How to create a cover page like this?
I saw my Chinese classmate reading a book whose cover page is really fancy, though I don’t know the Chinese characters on it.
How could I create a cover page in my own classnotes like that?
tikz-pgf covers bookcover
add a comment |
I saw my Chinese classmate reading a book whose cover page is really fancy, though I don’t know the Chinese characters on it.
How could I create a cover page in my own classnotes like that?
tikz-pgf covers bookcover
4
What have you tried so far?
– manooooh
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago
add a comment |
I saw my Chinese classmate reading a book whose cover page is really fancy, though I don’t know the Chinese characters on it.
How could I create a cover page in my own classnotes like that?
tikz-pgf covers bookcover
I saw my Chinese classmate reading a book whose cover page is really fancy, though I don’t know the Chinese characters on it.
How could I create a cover page in my own classnotes like that?
tikz-pgf covers bookcover
tikz-pgf covers bookcover
asked 3 hours ago
user450201user450201
6713
6713
4
What have you tried so far?
– manooooh
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago
add a comment |
4
What have you tried so far?
– manooooh
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago
4
4
What have you tried so far?
– manooooh
3 hours ago
What have you tried so far?
– manooooh
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
Can one do something like this? Yes. Most likely the curves in the upper right part are are some sort of Apollonius (Golden Ratio?) circles but I was too lazy to look them up.
documentclass{article}
usepackage{tikz}
usetikzlibrary{intersections,decorations.text}
definecolor{c1}{RGB}{62, 97, 127}
definecolor{c2}{RGB}{104, 182, 182}
definecolor{c3}{RGB}{107, 190, 190}
definecolor{c4}{RGB}{100, 172, 174}
begin{document}
thispagestyle{empty}
begin{tikzpicture}[overlay,remember picture,font=sffamilybfseries]
draw[very thick,c4,name path=big arc] ([xshift=-2mm]current page.north) arc(150:285:11)
coordinate[pos=0.225] (x0);
begin{scope}
clip ([xshift=-2mm]current page.north) arc(150:285:11) --(current page.north
east);
fill[c4!50,opacity=0.25] ([xshift=4.55cm]x0) circle (4.55);
fill[c4!50,opacity=0.25] ([xshift=3.4cm]x0) circle (3.4);
fill[c4!50,opacity=0.25] ([xshift=2.25cm]x0) circle (2.25);
draw[very thick,c4!50] (x0) arc(-90:30:6.5);
draw[very thick,c4] (x0) arc(90:-30:8.75);
draw[very thick,c4!50,name path=arc1] (x0) arc(90:-90:4.675);
draw[very thick,c4!50] (x0) arc(90:-90:2.875);
path[name intersections={of=big arc and arc1,by=x1}];
draw[very thick,c4,name path=arc2] (x1) arc(135:-20:4.75);
draw[very thick,c4!50] (x1) arc(135:-20:8.75);
path[name intersections={of=big arc and arc2,by={aux,x2}}];
draw[very thick,c4!50] (x2) arc(180:50:2.25);
end{scope}
path[decoration={text along path,text color=c4,
raise = -2.8ex,
text along path,
text = {|sffamilybfseries|02/18/2019},
text align = center,
},
decorate
] ([xshift=-2mm]current page.north) arc(150:245:11);
%
begin{scope}
path[clip,postaction={fill=c3}]
([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (4.2,7.7);
fill[c2] ([xshift=0.5cm,yshift=-8cm]current page.center)
([xshift=0.5cm,yshift=-8cm]current page.center) arc(180:60:2)
|- ++ (-3,6) --cycle;
draw[very thick,c4] ([xshift=-1.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=0.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=2.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=4.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
fill[red] ([xshift=2.5cm,yshift=-8cm]current page.center) +(60:2) circle(1.5mm)
node[above right=2mm]{$displaystylerho=frac{1+sqrt{-3}}{2}$};
end{scope}
%
fill[c1] ([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (-12.7,7.7);
node[text=white,anchor=west,scale=5,inner sep=0pt] at
([xshift=-8cm,yshift=-3.25cm]current page.center) {Some text};
node[text=white,anchor=west,scale=2.5,inner sep=0pt] at
([xshift=-8cm,yshift=-6cm]current page.center) {Some text};
%
draw[gray,line width=5mm]
([xshift=2mm,yshift=-1mm]current page.south west) rectangle ([xshift=-2mm,yshift=1mm]current
page.north east);
end{tikzpicture}
end{document}
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of usingdisplaystyle
you could usedfrac
fromamsmath
.
