LaTeX command for unknown symbol

Question

I am trying to type the shown symbol in LaTeX and I am hoping to find a command producing this symbol. This is just a very crude sketch. The three lines are supposed to be equal length with 120 degrees between them. It is quite large here, but I would like it to be of "normal character size", comparable to a + or so. Is there any direct way of obtaining this without including it as a graphic or perhaps drawing it in TikZ?

Answer 1

Within the package \usepackage{MnSymbol} you can find

$A \downY B \upY C$

which renders like:

I took that from comprehensive symbols, page 23f.

Unfortunately (thanks for the hint @egreg) the use of MnSymbol does not only add additional commands but altering some as well. This might produce ugly side-effects. Good for us there is a step by step tutorial How can I use a symbol provided by a package without changing the entire mathematics font?

Following this is fairly easy (give Leo Liu as much credit as you can) and so you can circumvent the usage of the hole package as follows:

\documentclass{article}

\DeclareFontFamily{U}{MnSymbolC}{}
\DeclareFontShape{U}{MnSymbolC}{m}{n}{
    <-6>  MnSymbolC5
   <6-7>  MnSymbolC6
   <7-8>  MnSymbolC7
   <8-9>  MnSymbolC8
   <9-10> MnSymbolC9
  <10-12> MnSymbolC10
  <12->   MnSymbolC12}{}
\DeclareFontShape{U}{MnSymbolC}{b}{n}{
    <-6>  MnSymbolC-Bold5
   <6-7>  MnSymbolC-Bold6
   <7-8>  MnSymbolC-Bold7
   <8-9>  MnSymbolC-Bold8
   <9-10> MnSymbolC-Bold9
  <10-12> MnSymbolC-Bold10
  <12->   MnSymbolC-Bold12}{}

\DeclareSymbolFont{MnSyC}{U}{MnSymbolC}{m}{n}
\SetSymbolFont{MnSyC}{bold}{U}{MnSymbolC}{b}{n}
\DeclareMathSymbol{\upY}{\mathbin}{MnSyC}{41}
\DeclareMathSymbol{\downY}{\mathbin}{MnSyC}{43}

\begin{document}
\[A \downY B \upY C \]
\boldmath
\[ A \downY B \upY C\]
\end{document}

Which (thanks to egreg again) renders even in \bold

Answer 2

You can use the picture environment:

\documentclass{article}
\usepackage{pict2e,picture}
\newcommand{\triplus}{\mathbin{
  \sbox0{$+$}
  \begin{picture}(\wd0,\ht0)
  \put(0,0){\line(1,.577){.5\wd0}}
  \put(.5\wd0,.289\wd0){\line(-1,.577){-.5\wd0}}
  \put(.5\wd0,.289\wd0){\line(0,1){\dimexpr\ht0-.289\wd0\relax}}
  \end{picture}
  }
}

\begin{document}
$a+b\triplus c$
\end{document}

---

Adding \roundcap, as suggested by Heiko Oberdiek, might be even better:

\documentclass{article}
\usepackage{pict2e,picture}
\newcommand{\triplus}{\mathbin{
  \sbox0{$+$}
  \begin{picture}(\wd0,\ht0)
  \roundcap
  \put(0,0){\line(1,.577){.5\wd0}}
  \put(.5\wd0,.289\wd0){\line(-1,.577){-.5\wd0}}
  \put(.5\wd0,.289\wd0){\line(0,1){\dimexpr\ht0-.289\wd0\relax}}
  \end{picture}
  }
}

\begin{document}
$a+b\triplus c$
\end{document}

Answer 3

Here's my TikZ shot. The length of the line was chosen to be as close to the length of the line in a + symbol as I could get it (interestingly, the ends of the lines on the + are not perfect half circles). And the height was so that the intersection was the same as the centre of the cross of the +.

Edit Looking at egreg's answer, I see that I can compute the dimensions automatically.

