[Tex/LaTex] Writing exercices in a specific style

tcolorboxtikz-pgf

Addendum:

I've written exercises with a simple box but I want to turn in the picture below style.Here's the code:

\documentclass[11pt]{book}
\usepackage[top=2cm, bottom=2.5cm, left=2.5cm, right=2.5cm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[most]{tcolorbox}
\usepackage{eso-pic}
\usepackage{enumerate}
\usepackage{fancyhdr}
\usepackage{mathrsfs}
\usepackage{cancel}
\usepackage{xcolor}
\usepackage{graphicx}
\usepackage{pgf,tikz,pgfplots}
 \usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{tikz,tkz-tab}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usetikzlibrary{arrows}
\usepackage{mdframed}
\usepackage{hyperref}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{tikz,lmodern}
\begin{document}
\section{EXERCICES}
\begin{tcolorbox}[enhanced,breakable,colback=white,colframe=green!50!white,
colbacktitle=white!15!pink,
coltitle=pink!50!black,
borderline={0.5mm}{0mm}{green!15!white},
borderline={0.5mm}{0mm}{green!50!white,dashed},
attach boxed title to top center={yshift=-2mm},
boxed title style={boxrule=0.4pt},
title=EXERCICES]


\underline{\textbf{Exercice 1}}
\\
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\dfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \item $B=\dfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
    \item $C=\dfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
    \item $D=\dfrac{\left[\left(\dfrac23\right)^2\right]^6\times \left[\left(\dfrac35\right)^{-2}\right]^3\times\left[\left(\dfrac52\right)^2\right]^{-6}}{\left(\dfrac46\right)^6}$
\end{enumerate}
\underline{\textbf{Exercice 2}}
\begin{enumerate}
    \item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.\\
Montrer que $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:
\\ $\dfrac{a}{ab+a+1}+\dfrac{b}{bc+c+1}+\dfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:
\\
$\dfrac{4a^2-1}{(a-b)(a-c)}+\dfrac{4b^2-1}{(b-c)(b-a)}+\dfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}
\underline{\textbf{Exercice 3}}
\\
\\
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\dfrac{a}{b}=\dfrac{c}{d}$.
\begin{enumerate}
    \item 
    \begin{enumerate}
        \item Montrer que: $\dfrac{7a+8c}{7b+8d}=\dfrac{a}{b}$
        \item Montrer que: $\dfrac{a^2+b^2}{ab}=\dfrac{c^2+d^2}{cd}$
        \item Montrer que: $\sqrt{\dfrac{a^2+c^2}{b^2+d^2}}=\dfrac{a}{b}$
        \item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
    \end{enumerate}
    \item 
    \begin{enumerate}
        \item Vérifier que si $\dfrac{a}{b}=\dfrac{c}{d}$ alors $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$.
        \item Démontrer la réciproque, à savoir: si $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$ alors $\dfrac{a}{b}=\dfrac{c}{d}$
        \\
    \end{enumerate}
\end{enumerate}

\underline{\textbf{Exercice 4}}
\\
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels

\end{enumerate}
\end{tcolorbox}
\end{document}

Here is a screenshot of what I have. What I want to do is:

1) Making the exercises into twocolumn format.

2) Writing exercises title and numbering like in the first picture or any other styled template for exercises specially with two column.

Best Answer

like this?

With enumitem instead of the enumerate package, enumerate list styles are changed only in tcolorboxses by help of `enumitem˙ package, considered @Leonardis comment:

\documentclass[11pt]{book}
\usepackage[margin=2.5cm]{geometry}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}
\usepackage{mathrsfs}

\usepackage{amsmath,amssymb}
\usepackage[most]{tcolorbox}
\newtcolorbox{exercisebox}%
    {
    enhanced,
    breakable,
    colback=white, colframe=green!15!white,
    colbacktitle=white!15!pink, coltitle=pink!50!black,
    left=0pt,right=0mm,top=3mm,bottom=3mm,
    pad at break=0pt,bottomrule at break=0pt,toprule at break=0pt,
    borderline={0mm}{0mm}{green!50!white,dashed},
    attach boxed title to top center={yshift=-2mm},
    boxed title style={boxrule=0.4pt},
    title=EXERCICES,
    }