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
@HenriMenke I believe that the cover is correct.rho
is the sixth root of unity, i.e.rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parametertau
withrho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)
– marmot
1 hour ago
|
show 2 more comments
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "85"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f475597%2fhow-to-create-a-cover-page-like-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Can one do something like this? Yes. Most likely the curves in the upper right part are are some sort of Apollonius (Golden Ratio?) circles but I was too lazy to look them up.
documentclass{article}
usepackage{tikz}
usetikzlibrary{intersections,decorations.text}
definecolor{c1}{RGB}{62, 97, 127}
definecolor{c2}{RGB}{104, 182, 182}
definecolor{c3}{RGB}{107, 190, 190}
definecolor{c4}{RGB}{100, 172, 174}
begin{document}
thispagestyle{empty}
begin{tikzpicture}[overlay,remember picture,font=sffamilybfseries]
draw[very thick,c4,name path=big arc] ([xshift=-2mm]current page.north) arc(150:285:11)
coordinate[pos=0.225] (x0);
begin{scope}
clip ([xshift=-2mm]current page.north) arc(150:285:11) --(current page.north
east);
fill[c4!50,opacity=0.25] ([xshift=4.55cm]x0) circle (4.55);
fill[c4!50,opacity=0.25] ([xshift=3.4cm]x0) circle (3.4);
fill[c4!50,opacity=0.25] ([xshift=2.25cm]x0) circle (2.25);
draw[very thick,c4!50] (x0) arc(-90:30:6.5);
draw[very thick,c4] (x0) arc(90:-30:8.75);
draw[very thick,c4!50,name path=arc1] (x0) arc(90:-90:4.675);
draw[very thick,c4!50] (x0) arc(90:-90:2.875);
path[name intersections={of=big arc and arc1,by=x1}];
draw[very thick,c4,name path=arc2] (x1) arc(135:-20:4.75);
draw[very thick,c4!50] (x1) arc(135:-20:8.75);
path[name intersections={of=big arc and arc2,by={aux,x2}}];
draw[very thick,c4!50] (x2) arc(180:50:2.25);
end{scope}
path[decoration={text along path,text color=c4,
raise = -2.8ex,
text along path,
text = {|sffamilybfseries|02/18/2019},
text align = center,
},
decorate
] ([xshift=-2mm]current page.north) arc(150:245:11);
%
begin{scope}
path[clip,postaction={fill=c3}]
([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (4.2,7.7);
fill[c2] ([xshift=0.5cm,yshift=-8cm]current page.center)
([xshift=0.5cm,yshift=-8cm]current page.center) arc(180:60:2)
|- ++ (-3,6) --cycle;
draw[very thick,c4] ([xshift=-1.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=0.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=2.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=4.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
fill[red] ([xshift=2.5cm,yshift=-8cm]current page.center) +(60:2) circle(1.5mm)
node[above right=2mm]{$displaystylerho=frac{1+sqrt{-3}}{2}$};
end{scope}
%
fill[c1] ([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (-12.7,7.7);
node[text=white,anchor=west,scale=5,inner sep=0pt] at
([xshift=-8cm,yshift=-3.25cm]current page.center) {Some text};
node[text=white,anchor=west,scale=2.5,inner sep=0pt] at
([xshift=-8cm,yshift=-6cm]current page.center) {Some text};
%
draw[gray,line width=5mm]
([xshift=2mm,yshift=-1mm]current page.south west) rectangle ([xshift=-2mm,yshift=1mm]current
page.north east);
end{tikzpicture}
end{document}
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of usingdisplaystyle
you could usedfrac
fromamsmath
.
