Problem mit Multicol und Big Picture

Question

Das Bild ist zu hoch, um in der ersten Spalte zu bleiben, aber der restliche Text erstreckt sich über die gesamte Höhe dieser Spalte. Die einzige Möglichkeit, die ich gefunden habe, um dieses Problem zu vermeiden, besteht darin, ein leeres Feld vboxin einer bestimmten Höhe zu platzieren. Dies ist kaum zufriedenstellend. Was ist die beste Methode?

\documentclass{scrartcl}
\usepackage{tikz}
\usepackage{multicol,lipsum}
\setlength{\columnsep}{24pt}
\setlength{\columnseprule}{0.4pt}

\begin{document}
\begin{multicols*}{2}
   
\section{Contruction with inversion} 

Let us consider the circle $\mathcal{T}$ of center $T$ point common to the circles $\mathcal{A}$ and $\mathcal{B}$ and tangent to the line $(EF)$. It intercepts the two circles at 4 points which define two lines $\mathcal{L_A}$ and $\mathcal{L_B}$ perpendicular to the line $(AB)$.
  
Consider the inversion of center $T$ with respect to the circle $\mathcal{T}$. The circles $\mathcal{A}$ and $\mathcal{B}$ have as inverses the lines $\mathcal{L_A}$ and $\mathcal{L_B}$.
     
The orange circle passing through $T$ and tangent to the circle $\mathcal{T}$ and to the line $(EF)$ is therefore tangent to the line $(EF)$. This circle is therefore tangent to the lines $\mathcal{L_A}$ and $\mathcal{L_B}$ since $(EF)$ is tangent to the circles $\mathcal{A}$ and $\mathcal{B}$.

%\vbox to 10cm{} 

 \begin{tikzpicture}
    \draw (0,0)--(7,15) (0,15)--(7,0);
 \end{tikzpicture} 
 
\lipsum[2]
 \end{multicols*}

\end{document}

Answer 1

Dadurch wird genügend Text hinzugefügt, um die erste Spalte zu füllen, und dann wird meine verknüpfte Lösung verwendet.

\documentclass{scrartcl}
\usepackage{tikz}
\usepackage{multicol,lipsum}
\setlength{\columnsep}{24pt}
\setlength{\columnseprule}{0.4pt}

\makeatletter
\newcommand{\split}[2]% #1 = text, #2 = insert
{\par\bgroup
  \setbox0=\vbox{\strut #1\strut}% measure height
  \sbox1{#2}% ditto
  \dimen0=\dimexpr \pagegoal-\pagetotal-\col@number\baselineskip\relax% fudge factor
  \dimen1=\pagegoal% compute height of one column
  \ifnum\col@number>1\relax
    \divide\dimen1 by \col@number
  \fi
  \advance\dimen1 by -\baselineskip% fudge factor
  \loop\ifdim\dimen0>\dimen1% space left in this column
    \advance\dimen0 by -\dimen1
  \repeat
  \dimen1=\dimexpr \ht1+\dp1+\textfloatsep\relax% space needed for insert
  \ifdim\dimen1>\dimen0
    \ifdim\ht0>\dimen0
      \setbox2=\vsplit0 to \dimen0
      \unvbox2
      \noindent\box1
      \vskip\textfloatsep
      \unvbox0
    \else% entire text will fit into column
      \unvbox0
      \vskip\textfloatsep% probably ignored
      \noindent\box1
      \vskip\textfloatsep
    \fi
  \else% insert will fit into column
    \vskip\intextsep
    \noindent\box1
    \vskip\intextsep
    \unvbox0
  \fi
\egroup\par}
\makeatother

\begin{document}
\begin{multicols*}{2}
   
\section{Contruction with inversion} 

Let us consider the circle $\mathcal{T}$ of center $T$ point common to the circles $\mathcal{A}$ and $\mathcal{B}$ and tangent to the line $(EF)$. It intercepts the two circles at 4 points which define two lines $\mathcal{L_A}$ and $\mathcal{L_B}$ perpendicular to the line $(AB)$.
  
Consider the inversion of center $T$ with respect to the circle $\mathcal{T}$. The circles $\mathcal{A}$ and $\mathcal{B}$ have as inverses the lines $\mathcal{L_A}$ and $\mathcal{L_B}$.
     
The orange circle passing through $T$ and tangent to the circle $\mathcal{T}$ and to the line $(EF)$ is therefore tangent to the line $(EF)$. This circle is therefore tangent to the lines $\mathcal{L_A}$ and $\mathcal{L_B}$ since $(EF)$ is tangent to the circles $\mathcal{A}$ and $\mathcal{B}$.\strut

\split{\lipsum[2-3]}{\noindent\begin{tikzpicture}
    \draw (0,0)--(\columnwidth,15) (0,15)--(\columnwidth,0);
 \end{tikzpicture}}

 \end{multicols*}

\end{document}

---

Wenn duwiegroße Lücken, können Sie verwenden

\split{\vfill}{\noindent\begin{tikzpicture}
    \draw (0,0)--(\columnwidth,15) (0,15)--(\columnwidth,0);
\end{tikzpicture}}