\usepackage{eso-pic}
\usepackage{etoolbox}
\usepackage{enumitem}
\newcommand\circitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[circle,draw=gray, fill=gray!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\newcommand\boxitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[fill=orange!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\AtBeginEnvironment{tcolorbox}{%
    \setlist[enumerate,1]{label=\protect\boxitem{\arabic*}}
    \setlist[enumerate,2]{label=\protect\circitem{\alph*}}
                          }
\usepackage{fancyhdr}
\usepackage{cancel}
\usepackage{xcolor}
\newcommand\Exercice[1]{\fboxrule1.5pt%
\fcolorbox{gray}% frame
          {cyan}% background
          {\large\itshape\textbf{Exercise~#1}}%
          \par\medskip}

\usepackage{graphicx}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{arrows}
\usepackage{tkz-tab}
\usepackage{mdframed}

\usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{hyperref}


\usepackage{lipsum}

\begin{document}
\section{EXERCICES}

\begin{exercisebox}
\Exercice{1}
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\mfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \item $B=\dfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
    \item $C=\dfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
    \item $D=\dfrac{\left[\left(\dfrac23\right)^2\right]^6\times \left[\left(\dfrac35\right)^{-2}\right]^3\times\left[\left(\dfrac52\right)^2\right]^{-6}}{\left(\dfrac46\right)^6}$
\end{enumerate}

\Exercice{2} 
\begin{enumerate}
    \item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.\\
Montrer que $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:
\\ $\dfrac{a}{ab+a+1}+\dfrac{b}{bc+c+1}+\dfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:
\\
$\dfrac{4a^2-1}{(a-b)(a-c)}+\dfrac{4b^2-1}{(b-c)(b-a)}+\dfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}

\Exercice{3} 
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\dfrac{a}{b}=\dfrac{c}{d}$.
\begin{enumerate}
    \item
    \begin{enumerate}
        \item Montrer que: $\dfrac{7a+8c}{7b+8d}=\dfrac{a}{b}$
        \item Montrer que: $\dfrac{a^2+b^2}{ab}=\dfrac{c^2+d^2}{cd}$
        \item Montrer que: $\sqrt{\dfrac{a^2+c^2}{b^2+d^2}}=\dfrac{a}{b}$
        \item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
    \end{enumerate}
    \item
    \begin{enumerate}
        \item Vérifier que si $\dfrac{a}{b}=\dfrac{c}{d}$ alors $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$.
        \item Démontrer la réciproque, à savoir: si $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$ alors $\dfrac{a}{b}=\dfrac{c}{d}$
        \\
    \end{enumerate}
\end{enumerate}

\Exercice{4}
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels
\begin{enumerate}
 \item Vérifier les identités suivantes:
\begin{enumerate}
\item $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
    \item $(ax+by)^2-(ay+bx)^2=(a^2-b^2)(x^2-y^2)$
    \item $(a^2+a+1)(a^2-a+1)=a^4+a^2+1$
    \end{enumerate}
    \item
    \begin{enumerate}
    \item Montrer l'identité de Gauss à savoir: \\
    $a^3+b^2+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
        \item Servez vous de cette relation pour montrer que pour tout $u,v$ et $w$ on a l'identité:
    \\ $(u-v)^3+(v-w)^3+(w-u)^3=3(u-v)(v-w)(w-u)$
    \end{enumerate}
\end{enumerate}

\Exercice{5}
Factoriser les expressions suivantes
\begin{enumerate}
    \item $ab+a+b+1$
    \item $a^2xy+aby^2+b^2xy+abx^2$
    \item $a^2+8+2a^2+4a$
    \item $ab(a+b)+bc(b+c)+ca(c+a)+2abc$
    \item $(ax+by)^2-(ay+bx)^2$
    \item $(a^2+b^2-9)^2-4a^2b^2$
    \item $(a+b)^2-a^3-b^3$
    \item $(a^4-b^4)+2ab(a^2-b^2)-(a^3-b^2)+(ab^2-a^2b)$
    \item $x^2y - xy^2 + yz^2 - xz^2 + x^2z - xyz + y^2z + xyz$
\end{enumerate}
\end{exercisebox}
\end{document}

Edit:

If you consider less fancy design of your exercises, than the package exam is right tool for you. For an example of it use see this answer.
With use of above proposed solution for questions labels is possible to arrange them in two column by manula insert each question in appropriate place in longtable:


\documentclass[11pt]{book}
\usepackage[margin=2.5cm]{geometry}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}
\usepackage{mathrsfs}