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
@HenriMenke I believe that the cover is correct.rho
is the sixth root of unity, i.e.rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parametertau
withrho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)
– marmot
1 hour ago
|
show 2 more comments
Can one do something like this? Yes. Most likely the curves in the upper right part are are some sort of Apollonius (Golden Ratio?) circles but I was too lazy to look them up.
documentclass{article}
usepackage{tikz}
usetikzlibrary{intersections,decorations.text}
definecolor{c1}{RGB}{62, 97, 127}
definecolor{c2}{RGB}{104, 182, 182}
definecolor{c3}{RGB}{107, 190, 190}
definecolor{c4}{RGB}{100, 172, 174}
begin{document}
thispagestyle{empty}
begin{tikzpicture}[overlay,remember picture,font=sffamilybfseries]
draw[very thick,c4,name path=big arc] ([xshift=-2mm]current page.north) arc(150:285:11)
coordinate[pos=0.225] (x0);
begin{scope}
clip ([xshift=-2mm]current page.north) arc(150:285:11) --(current page.north
east);
fill[c4!50,opacity=0.25] ([xshift=4.55cm]x0) circle (4.55);
fill[c4!50,opacity=0.25] ([xshift=3.4cm]x0) circle (3.4);
fill[c4!50,opacity=0.25] ([xshift=2.25cm]x0) circle (2.25);
draw[very thick,c4!50] (x0) arc(-90:30:6.5);
draw[very thick,c4] (x0) arc(90:-30:8.75);
draw[very thick,c4!50,name path=arc1] (x0) arc(90:-90:4.675);
draw[very thick,c4!50] (x0) arc(90:-90:2.875);
path[name intersections={of=big arc and arc1,by=x1}];
draw[very thick,c4,name path=arc2] (x1) arc(135:-20:4.75);
draw[very thick,c4!50] (x1) arc(135:-20:8.75);
path[name intersections={of=big arc and arc2,by={aux,x2}}];
draw[very thick,c4!50] (x2) arc(180:50:2.25);
end{scope}
path[decoration={text along path,text color=c4,
raise = -2.8ex,
text along path,
text = {|sffamilybfseries|02/18/2019},
text align = center,
},
decorate
] ([xshift=-2mm]current page.north) arc(150:245:11);
%
begin{scope}
path[clip,postaction={fill=c3}]
([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (4.2,7.7);
fill[c2] ([xshift=0.5cm,yshift=-8cm]current page.center)
([xshift=0.5cm,yshift=-8cm]current page.center) arc(180:60:2)
|- ++ (-3,6) --cycle;
draw[very thick,c4] ([xshift=-1.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=0.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=2.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=4.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
fill[red] ([xshift=2.5cm,yshift=-8cm]current page.center) +(60:2) circle(1.5mm)
node[above right=2mm]{$displaystylerho=frac{1+sqrt{-3}}{2}$};
end{scope}
%
fill[c1] ([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (-12.7,7.7);
node[text=white,anchor=west,scale=5,inner sep=0pt] at
([xshift=-8cm,yshift=-3.25cm]current page.center) {Some text};
node[text=white,anchor=west,scale=2.5,inner sep=0pt] at
([xshift=-8cm,yshift=-6cm]current page.center) {Some text};
%
draw[gray,line width=5mm]
([xshift=2mm,yshift=-1mm]current page.south west) rectangle ([xshift=-2mm,yshift=1mm]current
page.north east);
end{tikzpicture}
end{document}
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of usingdisplaystyle
you could usedfrac
fromamsmath
.
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
@HenriMenke I believe that the cover is correct.rho
is the sixth root of unity, i.e.rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parametertau
withrho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)
– marmot
1 hour ago
|
show 2 more comments
Can one do something like this? Yes. Most likely the curves in the upper right part are are some sort of Apollonius (Golden Ratio?) circles but I was too lazy to look them up.