\documentclass{article}
%\url{http://tex.stackexchange.com/q/103427/86}
\usepackage{tikz}

\newcommand\triad{%
  \mathbin{%
    \tikz[overlay,baseline=0 pt] {
    \pgfmathparse{(height("$+$") + depth("$+$") -
\the\pgflinewidth)/2}%
      \let\triht=\pgfmathresult
      \pgfmathparse{(height("$+$") - depth("$+$"))/2}%
      \let\tribs=\pgfmathresult
 \draw[yshift=\tribs pt,xshift=-6pt,red,line cap=round] (0,0) --
(0,\triht pt) (0,0) -- (-30:\triht pt) (0,0) -- (210:\triht pt);%
}
}%
}

\begin{document}

\[
  a + b a +\triad b
\]
\end{document}

Close up (overlaid on a +):

In "normal" use:

Answer 4

The following example constructs the symbol in the following way:

• The top part with the round line end of the vertical bar is used as branch line.
• The length of the branch is the same as in the plus symbol. But the length can be changed by redefining \UpDownYFactor to a value different than 1.
• The intersection point is the same as in the plus symbol at the mathematical center axis.
• The symbol adopts in size to the current math style.
• The side bearings have the same width as in the plus symbol.
• The names for the provided symbols are the same as in package MnSymbol: \upY and \downY.
• The symbols act as binary operators. It can be changed by replacing \mathbin.

Example file:

\documentclass{article}
\usepackage{graphicx}
\usepackage{trimclip}

\makeatletter
\providecommand*{\upY}{%
  \mathbin{%
    \mathpalette\@updownY{0}%
  }%
}
\providecommand*{\downY}{%
  \mathbin{%
    \mathpalette\@updownY{1}%
  }%
}
\providecommand*{\UpDownYFactor}{1}
\newcommand*{\@updownY}[2]{%
  % #1: math style
  % #2: 0 = up, 1 = down
  \sbox0{$#1+\m@th$}%
  \dimen2=.5\dimexpr\wd0-\ht0-\dp0\relax
  % => \dimen2: side bearing
  \sbox2{$#1\vcenter{}$}%
  \dimen4=\dimexpr\ht0-\ht2\relax
  % => \dimen4: branch length
  \setbox0=\hbox to 0pt{%
    \hss
    \clipbox{%
      0pt %
      {\dimexpr\totalheight-\UpDownYFactor\dimen4\relax} %
      0pt %
      -\dimen2%
    }{$#1|$}%
    \hss
  }%
  \ht0=\dimexpr\ht0-\dimen2\relax
  \kern\dimen2 %
  \raise\ht2\hbox{%
    \ifnum#2=0 %
      {\rotatebox{120}{\copy0}}%
      \copy0 %
      {\rotatebox{-120}{\copy0}}%
    \else
      {\rotatebox{60}{\copy0}}%
      {\rotatebox{180}{\copy0}}%
      {\rotatebox{-60}{\copy0}}%
    \fi
  }%
  \kern\dimen2 %
}
\makeatother

\begin{document}
\[\begin{array}{c}
  a \upY b \downY c
  \\
  \scriptstyle
  a \upY b \downY c
  \\
  \scriptscriptstyle
  a \upY b \downY c
  \\
  \rlap{$\upY$}{\downY}
\end{array}\]
\end{document}  

For analyzing the center:

\usepackage{pdfrender}
...
\[
  \rlap{%  
    \sbox0{$\vcenter{}$}%
    \sbox2{${\upY}{\downY}{\upY}$}%
    \raise\ht0\hbox{\vrule height.01pt depth.01pt width\wd2}%
  }%
  \pdfrender{TextRenderingMode=Stroke, LineWidth=.01pt}
  {\upY}\rlap{$\upY$}{\downY}{\downY}
\]

Answer 5

Second cut, trying to preserve the 120 degree angles, as well as getting proper alignment.

\documentclass{article}
\usepackage{scalerel}

\newcommand\triad{%
\let\savearraystretch\arraystretch
\renewcommand\arraystretch{0}
\begin{array}{c}
\rule{-.01ex}{.01ex}\rule{.05ex}{.8ex}\\
\stretchto{
    \scalerel*[\widthof{=}]{\wedge}
    {\rule{1ex}{3ex}}%
}{0.42ex}\\
\vphantom{\rule{1ex}{.01ex}}
\end{array}
\let\arraystretch\savearraystretch
}

\begin{document}

$x^2 \triad \begin{array}{c}a\\b\\c\end{array}$

\end{document}

Answer 6

You could also play around with the Mercedes star mark in TikZ:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{plotmarks}

\newcommand{\unkwn}{\tikz[scale=2] \draw plot [mark=Mercedes star] coordinates {(0,0)};}

\begin{document}

$a \unkwn b$

\end{document}

Answer 7

Since you want a normal character, an alternative would be to use the similar symbol \Yup of the package stmaryrd. The angles are about 135°, 135°, 90°. The picture shows $a \Yup b$.