\usepackage{nccmath,amssymb}
\usepackage[most]{tcolorbox}
\newtcolorbox{exercisebox}%
    {
    enhanced,
    breakable,
    colback=white, colframe=green!15!white,
    colbacktitle=white!15!pink, coltitle=pink!50!black,
    left=0pt,right=0mm,top=3mm,bottom=3mm,
    pad at break=0pt,bottomrule at break=0pt,toprule at break=0pt,
    borderline={0mm}{0mm}{green!50!white,dashed},
    attach boxed title to top center={yshift=-2mm},
    boxed title style={boxrule=0.4pt},
    title=EXERCICES,
    }

\usepackage{eso-pic}
\usepackage{enumitem}
\newcommand\circitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[circle,draw=gray, fill=gray!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\newcommand\boxitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[fill=orange!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}

\usepackage{fancyhdr}
\usepackage{cancel}
\usepackage{xcolor}
\newcommand\Exercice[1]{\fboxrule1.5pt%
\fcolorbox{gray}% frame
          {cyan}% background
          {\large\itshape\textbf{Exercise~#1}}%
          \par\medskip}

\usepackage{graphicx}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{arrows}
\usepackage{tkz-tab}
\usepackage{mdframed}

\usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{multicol}
\usepackage{longtable}
\usepackage{hyperref}



\usepackage{lipsum}

\begin{document}
\lipsum[66]
\setcounter{chapter}{7}
\section{EXERCICES}

\begin{exercisebox}
    \setlist[enumerate,1]{label=\protect\boxitem{\arabic*},leftmargin=*,resume}
    \setlist[enumerate,2]{label=\protect\circitem{\alph*}}
\begin{longtable}{@{}*{2}{p{\dimexpr0.5\linewidth-\tabcolsep}}}

\Exercice{1}
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\mfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \item $B=\mfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
    \item $C=\mfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
    \item $D=\mfrac{\left[\left(\mfrac23\right)^2\right]^6\times \left[\left(\mfrac35\right)^{-2}\right]^3\times\left[\left(\mfrac52\right)^2\right]^{-6}}{\left(\mfrac46\right)^6}$
\end{enumerate}
&
\Exercice{3}
\begin{enumerate}
    \item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.

Montrer que $\mfrac{b+c}{a}+\mfrac{c+a}{b}+\mfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:

$\mfrac{a}{ab+a+1}+\mfrac{b}{bc+c+1}+\mfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:

$\mfrac{4a^2-1}{(a-b)(a-c)}+\mfrac{4b^2-1}{(b-c)(b-a)}+\mfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}
\\
% new row
\Exercice{2}
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\mfrac{a}{b}=\mfrac{c}{d}$.
\begin{enumerate}
    \item
    \begin{enumerate}
        \item Montrer que: $\mfrac{7a+8c}{7b+8d}=\mfrac{a}{b}$
        \item Montrer que: $\mfrac{a^2+b^2}{ab}=\mfrac{c^2+d^2}{cd}$
        \item Montrer que: $\sqrt{\mfrac{a^2+c^2}{b^2+d^2}}=\mfrac{a}{b}$
        \item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
    \end{enumerate}
    \item
    \begin{enumerate}
        \item Vérifier que si $\mfrac{a}{b}=\mfrac{c}{d}$ alors $\mfrac{a^2+b^2}{ac+bd}=\mfrac{ac+bd}{c^2+d^2}$.
        \item Démontrer la réciproque, à savoir: si $\mfrac{a^2+b^2}{ac+bd}=\mfrac{ac+bd}{c^2+d^2}$ alors $\mfrac{a}{b}=\mfrac{c}{d}$
    \end{enumerate}
\end{enumerate}
&
\Exercice{4}
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels
\begin{enumerate}
 \item Vérifier les identités suivantes:
    \begin{enumerate}
    \item $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
    \item $(ax+by)^2-(ay+bx)^2=(a^2-b^2)(x^2-y^2)$
    \item $(a^2+a+1)(a^2-a+1)=a^4+a^2+1$
    \end{enumerate}
\item
    \begin{enumerate}
    \item Montrer l'identité de Gauss à savoir:

    $a^3+b^2+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
        \item Servez vous de cette relation pour montrer que pour tout $u,v$ et $w$ on a l'identité:

    $(u-v)^3+(v-w)^3+(w-u)^3=3(u-v)(v-w)(w-u)$
    \end{enumerate}
\end{enumerate}
\\
% new row
\Exercice{5}
Factoriser les expressions suivantes
\begin{enumerate}
    \item $ab+a+b+1$
    \item $a^2xy+aby^2+b^2xy+abx^2$
    \item $a^2+8+2a^2+4a$
    \item $ab(a+b)+bc(b+c)+ca(c+a)+2abc$
\end{enumerate}
&
\begin{enumerate}[start=5]
    \item $(ax+by)^2-(ay+bx)^2$
    \item $(a^2+b^2-9)^2-4a^2b^2$
    \item $(a+b)^2-a^3-b^3$
    \item $(a^4-b^4)+2ab(a^2-b^2)-(a^3-b^2)+(ab^2-a^2b)$
    \item $x^2y - xy^2 + yz^2 - xz^2 + x^2z - xyz + y^2z + xyz$
\end{enumerate}
\end{longtable}
\vspace{-2\baselineskip}
\end{exercisebox}
\lipsum[3]
\end{document}

Code:

\documentclass[12pt]{article}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}
\tikzset{>=latex} 

\begin{document}

\begin{tikzpicture}[scale=1.5,cap=0.5 ,color= blue] 

% grid
\draw[very thin,color = green,step = 0.5 cm, opacity= 0.9](0,0) grid (4, 4);

% axis 
\draw[->]  [blue,  thick] ( 0,0) --( 4,0) node[right] {$x$};
\draw[->]  [blue,  thick] ( 0,0) --( 0,4) node[above] {$y$};
% how to change the arrow tip style at specific path
% how to set up unique arrow head as a decoration at (1,0) coordinate and at (0,1) 
\draw[decoration={markings, mark=at position 0.5 with {\arrow [ ultra thick] {latex}}},
       postaction={decorate}]
        (0,0)--( 2,0);
\draw[decoration={markings, mark=at position 0.5 with {\arrow [ ultra thick] {stealth}}},
       postaction={decorate}]
        (0,0)--( 0,2);
\end{tikzpicture}
%
\begin{tikzpicture}[scale=1.5,cap=0.5 ,color= blue] 

% grid
\draw[very thin,color = green,step = 0.5 cm, opacity= 0.9](0,0) grid (4, 4);

% axis 
\draw[->]  [blue,  thick] ( 0,0) --( 4,0) node[right] {$x$};
\draw[->]  [blue,  thick] ( 0,0) --( 0,4) node[above] {$y$};
% how to change the arrow tip style at specific path
% how to set up unique arrow head as a decoration at (1,0) coordinate and at (0,1) 
\draw[decoration={markings, mark=at position 0.5 with {\arrow [ ultra thick] {>}}},
       postaction={decorate}]
        (0,0)--( 2,0);
\draw[decoration={markings, mark=at position 0.5 with {\arrow [ ultra thick] {>}}},
       postaction={decorate}]
        (0,0)--( 0,2);
\end{tikzpicture}
\end{document}

[Tex/LaTex] tcolorbox, subtitle style

I suggest to define a separate style for this, say mysubtitle.

(I've changed Arash Esbati's proposition 'slightly')

Since \tcbsubtitle itself is a tcolorbox, it accepts the same options like the outer box.

The most crucial option (in my point of view) is top=0pt (for example) and nobeforeafter for the subtitle.

Since tcbsubtitle is tcbox, fontupper instead of fonttitle has to be used, as well as colback.

\documentclass[]{article}
\usepackage[most]{tcolorbox} 
\usepackage{xcolor}

\definecolor{commentgreen}{HTML}{E0E5C1}

\newtcolorbox{commentbox}[2][]{
    frame hidden,
    boxrule=0pt,
    breakable,
    enhanced,
    before skip=8pt plus 1pt,
    toptitle=3mm,
    boxsep=0.25ex,
    left=8pt,
    right=8pt,
    arc=0mm,
    fonttitle=\fontfamily{fosj}\selectfont\scshape\bfseries\color{black},
    fontupper=\fontfamily{lmss}\selectfont,
    title=#2,
    parbox = false,
    colback=commentgreen,
    colframe=commentgreen,
    colbacktitle=commentgreen,
    after={\vspace{5pt plus 1pt}\noindent},
    #1
}

\tcbset{mysubtitle/.style={subtitle style={fontupper={\fontfamily{fosj}\selectfont\itshape\scriptsize\color{black}},nobeforeafter,top=0pt,colback={commentgreen}},top=0pt}}

\begin{document}

\begin{commentbox}[mysubtitle]{Title}
  \tcbsubtitle{Sub-title}
  Main Text is going to be here
\end{commentbox}
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Specific arrow tip style for specific path

Code:

[Tex/LaTex] tcolorbox, subtitle style

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