documentclass{article}
usepackage{tikz}
usetikzlibrary{intersections,decorations.text}
definecolor{c1}{RGB}{62, 97, 127}
definecolor{c2}{RGB}{104, 182, 182}
definecolor{c3}{RGB}{107, 190, 190}
definecolor{c4}{RGB}{100, 172, 174}
begin{document}
thispagestyle{empty}
begin{tikzpicture}[overlay,remember picture,font=sffamilybfseries]
draw[very thick,c4,name path=big arc] ([xshift=-2mm]current page.north) arc(150:285:11)
coordinate[pos=0.225] (x0);
begin{scope}
clip ([xshift=-2mm]current page.north) arc(150:285:11) --(current page.north
east);
fill[c4!50,opacity=0.25] ([xshift=4.55cm]x0) circle (4.55);
fill[c4!50,opacity=0.25] ([xshift=3.4cm]x0) circle (3.4);
fill[c4!50,opacity=0.25] ([xshift=2.25cm]x0) circle (2.25);
draw[very thick,c4!50] (x0) arc(-90:30:6.5);
draw[very thick,c4] (x0) arc(90:-30:8.75);
draw[very thick,c4!50,name path=arc1] (x0) arc(90:-90:4.675);
draw[very thick,c4!50] (x0) arc(90:-90:2.875);
path[name intersections={of=big arc and arc1,by=x1}];
draw[very thick,c4,name path=arc2] (x1) arc(135:-20:4.75);
draw[very thick,c4!50] (x1) arc(135:-20:8.75);
path[name intersections={of=big arc and arc2,by={aux,x2}}];
draw[very thick,c4!50] (x2) arc(180:50:2.25);
end{scope}
path[decoration={text along path,text color=c4,
raise = -2.8ex,
text along path,
text = {|sffamilybfseries|02/18/2019},
text align = center,
},
decorate
] ([xshift=-2mm]current page.north) arc(150:245:11);
%
begin{scope}
path[clip,postaction={fill=c3}]
([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (4.2,7.7);
fill[c2] ([xshift=0.5cm,yshift=-8cm]current page.center)
([xshift=0.5cm,yshift=-8cm]current page.center) arc(180:60:2)
|- ++ (-3,6) --cycle;
draw[very thick,c4] ([xshift=-1.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=0.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=2.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=4.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
fill[red] ([xshift=2.5cm,yshift=-8cm]current page.center) +(60:2) circle(1.5mm)
node[above right=2mm]{$displaystylerho=frac{1+sqrt{-3}}{2}$};
end{scope}
%
fill[c1] ([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (-12.7,7.7);
node[text=white,anchor=west,scale=5,inner sep=0pt] at
([xshift=-8cm,yshift=-3.25cm]current page.center) {Some text};
node[text=white,anchor=west,scale=2.5,inner sep=0pt] at
([xshift=-8cm,yshift=-6cm]current page.center) {Some text};
%
draw[gray,line width=5mm]
([xshift=2mm,yshift=-1mm]current page.south west) rectangle ([xshift=-2mm,yshift=1mm]current
page.north east);
end{tikzpicture}
end{document}
Can one do something like this? Yes. Most likely the curves in the upper right part are are some sort of Apollonius (Golden Ratio?) circles but I was too lazy to look them up.
documentclass{article}
usepackage{tikz}
usetikzlibrary{intersections,decorations.text}
definecolor{c1}{RGB}{62, 97, 127}
definecolor{c2}{RGB}{104, 182, 182}
definecolor{c3}{RGB}{107, 190, 190}
definecolor{c4}{RGB}{100, 172, 174}
begin{document}
thispagestyle{empty}
begin{tikzpicture}[overlay,remember picture,font=sffamilybfseries]
draw[very thick,c4,name path=big arc] ([xshift=-2mm]current page.north) arc(150:285:11)
coordinate[pos=0.225] (x0);
begin{scope}
clip ([xshift=-2mm]current page.north) arc(150:285:11) --(current page.north
east);
fill[c4!50,opacity=0.25] ([xshift=4.55cm]x0) circle (4.55);
fill[c4!50,opacity=0.25] ([xshift=3.4cm]x0) circle (3.4);
fill[c4!50,opacity=0.25] ([xshift=2.25cm]x0) circle (2.25);
draw[very thick,c4!50] (x0) arc(-90:30:6.5);
draw[very thick,c4] (x0) arc(90:-30:8.75);
draw[very thick,c4!50,name path=arc1] (x0) arc(90:-90:4.675);
draw[very thick,c4!50] (x0) arc(90:-90:2.875);
path[name intersections={of=big arc and arc1,by=x1}];
draw[very thick,c4,name path=arc2] (x1) arc(135:-20:4.75);
draw[very thick,c4!50] (x1) arc(135:-20:8.75);
path[name intersections={of=big arc and arc2,by={aux,x2}}];
draw[very thick,c4!50] (x2) arc(180:50:2.25);
end{scope}
path[decoration={text along path,text color=c4,
raise = -2.8ex,
text along path,
text = {|sffamilybfseries|02/18/2019},
text align = center,
},
decorate
] ([xshift=-2mm]current page.north) arc(150:245:11);
%
begin{scope}
path[clip,postaction={fill=c3}]
