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
          ((R-a)*cos(myalpha),a) circle[radius=a];
          endtikzpicture
          captionComputation of the center of the circle.
          labelfig:Computation
          endfigure
          enddocument


          enter image description here



          Of course, you can vary a.



          documentclass[tikz,border=3mm]standalone
          begindocument
          foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
          begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
          path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer



























          • Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

            – Benedito Freire
            Sep 19 at 21:57











          • @BeneditoFreire I added the derivation.

            – Schrödinger's cat
            Sep 19 at 22:01











          • Schrödinger's cat - Show!!!!

            – Benedito Freire
            Sep 19 at 22:44











          • Schrödinger's cat - It looks really good. Show!!!!

            – Benedito Freire
            Sep 19 at 22:47













          13















          13











          13









          documentclass[12pt]article
          usepackagetikz
          begindocument
          begincenter
          begintikzpicture[scale=0.3,declare
          function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          endcenter
          enddocument


          enter image description here



          Here is a possible derivation.



          documentclass[12pt,fleqn]article
          usepackagetikz
          begindocument
          Call the large radius $R$ and the small radius $a$. The condition that the
          circles ``touch'' means that the slopes of the circles at the touching point
          coincide. This implies that the circles touch at points with coinciding polar
          angle $alpha$ (see figure~reffig:Computation). Therefore,
          [ a+a,sinalpha~=~R,sinalpha]
          and thus
          [ alpha~=~arcsinfracaR-a;.]
          This means that the centers of the circles are at
          $(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

          beginfigure[!h]
          centering
          begintikzpicture[scale=0.8,declare function=R=6;a=1;]

          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          node at (-R,0) $bullet$;
          node at (R,0) $bullet$;
          node at (0,0) $bullet$;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[red] (0,0) -- (180-myalpha:R)
          (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a)
          -- ++ (180-myalpha:2)
          ((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
          node[midway,fill=white]$a$;
          draw (0,0) -- (40:R) node[midway,fill=white]$R$;
          draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
          ((R-a)*cos(myalpha),a) circle[radius=a];
          endtikzpicture
          captionComputation of the center of the circle.
          labelfig:Computation
          endfigure
          enddocument


          enter image description here



          Of course, you can vary a.



          documentclass[tikz,border=3mm]standalone
          begindocument
          foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
          begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
          path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer
















          documentclass[12pt]article
          usepackagetikz
          begindocument
          begincenter
          begintikzpicture[scale=0.3,declare
          function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          endcenter
          enddocument


          enter image description here



          Here is a possible derivation.



          documentclass[12pt,fleqn]article
          usepackagetikz
          begindocument
          Call the large radius $R$ and the small radius $a$. The condition that the
          circles ``touch'' means that the slopes of the circles at the touching point
          coincide. This implies that the circles touch at points with coinciding polar
          angle $alpha$ (see figure~reffig:Computation). Therefore,
          [ a+a,sinalpha~=~R,sinalpha]
          and thus
          [ alpha~=~arcsinfracaR-a;.]
          This means that the centers of the circles are at
          $(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

          beginfigure[!h]
          centering
          begintikzpicture[scale=0.8,declare function=R=6;a=1;]

          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          node at (-R,0) $bullet$;
          node at (R,0) $bullet$;
          node at (0,0) $bullet$;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[red] (0,0) -- (180-myalpha:R)
          (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a)
          -- ++ (180-myalpha:2)
          ((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
          draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
          node[midway,fill=white]$a$;
          draw (0,0) -- (40:R) node[midway,fill=white]$R$;
          draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
          ((R-a)*cos(myalpha),a) circle[radius=a];
          endtikzpicture
          captionComputation of the center of the circle.
          labelfig:Computation
          endfigure
          enddocument


          enter image description here



          Of course, you can vary a.



          documentclass[tikz,border=3mm]standalone
          begindocument
          foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
          begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
          path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
          draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
          pgfmathsetmacromyalphaasin(a/(R-a))
          draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
          ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
          draw[dashed] (L) -- (R);
          path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
          (180-myalpha:R) node[bullet] (myalpha:R) node[bullet];
          endtikzpicture
          enddocument


          enter image description here







          share|improve this answer















          share|improve this answer




          share|improve this answer








          edited Sep 19 at 22:25

























          answered Sep 19 at 21:46









          Schrödinger's catSchrödinger's cat

          27k2 gold badges43 silver badges67 bronze badges




          27k2 gold badges43 silver badges67 bronze badges















          • Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

            – Benedito Freire
            Sep 19 at 21:57











          • @BeneditoFreire I added the derivation.

            – Schrödinger's cat
            Sep 19 at 22:01











          • Schrödinger's cat - Show!!!!

            – Benedito Freire
            Sep 19 at 22:44











          • Schrödinger's cat - It looks really good. Show!!!!

            – Benedito Freire
            Sep 19 at 22:47

















          • Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

            – Benedito Freire
            Sep 19 at 21:57











          • @BeneditoFreire I added the derivation.

            – Schrödinger's cat
            Sep 19 at 22:01











          • Schrödinger's cat - Show!!!!

            – Benedito Freire
            Sep 19 at 22:44











          • Schrödinger's cat - It looks really good. Show!!!!

            – Benedito Freire
            Sep 19 at 22:47
















          Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

          – Benedito Freire
          Sep 19 at 21:57





          Schrodingesr's cat -I liked the design, but, excuse me, in the drawing it is unclear how to identify the center of each small circle. In fact, I would like the drawing to illustrate the segment by uniting these centers.

          – Benedito Freire
          Sep 19 at 21:57













          @BeneditoFreire I added the derivation.

          – Schrödinger's cat
          Sep 19 at 22:01





          @BeneditoFreire I added the derivation.

          – Schrödinger's cat
          Sep 19 at 22:01













          Schrödinger's cat - Show!!!!

          – Benedito Freire
          Sep 19 at 22:44





          Schrödinger's cat - Show!!!!

          – Benedito Freire
          Sep 19 at 22:44













          Schrödinger's cat - It looks really good. Show!!!!

          – Benedito Freire
          Sep 19 at 22:47





          Schrödinger's cat - It looks really good. Show!!!!

          – Benedito Freire
          Sep 19 at 22:47


















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