intersection coordinate systems

Question

I'm trying to write a \newcommand to draw a little stick-man in PGF. It isn't finished yet, but right now, it looks like this:

\newcommand{\man}[1] {
% #1 : The position of the head of the stick man
  \coordinate (X) at (#1);
  \coordinate (Y) at ($(#1)+(0,1mm)$);
  \draw [black,fill=green,thin] ($(#1)+(290:2mm)$) arc (-70:240:2mm)--++(-3mm,0)--++(0,-1mm)
  --++(3mm,0)--++(0,-3mm)--++(225:3mm)--++(315:1mm)coordinate(A)--++(45:1mm)coordinate(B)--
  (intersection of A--B and X--Y)--
  (#1)--cycle;
 }

The function input #1 is the center of his head. The path starts off on the right side of his head, draws an arc (his head) to the left side, then traces his left arm, and the left side of his body. When it comes to his leg, I switch to using polar coordinates, because I want his leg to extend out at 45 degrees. The question arises on the way back from drawing his left leg. I want his crotch (for lack of a better word) to be directly below the center of his head. So I'm trying to find a way to define the point that lies on the intersection of the vertical line through (#1) and the line at 45 degrees through (A).

Right now I have to define coordinates (X) and (Y) ahead of time, which define the vertical line through the center of his head. and I have to define (A) and (B) on the path to define the line at 45 degrees along his leg. I can then use (intersection of A--B and X--Y). My question is, can I do this more elegantly without needing to define these extraneous coordinates (X), (Y) and (B)?

I tried things like

(intersection of A--$A+(45:1mm)$ and X--$X+(0,1mm)$)

but that didn't work.

As per Peter's comment, here is a minimum working example:

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\man}[1] {
% #1 : The position of the head of the stick man
  \coordinate (X) at (#1);
  \coordinate (Y) at ($(#1)+(0,1mm)$);
  \draw [black,fill=green,thin] ($(#1)+(290:2mm)$) arc (-70:240:2mm)--++(-3mm,0)--++(0,-1mm)
  --++(3mm,0)--++(0,-3mm)--++(225:3mm)--++(315:1mm)coordinate(A)--++(45:1mm)coordinate(B)--
  (intersection of A--B and X--Y)--
  (#1)--cycle;
 }

\begin{document}
\tikz{\man{3,3}}
\end{document}

Answer 1

This is as elegant as I get (without pre-calculating anything).

First: no calc library. The ($(#1)+(#2)$) syntax translates to ([shift={(#1)}] #2) = ([shift={(#2)}] #1).

The pelvis area has been created with two auxiliary coordinates and the intersection of syntax.

Also, I find it more elegantly to use TikZ possibilited and created an insert path style. (You can of course simply translate it back to an own command.

Code

\documentclass[tikz,convert=false]{standalone}
\tikzset{
  man/.style={
    insert path={
      ([shift={(+-60:+2mm)}]#1) arc[radius=+2mm, start angle=+-60, end angle=+240] --
      ++(left: +3mm) --
      ++(down: +1mm) --
      ++(right:+3mm) --
      ++(down: +3mm) --
      ++(225:  +3mm) --
      ++(315:  +1mm) coordinate (@aux1) --
      ++(45:   +1mm)  -- 
        (intersection of #1--[shift={(down:+1cm)}] #1 and @aux1--[shift={(+45:+1cm)}]@aux1)
                     coordinate (@aux2) --
        (intersection of @aux1--[shift={(right:+1cm)}]@aux1 and @aux2--[shift={(+-45:+1cm)}]@aux2) --
      ++(45:   +1mm) --
      ++(135:  +3mm) --
      ++(up:   +3mm) --
      ++(right:+3mm) --
      ++(up:   +1mm) -- cycle
    }
  }
}
\begin{document}
\tikz \path[draw=black,fill=green,thin,line join=round] [man={3,3}];
\end{document}