([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (4.2,7.7);
fill[c2] ([xshift=0.5cm,yshift=-8cm]current page.center)
([xshift=0.5cm,yshift=-8cm]current page.center) arc(180:60:2)
|- ++ (-3,6) --cycle;
draw[very thick,c4] ([xshift=-1.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=0.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=2.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
draw[very thick,c4] ([xshift=4.5cm,yshift=-8cm]current page.center)
arc(180:0:2);
fill[red] ([xshift=2.5cm,yshift=-8cm]current page.center) +(60:2) circle(1.5mm)
node[above right=2mm]{$displaystylerho=frac{1+sqrt{-3}}{2}$};
end{scope}
%
fill[c1] ([xshift=2cm,yshift=-8cm]current page.center) rectangle ++ (-12.7,7.7);
node[text=white,anchor=west,scale=5,inner sep=0pt] at
([xshift=-8cm,yshift=-3.25cm]current page.center) {Some text};
node[text=white,anchor=west,scale=2.5,inner sep=0pt] at
([xshift=-8cm,yshift=-6cm]current page.center) {Some text};
%
draw[gray,line width=5mm]
([xshift=2mm,yshift=-1mm]current page.south west) rectangle ([xshift=-2mm,yshift=1mm]current
page.north east);
end{tikzpicture}
end{document}
edited 1 hour ago
answered 2 hours ago
marmotmarmot
101k4117226
101k4117226
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of usingdisplaystyle
you could usedfrac
fromamsmath
.
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
@HenriMenke I believe that the cover is correct.rho
is the sixth root of unity, i.e.rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parametertau
withrho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)
– marmot
1 hour ago
|
show 2 more comments
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of usingdisplaystyle
you could usedfrac
fromamsmath
.
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
@HenriMenke I believe that the cover is correct.rho
is the sixth root of unity, i.e.rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parametertau
withrho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)
– marmot
1 hour ago
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
You’re so great!! Thanks a lot!!!
– user450201
2 hours ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
Such a nice answer...great...
– MadyYuvi
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of using displaystyle
you could use dfrac
from amsmath
.– Henri Menke
1 hour ago
sqrt{-3}
I have the feeling that there is a typo on the original cover. Also, instead of using displaystyle
you could use dfrac
from amsmath
.– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
I don't think it's Apollonian circles, because they don't intersect: en.wikipedia.org/wiki/Apollonian_gasket
– Henri Menke
1 hour ago
1
1
@HenriMenke I believe that the cover is correct.
rho
is the sixth root of unity, i.e. rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parameter tau
with rho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)– marmot
1 hour ago
@HenriMenke I believe that the cover is correct.
rho
is the sixth root of unity, i.e. rho=(1+sqrt{-3})/2=(1+mathrm{i}sqrt{3})/2=exp(2pimathrm{i}/6)
. I agree that these are not the standard Apollonius circles, which is why I wrote "some sort of Apollonius circles". While I believe to understand the inlay figure (which is the fundamental domain of the torus parameter tau
with rho
being the nontrivial selfdual point, I do not remember what the circles are even though I should.)– marmot
1 hour ago
|
show 2 more comments
Thanks for contributing an answer to TeX - LaTeX Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f475597%2fhow-to-create-a-cover-page-like-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
What have you tried so far?
– manooooh
3 hours ago
There is a rather straightforward part, the graphics, which can be done with TikZ (for instance) and a part which requires familiarity with the Chinese characters. It seems to me that anyone trying to answer this will have to know TikZ and these characters.
– marmot
3 hours ago
@marmot I think these characters is book name, these characters are not important.
– user450201
3 hours ago
Thank you for a really good question. I'm going to be up all night, trying to recreate what @marmot has done.
– GermanShepherd
1 hour ago